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Thread: Topology primer

  1. #1 Topology primer 
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    I'm beginning to wonder if I was wise to relay so heavily on point set topology to explain some things about sets. So let's do some topology here. By the way, I'm working (virtually) without sources, so if there are errors, someone shout

    Consider a point set S.

    Definition
    A collection T of subsets of S is said to a be topology on S iff the following are true:

    finite intersection of elements of T (aka subsets of S) are in T;
    arbitrary union of elements of T (aka subsets of S) are in T;
    S is in T;
    Ø is in T.

    The pair (S,T) is referred to as a topological space. (Note that it is usual to abuse notation and assert, for example, that "S is a topological space" (when it is, of course), so take note: it means S = (S,T) in context)

    I'll give some examples in a minute, but let's press on a bit. The elements of T are said to be the open sets in (S,T). Subsets of S which are in the complement of T are the closed sets in (S,T). (the complement of any set A in T is S - A)

    So let's have a couple of examples. Consider the subsets S and Ø of S. The topology T = {Ø, S} is referred to as the indiscrete (or trivial, or concrete) topology on S. Note that the complement of S is Ø, the complement of Ø is S, so these guys are both open and closed in this topology (well all topologies actually).

    Now consider the power set on S, that is all elements and their combinations. (The power set P(S) for any set with n elements, has cardinality 2ⁿ e.g. for S = {a, b, c}, P(S) = {{a}, {b}, {c}, {a,b}, {a,c}, {b,c}, {a,b,c}, Ø}). This is in fact a topology on S and is referred to as the discrete topology on S: it is the finest possile topology on S. Note that for each set in T = P(S), it's complement also appears in T, so all sets in this topology are both open and closed.

    In fact it is the case that any topology T on S is a subset of P(S).

    Let R be the set of real numbers. In the standard topology on R, elements of T (aka open sets in (R,T)) are open intervals of the form (a, b). The union of all such intervals is the real line.

    Let (S,T) be a topological space. For some point s in S, an open set in (S,T) containing s is referred to as a neighbourhood of s.

    Let X and Y be topological spaces (note the abuse of notation I referred to earlier). A map f: X → Y is said to be continuous at x if, for each x in the domain of f there is a neighbourhood U of f(x) in Y whose pre-image f<sup>-1</sup>(U) is open in X. It is (relatively) easy to show that this corresponds to the usual ε - δ definition of continuity we learn in school.

    Obviously, if f is a bijection, we can think of f<sup>-1</sup> as a continuous inverse. Under these circumstances, f:X → Y is referred to as a homeomorphism. Note that the above places neighbourhoods of X and Y in one-to-one correspendence. But neighbourhoods are open sets i.e. elements in the topology, so X and Y are topologically equivalent, which is the meaning of homeomorphic. It is the topological equivalent of an isomorphism (and gives rise to the tired old joke about topologists not knowing the difference between a donut and a coffee cup)

    If homeomorphic spaces are topologically equivalent this means they must share what are known as topological properties. I'l briefly mention the three that seem, to me, to be the most important.

    The Hausdorff property.
    Let X be a topological space. If, for any pair of points x notequal y in X there exist neighborhoods U and V of x and y respectively such that U ∩ V = Ø, X has the Hausdorff property ("is Hausdorff")

    The Connected property.
    If, in a topological space X the only sets both open and closed are X and Ø, X is connected. Equivalently, X is connected if it is not the union of disjoint non-empty open sets.

    The Compact property.
    A collection of subsets of a topological space X whose union is X is called a cover of X. If these subsets are all open it is an open cover. And if there is a subclass of this collection similarly covering X, it is a subcover. X is compact if every open cover has a finite subcover.

    Finally, although for any set there are a number of topologies that can be placed on it, it is often of no real interest what the actual topology is. But it is always necessary to list the topological properties of the space.


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  3. #2  
    Forum Freshman omerta's Avatar
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    I love this forum for this reason. I have a book on topology that goes into all this but reads like ingrediants off of a soup can at times, lol. Its a chor to read it cause I loose interest. When you guys post it is so much easier to understand that you dont even realize your picking it up. A whole lot of 'ah ha' and ' oh, I see now' moments.


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  4. #3  
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    It occurs to me my description of topological properties was perhaps too abstract, but that's the price you pay for generalizaion. So. Let me ramble on a bit more.

    I mentioned the Hausdorff property. This is in fact one of several so-called seperation axioms, which go by the cuddly names T0, T1, T2 etc. Intuitivley, these tell us what it means for 2 points to be the same or different in any particular topology. These axioms are of increasing degrees of stringency, in the sense that, a T2 space is necessarily also a T0 and a T1 space, but not conversely. A Hausdorff space is a T2 space, which is the minimal seperation most (normal!!) people care about.

    But, what does it mean? It generalizes our intuition that the line connecting any 2 points in R is infinitely subdividable, and that, in a Hausdorff space, if I cannot find neighbourhoods of x and y that are disjoint, I must regard them as being the same point.

    Connectedness is easier in a way, it codifies our intuition that the line connecting any 2 points in R is continuous. In higher-dimensional spaces, things get a little more interesting. Consider a planar surface. If, for any 2 points I can find an unbroken path between them, my plane is path-connected, even though I may have to skirt round some holes. Now consider a plane with a hole in it. Although I may have many paths between 2 points, I cannot "transform" each into the other, because there is a hole in the way. Where I can do this, my space is arc-conncted, and clearly this is a stronger requirement than path-connectedness.

    Compactness is irritating, which is a shame, because it is hugely important, and not just in topology. I gave a definition before which, as far as I know, is the currently accepted one. But there is another (historically earlier, I think?) one. A space X is compact if every sequence in X has a limit in X. In a compact space, every subset has a convergent sequence, and that sequence has a limit in the subset.

    It follows that compact subsets of a compact space are closed. There is in fact a theorem (Heine-Borel) which connects the two definitions of compactness (at least in the case on R<sup>n</sup>), but it's far from trivial to prove.

