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Thread: Set theory axiom

  1. #1 Set theory axiom 
    Forum Masters Degree bit4bit's Avatar
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    I've decided to start reading into topology, which is apparently based on Set theory. My set theory is pretty useless, as I've never gone very far with it, but it seems topology is based on it, and can have some useful applications. So, I've started reading about set theory again, and I'm having trouble with this axiom: (...The first axiom :P)

    "Using the notion of equivalence of sentences we may rewrite the first axiom as:

    Axiom 1b:

    Consider two sets A and B. Then A = B if and only if x ∈A
    ⇔x ∈B."

    Am I first right in thinking that for, x ∈A⇔x ∈B to be true, then both x ∈A AND x ∈B must BOTH be true, or BOTH be false? so if one is true and the other is false, then the statement is false?...In which case the operator '⇔' is much like an "XOR" (exclusive OR) operator?

    If thats right, then what about the case where x DOES NOT belong to A, and x DOES NOT belong to B?...in which case x ∈A is false, and x ∈B is false, meaning the statement is true that A=B. but how can this be so? A, and B could be anything....in fact, because of that, what if we had:

    A={x, 1, 3, 5}
    B={x, 2, 4, 6}

    even if x ∈A AND x ∈B, it doesn't mean A=B. I think I'm missing something important here. :?


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  3. #2  
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    A = B if and only if x ∈A
    ⇔x ∈B.

    i think this means for all values of x, where x is an element of the set. so we have that if all values of x are in A then the sets A and B are equal iff all the values in the set A are found in B and all the elements of set B are in A.

    but that's just what me thinks


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  4. #3  
    Forum Professor serpicojr's Avatar
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    Wallaby's completely right--the axiom is just saying two sets are equal iff they contain the same elements. x is just supposed to represent an arbitrary element in the axiom. You'll be applying this axiom in topology mostly in the following sorts of fashions:

    -Let a be in A. Show that a is in B. Now let b be in B. Show that b is in A. Then A = B.

    -Show that A is contained in B. Now show B is contained in A. Then A = B. (This uses the definition of containment, i.e. that A is contained in B iff for all a in A, a is in B.)

    You don't need formal set theory to be able to do topology. There are just a handful of ideas you have to be comfortable with--set equality, containment, unions, intersections, and complements and how these things interact with each other. I'm not trying to discourage you from learning set theory--I'm just telling you that you can (and should) play it fast and loose with sets when doing topology.

    (Having an understanding of abstract functions is also helpful--you should learn about injective, surjective, and bijective functions, images and preimages of elements and sets, and how set operations and relations behave under images and, more importantly, preimages. The fact that preimages are more important than images may seem odd, but it will make sense when you learn about continuous functions.)
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  5. #4 set theory axiom 
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    as you have mentioned if x belongs to A and x belongs to B also then we do not have set A = set B. We have x common to both sets that is A intersection B gives us x.in all other cases set A is not equal to set B. if i am not mistaken set A = set B when all members of set A are also members of set B
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  6. #5  
    Forum Masters Degree bit4bit's Avatar
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    Oh so x is just an arbitrary representation of the elements in a set? This makes much more sense. If x was considered to be only a single element of the set then I think we would simply have some intersection of the sets in the case above, rather than the sets being equal. but if all the elements in A are in B, and all the elements in B are in A, then it makes much more sense that they'd be equal.

    Sepicojr: The text I am reading on set theory is more-a-less a chapter at the beginning of a topology book, so I'm assuming it will stick to the essentials needed for topology rather than an in depth set theory course. This is good, cause at the end of the day all I wanna do is learn the topology. I'm guessing the abstract functions will be introduced too at some point in the beginning of the book, but thanks for suggesting them anyway, because alot of texts always assume prior knowledge when jumping in at the deep end.

    Thanks for the replies.
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  7. #6  
    Forum Professor serpicojr's Avatar
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    Okay, that sounds good. What topology book are you using?

    I realized that I left out a few set theoretic things which are important to topology--for example, power sets, Cartesian products, cardinality... I imagine this is all covered in that chapter.
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  8. #7  
    Forum Masters Degree bit4bit's Avatar
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    Thanks, the book I'm using (if you can call it a book) is this:

    http://ocw.mit.edu/OcwWeb/Mathematic...otes/index.htm

    The first pdf is nice, but the rest are just scans of written work, which isn't the best to work from. It doesn't look as though the first chapter talks about abstract functions, but they might be somewhere later on in the notes.

