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Metric space and subsets of Euclidean space

  1. Sep 10, 2007 #1
    I am having some troubles understanding the following, any help to me will be greatly appreciated.


    1) Let S1 = {x E R^n | f(x)>0 or =0}
    Let S2 = {x E R^n | f(x)=0}
    Both sets S1 and S2 are "closed"

    >>>>>I understand why S1 is closed, but I don't get why S2 is closed, can anyone explain?<<<<<



    2) "A set X is a "metric space" if it admits a function d: X x X -> R such that:
    (i) d(x,y)>0 or =0
    (ii) d(x,y) = 0 iff x=y
    (iii) d(x,y) = d(y,x)
    (iv) d(x,z) < or = d(x,y) + d(x,z)"

    >>>>>Now, I just don't get the "d: X x X -> R" part at all. How can you multiply two sets together? I don't get the notation, can someone please explain?<<<<<



    3) "The mapping f: A->B is "one-to-one" if f(x)=f(y) implies x=y, and f is said to map A "onto" B if f(A)=B. A mapping f: A->B is said to be "invertible" if there is another mapping g: B->A such that g(f(x)) = x for all x E A and f(g(y)) = y for all y E B.
    ===================
    The equation g(f(x)) = x can be valid for all x E A only if f is one-to-one, and the equation f(g(y)) = y can be valid for all y E B only if f maps A onto B. Conversely, if these two conditions are satisfied, then f is invertible." <<<<<-These are quoted from my textbook, but I don't understand how they come up with the conclusions in these last two sentences. Could someone kindly explain the reasonings?>>>>>

    Thank you!
     
  2. jcsd
  3. Sep 10, 2007 #2
    1) Define f(x) to =0 if x lies inside some non-closed set A and -1 if x does not lie inside A. Both, S1 and S2 are not closed (because they are A). The only way for S1 and S2 to be non-closed would be that there is no non-closed set A which is a subset of R^n. But I think that except for the empty set, all open sets are non-closed. And open subsets of R^n do exist.
    2) "f : X x X -> R" means that f maps two elements from X on a real number. More specifically: You take two elements/numbers x and y from X, "multiply" them with the metric d (metric is just the name for the function d) and get a real number as a result. That's pretty much the same as the scalar product for vectors from R³: You take two vectors, multiply them and get a real number as result.
    3)
    - Assume [tex] x \neq y [/tex] and f(x) = f(y). What does g(f(x)) equal? What does g(f(y)) equal? Can that fit both conditions?
    - Assume f does not map A onto B. Then, there is (I think) some y in B which is not mapped by A. What does f(g(y)) equal, then?
     
    Last edited: Sep 10, 2007
  4. Sep 11, 2007 #3

    matt grime

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    1) If you see why the first set is closed, why don't you see that the second set is closed? I have no idea what Timo attempted to say.

    You're missing the assumption that f is continuous, or the statement is false. That is a big hint.


    2) You are not multiplying anything (Timo, stop implying he is!). XxX is the cartesian product of X with X, i.e. the set of all pairs (x,y) with x,y in X. A metric takes two points in X (i.e. a point in XxX) and tells you how far apart they are in some sense.


    3) Well, why don't you write out some non injective and non surjective functions from {1,2} to the set {1,2} and figure ouw what's going wrong?
     
  5. Sep 11, 2007 #4

    HallsofIvy

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    By the way, you say
    but don't say anything about f! Are you given any information about f? In particular is it assumed to be continuous?
     
  6. Sep 11, 2007 #5
    I tried using english sentences ("because it's less opaque") to say:
    Let [tex]A \subseteq R^n[/tex], A is non-closed,
    and [tex] f(x) := f_A(x) = \left\{ \begin{array}{rcl} 0 &:& x \in A \\ -1 &:& x \notin A \end{array} \right. [/tex]
    [tex] \Rightarrow S1 = S2 = A [/tex]

    What exactly was the problem understanding that? That I didn't explicitely state that [tex] A \subseteq R^n [/tex]?
     
    Last edited: Sep 11, 2007
  7. Sep 11, 2007 #6

    D H

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    The statement is false for many functions f, continuous or not. Example: [itex] f(x)\equiv0\;\forall x\in\mathcal R^n [/itex]
     
  8. Sep 11, 2007 #7

    HallsofIvy

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    ?? {x| f(x)= 0} = Rn which is closed!

    If f is continuous then f-1(A) is closed whenever A is closed. {0} is closed (all singleton sets in a metric space are closed) so {x|f(x)= 0}= f-1({0}) is closed.
     