    Hmm. I was going to go on and talk about metric topological spaces, but this post is already over-long.

    And anyway, it's Friday night, I'm going for a beer. Coming?
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  5. #4  
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    And anyway, it's Friday night, I'm going for a beer. Coming?
    I'm already there! :)

    But sorry guitarist, I just don't understand any of what you've written. It's too much of a leap for me at the moment (though conceptually I'm still intersting in the link beteween sets and topology).

    I've just got Penrose's book 'Road To Reality' out of the library and I'm going through that, though I don't fully agree with his philosophical position (or at least don't entirely trust it) - but who am I to argue, he's clearly a very good mathematician and explainer.

    I read something intersting in it about equivalence classes, seemed to spark something in my head.

    Anyway, getting a bit OT

    g
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  6. #5  
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    Sequential compactness (i.e. the existence of convergent subsequences) only implies compactness under very special circumstances Guitarist (like those found in a metric space for example).

    Arent you also thinking of the Bolzano-Weierstrass thm, not the Heine-Borel thm?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  7. #6  
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    Hey river_rat.

    Yes, I shouln't have said "convergent", that requires a metric as you say. Would you accept slip of the tongue?

    You are right that B-W deals with sequentially compact spaces. But look, it's true my notes are a little garbled here (that's why I said compactness is "irritating"!), but this is what I have; see if any of you out there can agree with any of it.

    We can define limits of sequences in any topological space. A sequence is a map N → X. A point x in X is a limit of the sequence if, for every neighborhood U of x, there is some n ∈ N above which all elements of the sequence belong to U.

    This doesn't require a metric, nor does it require that the space X have any order. We do have an order on N though. I think we can generalize this: if I have a map from some more general set with any order, then that's a net. One difference between nets and sequences is that whereas, when a sequence has a limit, it is unique, but a net may have many different limits.

    So, a set is bounded if every sequence in it has a limit. If a set is closed the limit of any contained sequence is in it. Note that the converse is not true in general; this is not the topological definition of a closed set.

    The Heine-Borel theorem says that every closed and bounded subset of R<sup>n</sup> is compact. This is what I was referring to.

    Let me now just quickly touch on metric spaces. But first I need to mention the Cartesian product.

    For many spaces, topological or not, we might have a combining operation (like plus, times etc) the result of which may or may not be in the space itself. But there are other sorts of operations, like comparing elements in the space, for example, where we need to keep the elements involved seperate. For this, we are strictly speaking using two (or more) copies of the same space. This is called forming the Cartesian product, and is written XxX, say. (Hmm. That's X times X) Then for x, y in X, this gives me what's called an ordered pair {x,y}.

    So any space, topological or not, is called a metric space if the following are true:

    There is a map d: XxX → N where
    d(x,y) = d(y,x) (symmetry)
    d(x, y) ≥ 0 (positiveness)
    d(x,y) = 0 iff x = y (definiteness)
    d(x,z) ≤ d(x,y) + d(y,z) (triangle equality).

    We can think of the "d" here being some sort of distance function. Every metric space carries a topology, where the open sets are just the open balls d(a,x) < ε. That's called the metric topology.

    And for a given topology, if it is the metric topology for some metric, then that topological space (with the topology we've given it) is called metrizable. Not every topological space is metrizable, meaning not every space admits a metric which produces the given topology.

    For example, the discrete metric always induces the discrete topology. If the discrete metric is the only metric we can find for a space, then unless you started with a discrete space, our space isn't metrizable.
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  8. #7  
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    Ill sort out the compactness stuff when i have a spare moment, but most of the time the different forms of compactness (psuedo, local, countable, normal etc) are independant of each other (except most of them imply countable compactness, bar that i mean) for a general topological space.

    Sequence convergence does not need a metric, the neighbourhood idea is the standard one for more abstact spaces as you showed (except it should read: A sequence {a<sub>i</sub>} converges to a point x if for any nhood U of x there exists a natural number n s.t. {a<sub>m</sub> : m > n} is contained in U.

    Regarding nets, the main reason for using them as the form of compactness you refered to in an earlier post (i.e. that every sequence has a convergent subsequence <=> space is compact) holds true for nets (every net has a convergent subnet <=> space is compact). BTW you need a directed preorder (i guess by order you meant total order).

    If a space is hausdorff then any sequence (or limit) will have a unique limit if it converges. However, in the case of a non-hausdorff space all bets are off as both cases are possible in that a sequence (or net) can converge to more then two points or to one unique point or even to every point in the space.

    One more thing, the open balls form a base for the standard topology on a metric space, but they dont exhaust all the open sets.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  9. #8  
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    Fair enough, river_rat. Obviously I am not the right person to be "teaching" topology. Shame, I thought I had the subject nailed. Ho, hum

    -b-
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  10. #9  
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    Nah, just topology is my area of "expertise" if you can call it that - so im a bit picky about it.

    Well here are a few forms of compactness (at least the ones i know)

    A space is compact if every open cover has a finite subcover. This finiteness condition is very, very, very handy! For example, every real valued continuous function (i.e. map) on a compact space is bounded and it obtains those extrema. You can prove this using Heine-Borel or as a good exercise in compactness from first principles. It is interesting to note that compact sets behave as if they where points in a very real way.

    A space is countably compact if every countable open cover has a finite subcover. This is a weaker from of compactness but it is still quite handy and for metric spaces this is the same thing as normal compactness.

    A space is pseudo-compact if every real valued map on that space is bounded.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  11. #10  
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    OK, rather than spend the rest of this thread sitting in the naughty corner, I'm going to press on. As before, I welcome (rather than resent) corrections

    So, how about local compactness and paracompactness?

    Let X be a topological space. X is locally compact if every point x ∈ X has a compact neighbourhood (aka open set containing x). Meaning that any cover of a neighbourhood of x has a finite subcover (easily shown for R)

    Recall that a cover C for X (or covering of X) is a collection of sets whose union is all X. The cover C is said to be locally finite if, for each point x ∈ X there is a neighbourhood of x which has non-empty intersection with only finitely many sets in C.