    This was the best online resource I could find for topology...besides a few introductory pages I read...so please let me know if you have something better....I'm too broke to even pay my outstanding fines off at the library and get a new book out at the moment. :P

    Cardinalitiy I've heard of before, as being like "a measure of infinities" or something like that. I'll have to look at it again along with power sets and cartesian products as you say.

    Also, just wondering:

    -Let a be in A. Show that a is in B. Now let b be in B. Show that b is in A. Then A = B.
    Can I say:

    A⊂B ⇔ [a∈A ⇔ a∈B]
    B⊂A ⇔ [b∈B ⇔ b∈A]
    A=B ⇔ [A⊂B ⇔ B⊂A]

    ...does that make sense / is it correct?

    Could I also use the third expression to say:

    -Show that A is contained in B. Now show B is contained in A. Then A = B. (This uses the definition of containment, i.e. that A is contained in B iff for all a in A, a is in B.)
    Thanks
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  9. #8  
    Forum Professor serpicojr's Avatar
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    Okay, you're not going to get very far with just the notes on that page. Note that a whole five weeks of material is skipped from the first to the second set of notes. This is bad. By the time you get to week 7, a lot of topology has already been covered, and you'll have no idea what they're talking about. Unless you have Munkres at your side (and it sounds like you don't and can't make it happen in the near future), you're really in a bind.

    I really like the first set of lecture notes, though! It's basically introducing the very basic set theory and logical arguments that you need to do topology (and, really, most abstract math), and it also serves as a nice way to introduce proofs.

    As for:

    A⊂B ⇔ [a∈A ⇔ a∈B]
    B⊂A ⇔ [b∈B ⇔ b∈A]
    A=B ⇔ [A⊂B ⇔ B⊂A]

    Well, the axiom of set equality states that "A=B ⇔ [x∈A ⇔ x∈B]". Thus, in the first two, I can replace each of the sentences in brackets with "A = B"--the fact that you're using a or b or x as the name for an arbitrary element doesn't matter. So the first two become:

    A⊂B ⇔ A=B
    B⊂A ⇔ A=B

    These don't make sense--A is a subset of B iff A is equal to B?

    As for the third sentence, one implication is true: if A=B is true, then A⊂B ⇔ B⊂A is true, since both A⊂B and B⊂A are true. But the other implication is not true. Let A and B be nonempty, disjoint sets (i.e., their intersection is the empty set). Then both A⊂B and B⊂A are false, making A⊂B ⇔ B⊂A true. But A=B is false.

    If we want to formalize what I was saying, we could say this:

    A is equal to B iff whenever x is in A, x is in B and whenever x is in B, x is in A: A=B ⇔ [[x∈A ⇒ x∈B] and [x∈B ⇒ x∈A]]

    A is equal to B iff A is contained in B and B is contained in A: A=B ⇔ [A⊂B and B⊂A]

    A is contained in B iff whenever x is in A, x is in B: A⊂B ⇔ [x∈A ⇒ x∈B]

    What I was trying to say is that the notes you're reading are not trying to prime you for reading complex logical statements with tons of ⇔'s and ⇒'s floating around, so don't focus on this. Rather, the notes are trying to prime you to understand the basic arguments and strategies you'll be seeing and using in proofs. Focus on reading and understanding the proofs and on doing the exercises. As you read more proofs and prove more things on your own, the basic set and logical operations will become second nature, so that you'll immediately be able to whip out things like, "To prove P implies Q, I can show not Q implies not P," and, "To show A and B are disjoint, I'll show B lies in the complement of A."
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  10. #9  
    Forum Masters Degree bit4bit's Avatar
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    Ahh thanks I think I understood that....I keep forgetting that the iff statement means BOTH have to be true for the statement to be true (or BOTH false?). I'll try not to get to bogged down with long expressions anyway, and carry on with that pdf.

