OK, so it's my turn to liquefy your brains!
Suppose that and that is compact. Let be continuous, prove that if then has a fixed point in .
PS. I solved this on another forum, just so you know

OK, so it's my turn to liquefy your brains!
Suppose that and that is compact. Let be continuous, prove that if then has a fixed point in .
PS. I solved this on another forum, just so you know
Let X={0,1) Define f as follows f(0)=1, f(1)=0. X is finite, hence compact. X is discrete in the relative topology, hence any function is continuous. In particular f is continuous on X. f(X)=X. f has no fixed point.Originally Posted by Guitarist
Are you sure about this? The proof I have in mind accommodates your "inversion" function.Originally Posted by DrRocket
If anyone else is interested (which I start to doubt), all that DrRocket's function does is pick up a line segment and turn it around. Intuitively we see that there must be some "pivot point" which is fixed under this function
BTW, compactness and continuity are data, there is no need to reargue them
DrRocket's function is on the two point set Guitarist, not a line segment.
River rat is correct. My function is defined on the set consisting of just two points, 0 and 1. It is a counterexample to the purported theorem.Originally Posted by Guitarist
If the set were a segment, one could use the connectedness to show the existence of a fixed point, basically the intermediate value theorem of elementary calculus.
I did not "reargue" compactness or continuity. I observed that the set X is compact and the function f is continuous, showing that they met the hypothesis of your theorem, but that the conculsion was not met thereby producing a counterexample.
Well, well. {0,1) is a 2point set? In fact it is a set of any sort? Then I must have slept through several topology tutorials.
Just to be clear, in case it wasn't blindingly obvious  I was talking about subsets of with the standard topology.
In fact, don't bother, this exercise has gone sour on me.
Nice work......
Hey Guitarist
All DrRocket is trying to show is that you must have used more then you think you used in your proof  as the result doesn't follow from the given assumptions. {0, 1} is a compact set in the standard topology and it serves as a counterexample. So at least we know we must be looking at infinite compact subsets of [tex]\mathbb{R}[\tex]. So double check your proof and repost the question, as it does sound interesting...
Look. Anyone with even a passing familiarity with notation will have recognized "{0,1)" as a typo. You apparently interpreted this as "{0,1}"  I did the other thing, that is I interpreted it as "(0,1)". Since either interpretation is valid, I make no apology for that.
What I do hold my hands up to, though, is this: I should have specified that the set is connected (by your interpretation, {0,1} is not since it is the union of disjoint sets {0} and {1}). So, bad boy me.
But what gets my goat is this: This is basically a chatroom, where we come to have fun; it is not fun to be sat upon by anyone whether one accepts they are more knowledgeable than one's self or not.
In short my exercise was supposed to be entertainment for we sad bastards who get our kicks that way. Given the above, would not a better, more friendly, response to have been something like "The thm is true iff is connected, so for the sake of entertainment, let us assume this is the case".
Obviously this would fail us in our Cambridge Tripos  so what, this is an effing chat room
Yes, it was a typo. But DrRocket also wrote:Originally Posted by Guitarist
If you had read that, you ought to have known that DrRocket couldn’t possibly be referring to the open interval since that is neither finite nor compact.Originally Posted by DrRocket
So DrRocket was taking the twoelement set with the discrete topology, which is compact but not connected. This fails to satisfy the result you want. Thus, for that result to hold, you might need to be connected as well as compact.
Okay, I believe the problem comes down to this. We want to be a compact, connected subset of with the usual topology. That means (by the Heine–Borel theorem) that is a closed and bounded interval. Hence we may take . If any other closed and bounded interval, simply transform to via the homeomorphism .
So, the problem can be restated as follows:
Here’s my attempt.Prove that any continuous function where has a fixed point.
If or , then 0 or 1 is our fixed point. Assume that and .
Since the domain of is a subset of its range, there exist such that and . And as we are assuming that and , we have and .
Either or . In the former case, for all ; in the latter for all . In either case, and so makes sense for all .
Now define a new function by . Then is continuous and
Hence, by the intermediatevalue theorem, there exists such that . In other words, is a fixed point of .
So unless you want me to prove the intermediatevalue theorem, I should think that wraps it up.
Yes, I guess that'll do it, though your function gave the willies for a while.
Here's my somewhat more pedestrian approach, again assuming connectedness;
By the HeineBorel Thm, compact subsets of are closed and bounded.
Let , so that
We know that the continuous image of a compact set is compact. Thus .
Since , we may assume that . Let us assume that , otherwise there's nothing to prove.
Define the continuous function by . Now since then . Similarly, since then .
Then by the Intermediate Value Thm, for some we must have which from the above implies i.e. for some .
But since , this can only be true when . So is a fixed point under
I was pointing out in my post you just deleted that you did not make use of the assumption of connectedness in your proof. You stated your assumption, but that is not the same as using it in your proof! You did not do the latter.
Besides, this
does not seem right. Since , you are assuming that . Why if ? That is possible too.Originally Posted by Guitarist
Right, this is where you went astray.
From , you can only say that there exist such that and . Then since and are the inf and sup respectively. and have nothing to do with this part of the proceedings.Originally Posted by Guitarist
The next step is to assume that the inequalities are strict. If or , then we have a fixed point and there would be nothing to prove. So assume and .
Now, any one of these is a possibility: , , . We don’t know which. However, we can be sure of one fact:
for all
This is the important point where the connectedness assumption comes in. Since is connected, . Hence if we set , we have for all . This means that makes sense for all . If we define , we can then show that for some – then is a fixed point of .
If were not connected, then would not be the interval (it would be a finite union of disjoint intervals); then we can’t say that for all . That would make a difference in our being able to proceed with our proof.
Here is a slight variation of the problem. As before, let be compact and connected and be continuous. But now, assume that (instead of the other way round).
Prove that still has a fixed point.
If you make the assumption that the set X is connected as well as compact then X is simply an interval. and the proof basically boils down to showing that if the graph of a continuous function is contained in the unit box then it must intersect the diagonal somewhere. This is basically the intermediate value theorem from elementary calculus which is a reflection of the fact that continuous functions preserve connectedness.Originally Posted by JaneBennet
The real point here is not only that connectedness is necessary, but that the theorem is rather unique to the real line. If you go to even slightly more general spaces then the theorem fails. For instance if you consider the unit circle, the the function of rotation by any angle other than a multiple of 2 pi has no fixed points.
Of course {0,1} is a set. It is a set with precisely two elements  0 and 1. It is also in this example a subset of with the standard topology. That relative topology happens to be the discrete topology on that twopoint set. This is in fact very standard topology, about page 2 in most books.Originally Posted by Guitarist
I have no idea what your complaint is addressing.
The counterexample serves to show that stronger hypotheses are required for your theorem. There are rather a lot of compact subsets of the real line and the theorem appears to apply to only connected compact sets, which are just intervals, and a rather small subclass of the compact sets.
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