    A refinement of a cover C is a cover B of X where each set in B is a subset of at least one set in C. (Contrast this to a subcover, where we simply threw out some sets in C).

    So, X is said to be paracompact if every open cover of X has an open refinement which is locally finite.

    NOTE: This post has been edited to remove an error pointed out by river_rat below.

    I'm minded to go on and talk about some elements of homotopy theory, a subject in algabraic topology, but first I must alert you to an error of mine I just detected. I gave as an alternative definition of connectedness that, if a space is the union of disjoint non-empty sets, it is not connected. I should have said "if a space can be written as the union of two disjoint non-empty sets then it is not connected". Maybe the difference is subtle, but it is crucial.

    I bring this up, because I went on to talk about path-connectedness vs. arc-connectedness, which is a nice entry into homotopy theory. Obviously, there is tons more to say about point set topology, but it is a rather arid subject. Algebraic topolgy uses some more familiar language (I did say some)

    Gosh, what a wind-bag I am!
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  12. #11  
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    Im not too sure about your definition of paracompactness - im sure you need to have open covers and refinements. Your example is still paracompact even if you allow for open covers and refinements so im not sure what it shows.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  13. #12  
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    I rechecked my notes, and you are right - we do need refinements to be open. I've amended the post accordingly.
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  14. #13  
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    Well why don't we move on to partitions of unity while we are talking about paracompactness?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    The stage is yours!
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  16. #15  
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    Aargh - thought i could weasel out of that

    Okay, first we need Urysohn's Lemma and for that we need to know what a normal space is. Now a normal space is a T1 space where we can seperate closed disjoint subsets by open sets. Metric spaces are normal, as are paracompact hausdorff spaces.

    Okay, now Urysohn's Lemma tells you that a normal space has quite a few continuous maps - infact there are enough continuous maps to seperate every disjoint closed set. i.e. given two disjoint closed sets A, B there exists a function f:X->R such that f(A) = 0 and f(B) = 1. If you want i can sketch the proof (known as the onion skin method) as it is quite clever, or you can take it on faith. This property is also quite special, for example the only continuous function from a discrete space to the reals are the constant functions.

    Now on to partitions of unity A space admits partions of unity if for every open cover of the space we can find a collection of functions {f<sub>a</sub>:X->[0, 1]} such that the set A<sub>a</sub> = {x : f<sub>a</sub>(x) is not 0} is contained in some open set of the open cover and for every point x we can find a neighbourhood of that point such that all but a finite number of those functions are zero and the rest sum to 1 for every other point. Every normal space admits a partion of unity, so that is the connection between hausdorff paracompact spaces and partitions of unity.

    Now for some uses, partitions of unity allow you to take a local construction (like integration on R<sup>n</sup>) and extend it to an entire space (like a manifold, which locally looks like R<sup>n</sup>).
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  17. #16  
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    Ah found the thread - consider it resurrected (cue lightning)
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  18. #17  
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    Quote Originally Posted by river_rat
    Ah found the thread - consider it resurrected (cue lightning)
    We await the thunder!
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  19. #18  
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    Sorry for the storm being delayed for a few days but work has not been playing ball - that and the heat here in joburg has officially fried my brain to a crisp!

    I thought i would carry on our discussion of paracompact spaces and prove that any hausdorff paracompact space is actually a normal space and then finish off by construction partitions of unity for any normal space before moving on to manifolds as these constructions are important in that setting.

    So lets recap a few definitions:

    1. A space X is called Hausdorff if for any two distinct points x, y in X you can find disjoint open neighborhoods of x and y. i.e. distinct points are topologically distinct
    2. A space X is called paracompact if every open cover of X has a locally finite refinement
    3. A space is called regular if for every closed subset A of X and every point x not in A you can find disjoint open subsets U, V of X such that x is in U and A is contained in V
    4. A space X is called normal if for all disjoint closed subsets A, B of X you can find disjoint open subsets U, V of X such that A is contained in U and B is contained in V

    We also assume that all our spaces are at least T1 so we can assume that singletons are closed subsets of our space.

    Our plan of attack here is to climb the separation axioms ladder by first proving that paracompact+hausdorff => regular and that paracompact + regular => normal. I cant remember if we cover the closure and interior operators anywhere and they are needed for this construction so i think it will be helpful to cover them first.

    1. The closure of a subset A of a space X is the smallest closed set that contains A and is denoted by cl(A).
    2. The interior of a subset A of a space X is the largest open set contained in A and is denoted by int(A)

    Lemma 1: x is an element of the int(A) iff there exists an open nhood U of x which is contained in A.

    Lemma 2: X \ cl(A) = int(X\A)

    Lemma 3: x is an element of the cl(A) iff every open nhood U of x has non-empty intersection with A

    Each of these lemmas are easy to prove so ill leave them to the reader if they feel industrious Using these properties we can show that the closure of the union of any finite number of subsets of X is just the finite union of the closures of those sets (once again the proof is left to the reader). The important point here is that this only holds for finite unions in general! However, we have the following nice result if our subsets are locally finite (which should make you think of paracompact spaces!):

    Lemma 4: If X be a topological space and {V<sub>i</sub>} a family of locally finite subsets of X then cl(U V<sub>i</sub>) = U cl(V<sub>i</sub>).

    Now i have left a few things for you guys to prove so does anyone have any questions before i start proving the main lemma to show that every paracompact hausdorff space is normal?
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    Nice post, river_rat. I'll help you out (!) with your problems if nobody else cares to try, but let's see first if someone else wants a go.

    And if they do, here's a coupla alternative definitions of the interior and closure operations they might find helpful:

    The interior of the set A, intA, is the union of all open sets contained in A;

    The closure of A, clA, is the intersection of all closed sets containing A.