    I've also had a look at the abstract functions you mentioned, and realised I have actually breifly learned about them before...though when I learned about them, we called them one-to-one, one-to-many, many-to-one, and many-to-many functions. I don't remember much of that to be honest, but I thought these categories were relating the variables of a given function, for example:

    y=f(x) ......one-to-one
    y=f(x,y) ........many-to-one
    Φ(u) = (f(u),g(u)) .........one-to-many
    Φ(u,v) = (f(u,v),g(u,v),h(u,v)).......many-to-many

    Well thats what I thought, but looking at them again it,s something to do with functions relating the elements of one set to the elements of another. I couldn't really get to grips with it. I also didn't understand the meaning of the expressions like:

    f: A → B

    From what I picked up this is a 'map'..showing how the elements of one set A are related to the elements of another, B, but I'm not quite sure.
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  11. #10  
    Forum Professor serpicojr's Avatar
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    The notation f: A -> B just means that f is a function from a set A to a set B. That is, you should really read f: A -> B as the sentence "f is a function from A to B" or the fragment "a function f from A to B". It's just like writing a∈A, which means "a is an element of A" or "an element a of A".

    So given f: A -> B, let's go over some definitions:

    f is injective or one-to-one if no two distinct elements of A are mapped to the same element of B. This is formally written as, "If x, y∈A and f(x) = f(y), then x = y," or, "If x, y∈A, x ≠ y, then f(x) ≠ f(y)."

    f is surjective or onto if every element of B is the image of an element in A. This is formally written as, "If x∈B, then there exists y∈A such that f(y) = x."

    f is bijective if f is injective and surjective.

    You'll begin to think of multivariable functions as just functions of one variable as you learn abstract math. For example, a function f of two variables x and y from sets A and B can be thought of as a function of one variable z from the set C of pairs of elements of A and B.
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  12. #11  
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    Quote Originally Posted by serpicojr
    So given f: A -> B, let's go over some definitions:

    f is injective or one-to-one if no two distinct elements of A are mapped to the same element of B. This is formally written as, "If x, y∈A and f(x) = f(y), then x = y," or, "If x, y∈A, x ≠ y, then f(x) ≠ f(y)."
    If x and y both belong to A then how can the function relate to any elements of B (f: A->B)? Isn't that just relating one element of A, with another different element of A? I'm finding this combination of sets and functions really hard to grasp...in particular seeing what possible applications ther could be to make functions between elements of different sets. It must have some, since I've seen all this stuff used before reading some of wikipedias over-complicated articles on things.
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  13. #12  
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    Well, higher math is pretty much just functions from sets to other sets.

    When I write f(x) and f(y), I mean the elements of B to which f sends x and y (elements of A). So the equality f(x) = f(y) is interpreted as equality of elements of B, and f injective implies x = y as elements of A.
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  14. #13  
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    So x and y are in A, but f(x) and f(y) are in B? I think I see what you mean.

    I'm finding it hard to see how this kind of thing can be at all useful, but theres a point in that pdf i'm reading about a similar thing as above, where x,y ∈R, and f(x),f(y)∈R or something like that. I also remember you saying ages ago as a reply to a thread of mine, something like "you need to compose a map:

    f: R<sup>3</sup> -> R"

    ...and I didn't have a clue what it meant.
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  15. #14  
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    It's useful to talk about a function without specifying it or without giving an exact formula for it. You already use this notation! Consider the function sin: R -> R! More generally, I can say something like:

    Let f: R -> R be a differentiable function. Then the integral of f'(x) from a to b is f(b)-f(a).

    Useful, eh?
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  16. #15  
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    Thanks, I can understand why it's important to talk about functions generally as being f(x), rather than specifying the exact relationship between f and x because you can see what quantities are being related to one another. I can't see why it's necesary to take it any further using the notation f: A -> B. In your example, why isn't it enough to just say:

    Let f(x) be a differentiable function. Then the integral of f'(x) from a to b is f(b)-f(a).
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  17. #16  
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    Because A and B aren't necessarily the real numbers or even subsets thereof. For example, if f is a complex-valued function of two real variables, I have to say so. And writing f: R<sup>2</sup> -> C is a lot more compact than the wordy description given above.
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  18. #17  
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    Oh yeh that is kind of useful I suppose, as kind of a shorthand for describing the general nature of function. Using this practically, is it always the special sets (I mean number sets like rational, real, complex, etc) that are the sets of interest (as your example above), or are the sets often found as representing other things too?

    I suppose the use is different depending on what area of maths it's being used in? What about Topology?..Are the sets used to represent different spaces, like R<sup>2</sup>, R<sup>3</sup> etc?

    If f(x) = x<sup>2</sup> + 3x + 5 Would I say f: R -> R, and the function is injective or one-to-one?

    And if f(x,y) = 3xy<sup>2</sup> + 4yx<sup>2</sup> Would I say f: R<sup>2</sup> -> R and the function is surjective or 'onto'?

    For your example above, you could have a function like f(x,y)=xy+xyi ?

    Thanks
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  19. #18  
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    Quote Originally Posted by bit4bit
    Oh yeh that is kind of useful I suppose, as kind of a shorthand for describing the general nature of function. Using this practically, is it always the special sets (I mean number sets like rational, real, complex, etc) that are the sets of interest (as your example above), or are the sets often found as representing other things too?
    The sets are not necessarily sets of numbers. They can be sets of sets, they can be sets of points on some geometric object, they can be sets of functions, they can be sets of purely abstract elements. So when we just write A and B as sets, we're just talking about abstract sets.

    I suppose the use is different depending on what area of maths it's being used in? What about Topology?..Are the sets used to represent different spaces, like R<sup>2</sup>, R<sup>3</sup> etc?
    When we're talking about specific kinds of sets, or if we're trying to emphasize particular aspects of these sets, we'll call them by different names. For example, in topology, we look at topological spaces, open sets, and closed sets, among others. In linear algebra, we look at vector spaces. The set R<sup>3</sup> may be looked at as a topological space or a vector space (or both) depending on what sort of structure you want to emphasize.

    If f(x) = x<sup>2</sup> + 3x + 5 Would I say f: R -> R, and the function is injective or one-to-one?
    Look at the definition of injective again and see if this function meets the criteria. (Specifically, can you find real numbers x ≠ y such that x<sup>2</sup> + 3x + 5 = y<sup>2</sup> + 3y + 5?)

    And if f(x,y) = 3xy<sup>2</sup> + 4yx<sup>2</sup> Would I say f: R<sup>2</sup> -> R and the function is surjective or 'onto'?
    The function here is onto, but I want you to explain to me why it is. Namely, for any real number r, I want you to find values x and y such that f(x,y) = r.

    For your example above, you could have a function like f(x,y)=xy+xyi
    Indeed!
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  20. #19  
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    Thanks!

    Look at the definition of injective again and see if this function meets the criteria. (Specifically, can you find real numbers x ≠ y such that x2 + 3x + 5 = y2 + 3y + 5?)
    I subtracted the 5's and factorized both sides of the equation to get:

    x(x+3)=y(y+3)

    If x(x+3)=y(y+3)=0, then y=0 or -3, and x = 0 or -3

    So when y=0 and x=-3, x ≠ y, but f(x)=f(y) so it can't be injective.

    The function here is onto, but I want you to explain to me why it is. Namely, for any real number r, I want you to find values x and y such that f(x,y) = r.
    Can't you simply say, if x=1, and y=2:

    f(x,y) = 3xy<sup>2</sup> + 4yx<sup>2</sup>
    =3*4+8=20

    since 3,4,20 ∈R, f:R<sup>2</sup>->R is surjective?
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  21. #20  
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    Good job showing that f(x) = x<sup>2</sup> + 3x + 5 is not injective!

    As for the surjective function, you showed that 20 is in the image of the function, but you've got infinitely more real numbers to go before you convince me that it's surjective. You need to show me that EVERY real number r is in the image. (As a side note, "for every" and "for any" are used synonymously in math. Think of, "For any x, blah is true," as meaning, "I can choose any x, and no matter what x is, blah is true.")
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  22. #21  
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    Thanks

    but you've got infinitely more real numbers to go before you convince me that it's surjective
    I knew you were gonna say that! I've been thinking about how I can show it for all real numbers, and I just can't fathom it. I'm sure I need to use some logic/set functions somehow like union, intersection, iff or even just belonging, but I can't think how.

    I wrote this, but I'm not sure it passes for a proof or even makes sense:

    f(x,y)∈R <==> x,y∈R

    therefore, since x and y are real numbers f(x,y) must also be a real number?
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  23. #22  
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    You have to convince me that if I give you a real number r, you can find real numbers a and b that I can plug into f and get r--i.e., f(a,b) = r. I think you're relegated to showing this in a fairly direct manner--i.e., you have to describe what a and b are depending on what r is. Let me give you a start.