    To show that these definitions are equivalent to river_rat's is child's play. So come on, guys, get your pencils out.
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  21. #20  
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    Ah no where was I? Oh yes, we need to prove our lemma that will help us show that every hausdorff paracompact space is regular => normal. So here it is:

    Lemma 5: Let X be a paracompact space and A, B be two closed subsets of X such that for all x in A there exists open disjoint subsets U', V' of X such that x is an element of U' and B is contained in V'. Then there exists open disjoint subsets U and V of X such that A is contained in U and B is contained in V.

    Proof:

    Let x be an element of A, so there exists disjoint open subsets U'<sub>x</sub> and V'<sub>x</sub> such that x is an element of U'<sub>x</sub> and B is contained in V'<sub>x</sub>. Consider the open cover T = {U'<sub>x</sub>}<sub>x in A</sub> U {X \ A} of X. Now X is paracompact so this cover has a locally finite refinement {T<sub>i</sub>}<sub>i in I</sub> Let I' = {i in I : T<sub>i</sub> intersect A is not empty} then each {T<sub>i</sub>}<sub>i in I'</sub> refines some U'<sub>x</sub>. Let U = U T<sub>i</sub> where i in I' so U is an open set with A contained in U. Now B is disjoint from the cl(T<sub>i</sub>) for all i in I' as if x is in B and x is in cl(T<sub>i</sub>) which is contained in cl(U'<sub>t</sub>) for t in A. But U'<sub>t</sub> is disjoint from V'<sub>t</sub> and B is contained V'<sub>t</sub> which is a contradiction. So B is not contained in U cl(T<sub>i</sub>) = cl( U T<sub>i</sub> ) Let V = X \ cl( U T<sub>i</sub> ) which is open, disjoint from U and contains B.

    With this lemma our result is quite easy but first, any questions?

    Corrections in red - i should really proof read these things properly and not swop the names of subsets as i start a proof!
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  22. #21  
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    This is all gibberish to me
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  23. #22  
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    Quote Originally Posted by Nevyn
    This is all gibberish to me
    Anything in particular or everything in general? You cant talk about modern physics with out understanding topology now a days - it comes part and parcel!
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  24. #23  
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    all of it, though i'm not sur at whicg level of study this comes under as i am only at GCSE
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  25. #24  
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    Ah thats why - dont feel down, you are not used to this level of technicality thats all. Topology is usually introduced 2nd year varsity, at least where i am from.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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  26. #25  
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    yay, i feel much better
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    EDIT: I see I missed a bit while eating supper, anyway, here it is (was?)

    OK, let's do it, hopefully others will follow. But first, a coupla remarks. It had been my understanding that a neigbourhood of the point x is by definition open, hence your stipulation "an open neighbourhood of x" is redundant. In case I'm wrong about this, I'll stick with your terminolgy (although I believe I can make my case).

    Second, I never did introduce the notation X\A for the complement in X of A. Anyway, there it is.

    Quote Originally Posted by river_rat
    Lemma 1: x is an element of the int(A) iff there exists an open nhood U of x which is contained in A.
    There are two parts here. First: we know that intA is the union of all open sets in A. Let U<sub>x</sub> be an open neighbourhood in A of x. Then U<sub>x</sub> is a set in the union of open sets which is intA. Then, as, by defintion, x ∈ U<sub>x</sub>, then x ∈ intA.

    Conversely, suppose U<sub>x</sub> ∉ A. Then x ∉ A and, as intA ⊆ A, x ∉ intA.

    Lemma 2: X \ cl(A) = int(X\A)
    clA is closed by definition, therefore x\clA is open. Now just as A ⊆ clA, so then X\clA ⊆ X\A. Then since clA is the smallest set containing A, X\clA is the largest set containing X\A., i.e. X\clA = int(X\A). Um, I don't like that, lemme think on.

    Lemma 3: x is an element of the cl(A) iff every open nhood U of x has non-empty intersection with A
    Again, we have two parts. The only shared elements of allopen neighbourhoods of x is x itself. Then the condition that U<sub>x</sub> ∩ A ≠ Ø for all U<sub>x</sub> ==> x ∈A. But A ⊆ clA, therefore x ∈ clA.

    Conversely, suppose some U<sub>x</sub> ∩ A = Ø. This implies x ∉ A. Suppose further that A ≠ clA. Then A is open, so A = intA. By the definition of an open neighbourhood U<sub>x</sub>, x ∈ intA, so x ∈ clA iff every U<sub>x</sub> ∩ A ≠ Ø.

    Using these properties we can show that the closure of the union of any finite number of subsets of X is just the finite union of the closures of those sets (once again the proof is left to the reader).
    Any closed set = its own closure, evidently' So, suppose X is closed, clX = X. Suppose Y is open then clY = cl(clY). So that in general it is true that, for A, open or closed, cl(clA) = clA.

    So for finitely many A, B, we know that clA and clB are closed , therefore clA ∪ clB is closed ( as river_rat points out this is ony true in the finite case). Anyway, by the above, cl(cl ∪ clB) = cl(A ∪ B).
    The important point here is that this only holds for finite unions in general!
    Yes, this follows from the axioms of a topology, or, if we prefer, the theorems of Kuratowski. But the locally finite condition guarantees that arbitrary union of closed sets in a locally finite space is closed after all, so I think the proof of this....

    Lemma 4: If X be a topological space and {V<sub>i</sub>} a family of locally finite subsets of X then cl(U V<sub>i</sub>) = U cl(V<sub>i</sub>).
    ...follows through. I'm running out of steam, so I may need to think on this a bit. But do continue.
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    Ah okay - time for a technical lemma, the infamous shrinking lemma

    Lemma 6 : Let X be writing as the union of two open subsets U and V and suppose X is a normal topological space. Then you can find a open set W such that W is contained in cl(W) which is contained in U and that the union of W and V still cover X.

    i.e. Open covers of a normal space have quite a bit of overlap!

    Proof:

    The proof is not hard, though for some reasons when i was sitting down proving this i ended up chasing my tail for a while! Once again i blame the incessant heat! (It was only 32 degrees today in the shade, and we had our first rain in a month last night - a whole 3mm). Anyway, on with the proof.