    Suppose r > 0. Let's let a = 1 for kicks. Then we need to find a value b such that:

    r = f(1,b) = 3*1*b<sup>2</sup> + 4b*1<sup>2</sup> = 3b<sup>2</sup> + 8b

    Well, we can do this using the quadratic formula. We're looking for a value of b that satisfies:

    3b<sup>2</sup> + 8b - r = 0

    So we can take:

    b = (-8±sqrt(64+12r))/6

    Note that, since r > 0, 64+12r > 0, and so b is real. Thus, given this choice of b, we have f(1,b) = r.

    What can you do when r ≤ 0?
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  24. #23  
    Forum Masters Degree bit4bit's Avatar
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    Thanks. Shouldn't it be:

    3b<sup>2</sup>+4b - r =0 ?

    In which case, the quadratic formula gives:

    b=(-4±sqrt(16-12r)) / 6

    Note that, since r > 0, 64+12r > 0, and so b is real.
    I don't understand this. Is it because the discriminant must be positive for b to be a real number?....because if it's negative then b would be complex having a square root of a negative? So all I need to do is to show that the discriminant is positive? but if r>0, then what if r=2? the discriminant would then be negative wouldn't it?
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  25. #24  
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    You're right, I changed the 4 to an 8. But the discriminant is still going to be positive--it's 16+12r (-4ac = -4*3*-r = 12r). But, yeah, if the discriminant is nonnegative, then the roots are real. If it's negative, they're complex. But this is just a proof that I can hit any positive number r by our function f(x,y). Now we need to show that we can do it for negative numbers and 0, too.
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  26. #25  
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    Oh yeh, I got my signs mixed up there. For when r=0,

    0=f(1,b)=3b<sup>2</sup>+4b
    0=b(3b+4b)

    therefore you can get r=0 when b=0 or b=-4/3

    when r<0:

    16+12r can be negative. therefore the roots are in the complex numbers. Doesn't this disprove it?

    Also, don't we also need to do the same for a?...and doesn't setting one variable constant while changing the other affect the outcome of the discriminant? Don't we really need a way to show that any combination of real numbers, a and b, can be used to get a real value r? For example we have shown that when a is set to one, then a real number r≥0 can be obtained with real values of b, but don't we then need to show that this is simultaneously the case for allal values of a? Thanks
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  27. #26  
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    Well, we don't have to choose x = 1. And the issue isn't that we have to show that f(a,b) is real for any real values a and b. This is clear: f(x,y) is a function which takes two real inputs and performs various arithmetic operations on them involving real numbers, so it must return a real value. The issue is whether can we obtain every real value by plugging real numbers into f(x,y). That's what onto/surjectivity is about.

    So we've already shown that the function maps onto nonnegative numbers. Now suppose r < 0. We wish to find values a and b such that:

    f(a,b) = 3ab<sup>2</sup> + 4ba<sup>2</sup> = r

    I'll accomplish this by letting a = -1. Then for any real value b, we have:

    f(-1,b) = -3b<sup>2</sup> + 4b

    So we are trying to solve:

    -3b<sup>2</sup> + 4b = r

    or:

    -3b<sup>2</sup> + 4b - r = 0

    Again, we use the quadratic formula:

    b = (-4 ± sqrt(16-12r))/-6

    But this time, r < 0, and so 16-12r > 0, and so the roots of this equation are real. So we can indeed find a value for b such that:

    f(-1,b) = r

    So every real number r is in the image of f(x,y), i.e. we can find real numbers a and b such that f(a,b) = 3ab<sup>2</sup> + 4ba<sup>2</sup> = r. This is why the function is onto.
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  28. #27  
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    From Wikipedia:


    Injective:


    Surjective:


    Bijective:
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  29. #28  
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    Thanks alot! I guess I was looking at it all wrong. So it's all about seeing whether the function spans the entire co-domain, whether we use all the values from the domain to get to it or not?
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    Exactly. The definition, "f: A -> B is surjective iff for every b in B, there exists an a in A such that f(a) = b," is definitely the most useful at your stage, and so understanding how this is saying the same thing as what you just said is definitely worthwhile.
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  31. #30  
    Forum Masters Degree bit4bit's Avatar
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    Jul 2007
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    Cheers again. I'm gonna try and make up some examples of functions for myself, and prove whether they are injective/surjective/bijective. I'll post my answers here if do.

    However I'm still working through my multivariable calc book amongst other things (including having a life), so I'm crammming alot in.

    Looking around at this stuff, it seems to be classified as pre-calculus, but when I learned about calculus, I'd never seen this stuff before...not in depth anyway.
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