    Now X is a normal space so there exists a set and X\U is a closed set disjoint from the closed set X\V so there exists disjoint open subsets A and B s.t. X\U is contained in A and X\V is contained in B. Then B is contained in X\A (closed) which is contained in U as A and B are disjoint. So we have that cl(B) is also contained in U. Also B union V = X, so let W = B.

    Now using this lemma and some horrible transfinite induction steps and the axiom of choice we can show the Shrinking lemma:

    Definition: A cover {T<sub>i</sub>} is termed point finite if for any point x of X we have that x lies in only a finite number of members of {T<sub>i</sub>}.

    The Shrinking Lemma: Let X be a Normal space and let {V<sub>i</sub>} be a point finite open cover of X. Then there exists an open cover {T<sub>i</sub>} such that T<sub>i</sub> is contained in cl(T<sub>i</sub>) which is contained in V<sub>i</sub> for all i in I.
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Whoa there, pony, not so fast! I have some points which you're not giving me time to articulate.

    First, your notation: I would tend to use U<sub>x</sub> to denote an open set containing x a.k.a a neighbourhood of x. In your proof of Lemma 5, you have U'<sub>x</sub> and V'<sub>x</sub> as disjoint, hence x ∉ V'<sub>x</sub>. You then compound my confusion be writing T= {V'<sub>x</sub>}<sub>x ∈ A</sub> ∪ X\B for your cover. This notation has me worried. What gives?

    Second, in the same lemma, I can see no reason to distinguish between V' and V, for if x ∉ B, x ∉ V', surely it suffices to assert that U and V' are disjoint?

    Third, it was my understanding that, in order to qualify as a paracompact space, the space needed to be Hausdorff? I only have it given as a definition, maybe you're using a different one? Or is it just over-kill?

    Anyway, yeah, a regular paracompact space is normal (if we assume Hausdorff-ness in the definition of paracompact): a space is regular iff every neighbourhood U<sub>x</sub> of x contains a closed set containing x.

    A space is normal iff, for any two disjoint closed sets A, B, there are 2 disjoint open sets U, V, where A ⊆ U and B ⊆ V.

    Then if X is Hausdorff and regular, it follows that X is normal (I think immediately from the definitions above and that of the Hausdorff property, or am I just being lazy?)

    Finally, the shrinking lemma. The intuitive meaning is clear (I think): we can take a space X to be as fine-grained as we like i.e. points may be arbitrarily close to each other but still distributed through the space. But: if for each x ∈ U<sub>x</sub> (open) there is some contained closed set containing x, and as each closed set has an open interior, this seems to imply an infinite regression terminating on {x}. Surely this isn't what you intended?

    Ah, maybe you intended it to terminate on an element of the refinement of the cover, of which {x} is one only in the discrete topology, as far as I know?
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    Hey guitarist, i have always known a neighbourhood of a point x as a subset of X which contains an open set which has x as an element. This definition allows you to talk unambiguously about the neighbourhood filter of a point (its just the collection of all neighbourhoods of that point) but forces you to specify when you are assuming your neighbourhood is open. Its a trade between the two ideas, either your filter is easy to define or your neighbourhoods are always open and as i tend to work with filters a lot in my work i lean towards this definition.

    Quote Originally Posted by Guitarist
    First, your notation: I would tend to use U<sub>x</sub> to denote an open set containing x a.k.a a neighbourhood of x. In your proof of Lemma 5, you have U'<sub>x</sub> and V'<sub>x</sub> as disjoint, hence x ∉ V'<sub>x</sub>. You then compound my confusion be writing T= {V'<sub>x</sub>}<sub>x ∈ A</sub> ∪ X\B for your cover. This notation has me worried. What gives?
    Okay, i need to index the family of sets according to the set A and that is where the notation comes from. The U'<sub>x</sub>'s are the neighbourhoods but as we are not sure that the intersection of all these sets is still open (as it is not in general) we have to go the other way around. Ah shoot i see an error in my proof (I swopped subsets A and B - no wonder it doesn't make sense!), let me correct it and then lets continue
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    Hey, that's much better. Moral: always proof your proofs! You can blame the heat, if you like. (A beautiful Spring day here, all green and lovely with Spring bulbs in bloom - read that and weep!).

    I'm happy to go with your definition of a neighbourhood, though I think it would cause problems in the trivial (concrete) topology, but it's a pathological beast anyway, so let's not worry.

    EDIT: That last was a silly comment, forget it (because in the concrete topology, X and Ø are both open and closed. Any point in the neighbourhood X (closed) is also a point in X (open), and X subset X always!)
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    Lol - i can blame the heat or the rain it seems, we had 28mm fall in just under 10 minutes today! I hope that means the end of this month long heat wave, now come on cricket world cup

    To answer your other questions:

    Third, it was my understanding that, in order to qualify as a paracompact space, the space needed to be Hausdorff? I only have it given as a definition, maybe you're using a different one? Or is it just over-kill?
    Depending on whose text you are reading paracompact spaces need and need not be hausdorff. Many results need both properties so to save type space i guess some people just define a paracompact space to be hausdorff. Ill just play it safe and state all the criteria.

    Then if X is Hausdorff and regular, it follows that X is normal (I think immediately from the definitions above and that of the Hausdorff property, or am I just being lazy?)
    In general that is not true - for example the Moore plane is regular, hausdorff but not normal. In the paracompact case you need that lemma i proved twice lol. The steps go like this -

    Suppose X is a paracompact hausdorff space. Let B be a closed subset of X and A = {x} disjoint from B (i.e. x is not in B) then as the space is hausdorff there exists disjoint open subsets U'<sub>y</sub> and V'<sub>y</sub> such that x is in U'<sub>y</sub> and y is in V'<sub>y</sub> for all y in B. Applying the lemma shows that there exists open disjoint subsets U and V such that x is in U and B is contained in V i.e. X is regular.

    Now using the same argument, we can show that paracompact + regular => normal.

    Finally, the shrinking lemma. The intuitive meaning is clear (I think): we can take a space X to be as fine-grained as we like i.e. points may be arbitrarily close to each other but still distributed through the space. But: if for each x ∈ Ux (open) there is some contained closed set containing x, and as each closed set has an open interior, this seems to imply an infinite regression terminating on {x}. Surely this isn't what you intended?
    If im understanding your question correctly the problem lies with the part i placed in bold - not every closed set as a non-empty interior (consider the closure of any nowhere dense set for example). The shrinking lemma helps us find enough continuous functions where each functions support is subordinate to some member of our cover - we need all those spare closed sets to use Urysohn's lemma.

    Did i cover everything or have i missed a few?
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    Quote Originally Posted by river_rat
    now come on cricket world cup
    Amen to that, though I'm not a huge fan of the one-day game. Fancy your chances, then?

    i guess some people just define a paracompact space to be hausdorff. Ill just play it safe and state all the criteria.
    OK, I'll go with that.

    Suppose X is a paracompact hausdorff space. Let B be a closed subset of X and A = {x} disjoint from B (i.e. x is not in B) then as the space is hausdorff there exists disjoint open subsets U'<sub>y</sub> and V'<sub>y</sub> such that x is in U'<sub>y</sub> and y is in V'<sub>y</sub> for all y in B.
    Well, OK, I guess Ill just have to get used to your notation. But, I just hate the way you refer to U'<sub>y</sub> as a neighbourhood of x, where V'<sub>y</sub> is a neighbourhood of y. It's really confusing, but, ah well, notation is all arbitrary, as they tell me. EDIT: No, I'm feeling grumpy about this! U<sub>x</sub> ∩ V<sub>y</sub> = Ø is the Hausdorff criterion, right? It is, at the very least, unilluminating, to write U<sub>y</sub> ∩ V<sub>y</sub> = Ø.
    not every closed set has a non-empty interior (consider the closure of any nowhere dense set for example).
    How about the everywhere dense Guitarist? Yep, you're right, silly me.

    Did i cover everything or have i missed a few?
    Nope, you're done, thank you. And now.....?
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  34. #33  
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    Quote Originally Posted by Guitarist
    Amen to that, though I'm not a huge fan of the one-day game. Fancy your chances, then?
    Yeah, why not We are going in rated the best one day team in the world but our chances looked good the last three times and we still havent got that damn trophy. Maybe 4th times the charm.

    [quote]Well, OK, I guess Ill just have to get used to your notation. But, I just hate the way you refer to U'<sub>y</sub> as a neighbourhood of x, where V'<sub>y</sub> is a neighbourhood of y. It's really confusing, but, ah well, notation is all arbitrary, as they tell me. EDIT: No, I'm feeling grumpy about this! U<sub>x</sub> ∩ V<sub>y</sub> = Ø is the Hausdorff criterion, right? It is, at the very least, unilluminating, to write U<sub>y</sub> ∩ V<sub>y</sub> = Ø.[quote]

    The problem is that we have a whole host of neighbourhoods of x to contend with in these proofs and the best way to index them IMHO is to label them with respect to the point they have been chosen to miss. We cant find one nhood U<sub>x</sub> such that U<sub>x</sub> ∩ V<sub>y</sub> = Ø for all y and so this is the best notation i can think of.

    Ok i want to sketch the proof of the existence of partitions of unity and then move on to manifolds. Should i prove Urysohn's lemma or can i just state the result (though the proof is important!)
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    Quote Originally Posted by river_rat

    The problem is that we have a whole host of neighbourhoods of x to contend with in these proofs and the best way to index them IMHO is to label them with respect to the point they have been chosen to miss.
    Ya, but you were not doing that. If you had, it would be weird, but at least consistent; U<sub>x</sub> is neighbourhood not containing x etc. But that's not what you're writing. You have x ∈ U<sub>x</sub>, x ∉ V<sub>x</sub>. I have problem with this notation.

    Sorry to be so pedantic, but we seem to be developing a real communication problem here.

    Let me set out my stall. In a Hausdorff space, for any pair of points x ≠ y, I may find neighbourhoods of x and y s.t. U<sub>x</sub> ∩ V<sub>y</sub> = Ø. This is, course, the definition of Hausdorff-ness (yes, I know, it's not a real word). But, as you say, we may have as many neighbourhoods of x as we like, let's call them U<sub>x</sub>, V<sub>x</sub> etc. Evidently, these guys are not disjoint. But, unless I badly misunderstood you, you want x ∈ U<sub>x</sub>, y ∈ V<sub>x</sub> s.t. U<sub>x</sub> ∩ V<sub>x</sub> = Ø. That's a queer way of expressing it. If I have to live with your notation, I promise you I shall struggle,
    We cant find one nhood U<sub>x</sub> such that U<sub>x</sub> ∩ V<sub>y</sub> = Ø
    What do you mean? We can't find any, or we may find many? In a Hausdorff space, the first is, by definition, false, the second may be, but rarely is.

    PS by EDIT: I'm aware this post must seem a real turn-off for casual viewers, for which I apologize. But hang in there, quiescent readers, it will come good in the end, it feel sure!

    Should i prove Urysohn's lemma or can i just state the result (though the proof is important!)
    Well, you mentioned Urysohn, but I can't remember if you proved it. Anyway, you choose, I'm putty in your hands
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    Hey guitarist - sorry for the long radio silence on my side but work has not been playing ball.

    Okay, lets unpack my notation the way i see it. We have two disjoint closed sets A and B and we know that for every x in A we can find two disjoint open subsets U and V such that x is in U and B is contained in V. Now the U and V you choose here depend in a huge way on what x you are looking at - in that you have a function from the space to the Cartesian product of the topology that chooses these two open subsets and that is why i index with respect to the x's in A.

    What i was saying here is that there does not necc. exist one open subset that contains B such that all nhoods of points of A miss it - the open superset of B you choose depends on the point of A you are looking at. Perhaps i did not phrase that correctly - sorry.

    Does that help with the notation?

    Edit - what was that about proof reading my posts again? Lol
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    Yes, that's a help, thank you (I even forgave you your typo, line 2, para 1!)
    Quote Originally Posted by river_rat
    - in that you have a function from the space to the Cartesian product of the topology that chooses these two open subsets
    But this had me scratching my head a little. Isn't this equivalent to forming the disjoint union topology on X? On reflection, no it's not, precisely for the reason you gave - U<sub>x</sub> ∩ U<sub>y</sub> = Ø and V<sub>y</sub> ∩ V<sub>z</sub> = Ø does not imply U<sub>x</sub> ∩ V<sub>z</sub> = Ø!

    Yay! World Cup's started, Windies 54 for 1 at last look.
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    I think we are all on the same page (and by all i mean guitarist and me by the looks of things) so lets move on to the union skin proof i.e. Urysohn's lemma :

    Let X be a normal topological space and let A and B be two closed disjoint subsets of X then their exists a continuous function f:X -> [0, 1] such that f(A) = 0 and f(B) = 1 (i.e. we can separate closed sets by functions)

    Proof:

    As the rationals in the unit interval are countable, let {r<sub>n</sub>} be an enumeration of all the rationals in the unit interval with r<sub>0</sub> = 0 and r<sub>1</sub> = 1. Let U<sub>0</sub> = X\B and as X is normal we can find an open U such that A is contained in U which is contained in cl(U) which is contained in X\B. Define U<sub>1</sub> = U.

    The trick here is to define by induction a sequence of open sets such that cl(U<sub>s</sub>) is contained in U<sub>t</sub> if s < t (which is where the onion skin analogy comes from btw). So lets suppose we have defined our U<sub>r<sub>i</sub></sub> for i = 1..n and consider the rational number r<sub>n+1</sub>. Now there exists two rational numbers r<sub>m</sub> and r<sub>l</sub> which are closest to r<sub>n+1</sub> with m, l < n+1 (as we only have a finite set) and by the induction hypothesis we know that U<sub>r<sub>m</sub></sub> is contained in cl(U<sub>r<sub>m</sub></sub>) which is contained in U<sub>r<sub>l</sub></sub>. This implies that cl(U<sub>r<sub>m</sub></sub>) and X\U<sub>r<sub>l</sub></sub> are disjoint closed subsets of X and as X is normal we can find an open U such that cl(U<sub>r<sub>m</sub></sub>) is contained in U which is contained in cl(U) which is contained in X\U<sub>r<sub>l</sub></sub>. We define U<sub>r<sub>n+1</sub></sub> = U.

    Now that we have our onion skin we define a function f:X-> [0, 1] by

    f(x) = inf{r: x is an element of U<sub>r</sub>} if x is not an element of B and f(x) = 1 if x is an element of B. This function is well defined as our set of rationals is bounded below by 0. If we can prove that this function is continuous we are done

    Before i do that little hurdle, any questions?
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    Quote Originally Posted by river_rat
    I think we are all on the same page (and by all i mean guitarist and me by the looks of things) so lets move on to the union skin proof i.e. Urysohn's lemma :
    This looked pretty hairy until I drew it out on paper. Just one thing, though.

    Now there exists two rational numbers r<sub>m</sub> and r<sub>l</sub> which are closest to r<sub>n+1</sub> with m, l < n+1 (as we only have a finite set) and by the induction hypothesis we know that U<sub>r<sub>m</sub></sub> is contained in cl(U<sub>r<sub>m</sub></sub>) which is contained in U<sub>r<sub>l</sub></sub>.
    I take it you mean for m< l?
    If we can prove that this function is continuous we are done :)

    Before i do that little hurdle, any questions?
    On the face of it, this looks relatively straightforward, lemme take a crack at it, if I can find time.
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    Yep - sorry i should of added that

    If you want a hint highlight below:

    Remember that you need only check the base elements of the codomain for continuity and in [0, 1] the basic open sets are of the form [0, a) and (b, 1].

    End of Hint
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    Sorry for the delay chaps, been a bit manic round here, plus I freely confess I was slightly puzzled by river_rat's hidden hint. That much, at least, I have figured now, but I'm not too sure about the rest.

    Let R carry the usual topology T with order. Let [0, 1] subset R. Then the subspace topology T<sub>[0,1]</sub> is induced on [0,1] as T<sub>[0,1]</sub> = V ∩ [0,1] (for all V ∈ T). Suppose (p,q) ∈ T. Then the following are open sets in T<sub>[0,1]</sub>:

    [0,1] = (p,q) ∩ [0,1], where p ≤ 0, q ≥ 1.

    (p, 1] = (p,q) ∩ [0,1], where p ∈ [0,1], q ≥ 1

    [0,q) = (p,q) ∩ [0,1], where q ∈ [0,1], p ≤ 0

    (p,q) = (p,q) ∩ [0,1] where p > 0, q < 1.

    Then the [0, q) and (p, 1] form an open neighbourhood basis for [0,1].

    Now continuity on top. spaces X and Y is defined as: f: X → Y is continuous at x in X if, for every open neighbourhood ε of f(x) in Y, there is an open neighbourhood δ of x in X s.t. δ ⊆ f<sup>-1</sup>(ε). Or less abstractly (perhaps): f: X → Y is continuous iff, for every open set U in Y, f<sup>-1</sup>(U) is open in X, a map from open sets to open sets, in other words.

    Now the the meat of it, the bit I'm not entirely confident about.

    Let A, B be disjoint closed sets in a normal top. space X. Evidently A ∈ X\B and B ∈ X\A. Let's assume that X\A ∩ X\B ≠ Ø.

    Let there be a surjection f:X → [0,1] s.t. f(a) = 0, f(b) = 1 for all a in A, all b in B. To show continuity at all x in X, I will construct a family {U<sub>i</sub>} of open sets in X ordered by inclusion, centred, say, on A as follows:

    Let U<sub>A,x</sub> denote the open set containing all a in A and some x in X, not in A, likewise U<sub>A,x,y</sub> and so on. Then A subset U<sub>A,x</sub> subset X\B. Similarly, let U<sub>A,x,y</sub> subset X\B contain A, x and y, and so on.

    I note that U<sub>A,x,y</sub> - A = U<sub>x,y</sub>, U<sub>x,y</sub> - clU<sub>x</sub> = U<sub>y</sub> for all x, y in X.

    I can perform the same trick centred on B.

    Then it follows that, for all f(x) in [0,p), (p,q) and (q,1] in [0,1], there are open sets U<sub>A,x</sub>, U<sub>x,y</sub> and U<sub>y,B</sub> for all x in X

    Eek! I don't like that at all.
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    Okay, let me show you half of the proof and then leave the other half as an exercise:

    As you showed, the basic open sets of [0,1] are of the form [0, p) and (q, 0] so all we need to show that f<sup>-1</sup>( [0, p) ) and f<sup>-1</sup>( (q, 0] ) is open for all 0< p, q < 1 (as you showed as well).

    So lets consider when 0<f(x)<p<1 for some p in [0, 1). Now we know that f(x) < p iff we can find a rational number p' < p with x in U<sub>p'</sub> (as f(x) is defined by greatest lower bounds and the open sets are an increasing chain). This shows that f<sup>-1</sup>( [0, p) ) = union of all U<sub>r</sub> where r is a rational number less then p which is the union of open sets and thus is open

    Does that make sense?
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    Not sure. You're using Q to index the open sets in X, right? Or maybe Q ∩ [0,1], I guess it amounts to the same thing. I suppose that's OK, it hadn't occurred to me to focus on the rationals, but as they are countable and dense in [0,1] it makes sense. I think our approach is broadly the same, yours of course has the virtue of rigour!

    Anyway, you have x ∈ U<sub>p'</sub>, which implies f(x) ≤ p', so with p' strictly < p this implies f(x) strictly < p, so f(x) is bounded below in [0,p)?

    The least upper bound is found analogously, I guess. Let q' > q in (0, 1] with U<sub>q</sub> subset U<sub>q'</sub> . Suppose x ∈ X\clU<sub>q'</sub> (or some open subset of). This implies f(x) ≥ q' > q, so, for q < f(x) < 1, f(x) is bounded above in (q, 1]. And for all rational s > q, f<sup>-1</sup>((q, 1]) is the union of all U<sub>s</sub>, thus open
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  44. #43  
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    Quote Originally Posted by Guitarist
    Not sure. You're using Q to index the open sets in X, right? Or maybe Q ∩ [0,1], I guess it amounts to the same thing. I suppose that's OK, it hadn't occurred to me to focus on the rationals, but as they are countable and dense in [0,1] it makes sense. I think our approach is broadly the same, yours of course has the virtue of rigour!

    Anyway, you have x ∈ U<sub>p'</sub>, which implies f(x) ≤ p', so with p' strictly < p this implies f(x) strictly < p, so f(x) is bounded below in [0,p)?

    The least upper bound is found analogously, I guess. Let q' > q in (0, 1] with U<sub>q</sub> subset U<sub>q'</sub> . Suppose x ∈ X\clU<sub>q'</sub> (or some open subset of). This implies f(x) ≥ q' > q, so, for q < f(x) < 1, f(x) is bounded above in (q, 1]. And for all rational s > q, f<sup>-1</sup>((q, 1]) is the union of all U<sub>s</sub>, thus open
    That looks about right - we need to use the countability of the rationals here else the induction doesn't go through sadly (unless you assume choice etc)

    So thats Urysohns lemma - it is customary to show Tietze's extension theorem about now but im in the mood to prove Tychanoff's theorem. Any suggestions?
    As is often the case with technical subjects we are presented with an unfortunate choice: an explanation that is accurate but incomprehensible, or comprehensible but wrong.
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    Whatever you think. Brouwer's fixed point theorem, one of the most amazing in all topology, would be fun to do at some time, though I think the proof is quite hard.

    Anyway, are you going to do manifolds, after all? It would make tensor transformations easier for people to understand, I think.

    By the way, your new signature line - "not even wrong!" - is a contraction of a famous Pauli saying, not even eligible to be wrong. Apparently, Pauli was a great wit. My favourite of his is this, referring to Einstein's abortive attempt to find a unified field theory: "what god has torn asunder, let no man join", a parody on the Anglican marriage ceremony.
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  46. #45  
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    Quote Originally Posted by Guitarist
    Whatever you think. Brouwer's fixed point theorem, one of the most amazing in all topology, would be fun to do at some time, though I think the proof is quite hard.
    The general Brouwer Fixed point theorem uses homology theory - it boils down to showing that you cant retract a ball onto its boundary without tearing it and to do that you compare homology groups. The proof is not hard, but homology theory takes a while to set up.

    I can do the 1D case in a few lines though - Let f:[0,1] -> [0, 1] be continuous, then there exists a t in [0, 1] such that f(t) = t.

    Proof: Define a continuous function h(x) = f(x) - x, then if h(x) = 0 then x is the required fixed point. Define t = Sup{x in [0, 1] : f(x) > x} which exists as 1 bounds this set from above. I claim that h(t) = 0

    Suppose that h(t) > 0, then we can find an open nhood of t contained in [0, 1] (say (t-d, t+d) for some d>0) such that for all x in this nhood h(x) > 0. Let s be an element of (t, t+d) then s > t and f(s)>s which contradicts that t is an upper bound of the set {x in [0, 1] : f(x) > x}

    So suppose that h(t) < 0, then we can similarly find an open nhood of t contained in [0, 1] such that for all x in (t-d, t+d) h(x) < 0. Let s be an element in (t-d, t) then f(s) < s and s < t. I claim that s is an upper bound for the set {x in [0, 1] : f(x) > x} for suppose there is an x greater then s with f(x) > x then x lies in (s, t) which is contained in (t-d, t) so f(x) < x which is a contradiction. But then s is an upper bound strictly smaller then the least upper bound, a contradiction.

    So h(t) = 0 i.e. f(t) = t. QED

    Im planning the manifold thread as i type
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