  9. Sep 11, 2007 #8
    That's not a good example. What you probably mean is that then [tex]S1=S2=R^n[/tex] and that [tex]R^n[/tex] is open. That seems true, but: A set being open does (according to the definition I am familiar with) not exclude that it's closed. According to my definition, a set is closed if it's complement is open. The complement of R^n is {}. {} is open => R^n is closed. Sidenote: f(x)=-1 for all x would also not work; {} is open and closed, too.
     
    Last edited: Sep 11, 2007
  10. Sep 11, 2007 #9

    D H

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    I was taught that R^n is open. It appears it is both open and closed, as is the empty set. From http://mathworld.wolfram.com/EmptySet.html, "Strangely, the empty set is both open and closed for any set X and topology."
     
  11. Sep 11, 2007 #10
    a) is obviously true when f is continuous.

    a counterexample would be f:R->R s.t

    f=-1 for x nonpositive (-infinity,o] and f=0 for x positive ie for x in (0,infinity).

    Then S1=S2=(0,infinity) which is NOT closed.

    for part b) X x Y is a cartesian product....it describes the set of all ORDERED pairs (a,b) s.t. a is in X and b in Y. So R^2 can be described as R x R b/c R^2 ={(a,b), a and b real). What it means for (a,b) to be an ordered pair is that (a,b) DOES NOT equal (b,a) (unless of course b=a).

    for part c) the reasoning is simple, construct an example of such a function that satisfies one-to-one, like y=x and one that doesn't y=x^2. OnetoOne-ness ensures that your inverse function is properly defined. (functions have ONE output per input, whereas if we put the inverse x=squareroot(x), then if we input 4, we get BOTH 2 AND -2 which makes this inverse NOT a properly defined function.

    hope this helps.
     
  12. Sep 11, 2007 #11

    matt grime

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    It wasn't clear what you were trying to do. I now see you were attempting to provide a counter example to a true statement (albeit one clearly missing the obvious hypothesis that f is continuous).
     
  13. Sep 11, 2007 #12
    Not exactly. I was attempting to (and imho did) provide a counter example to the false statement which was false because it missed the constraint that f is continuous.
     
    Last edited: Sep 11, 2007
  14. Sep 11, 2007 #13

    HallsofIvy

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    In any metric space, the entire set X itself and the empty set are both open and closed. I'm not sure I consider that "strange". It's part of the definiton of "topology"!

    It is true that in a connected space, those are the only sets that are both open and closed (clopen!) but in non-connected sets there are others.
     
  15. Sep 12, 2007 #14

    matt grime

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    Then you should have said 'it is false, and here is a counter example' - explain what you're doing at all times when writing maths. I have gotten used to automatically correcting people's posts when they clearly misstate a question, which doesn't help here.
     
  16. Sep 12, 2007 #15
    1) Sorry, I missed the assumption that f(x) is continuous. Now assuming f(x) is continuous, I don't get why S2 is closed...
    The definition of "closed" in my textbook is as follows:
    Let S be a subset of R^n. S is called "closed" if it contains all of its boundary points.

    But what are the boundary points for S2?



    3) First of all, I just want to know something...
    Is there any difference between "only if" and "if and only if" ? Are they equivalent in meaning?


    Thanks!
     
  17. Sep 12, 2007 #16

    matt grime

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    1) What is your definition of continuous.

    3) They are not the same. That is just English, and not mathematics.
     
  18. Sep 12, 2007 #17
    1) Definition: f is continuous at a iff lim (x->a) f(x) = f(a)

    But I don't see why S2 is closed is f is assumed to be continuous...
    The definition of "closed" in my textbook is as follows:
    Let S be a subset of R^n. S is called "closed" if it contains all of its boundary points.

    But what are the "boundary points" for S2? I don't get this part...


    3) I am sorry to say that I still don't understand the logical difference between "only if" and "if and only if", would you mind explaining them a little more?


    Thanks!
     
  19. Sep 12, 2007 #18

    D H

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    The phrase "If x then y" in logic is called material implication. The statement is invalid in only one circumstance: x is true and y is not true. If x is false, the statement is valid regardless of whether y is true or false (e.g., if the moon is made of cheese then I'm a monkey is logically valid statement). "x only if y" is the other way around: It means the same is "If y then x". If and only if means x and y are equivalent. "x iff y" means that if y is true, so is x, and if y is false, so is x.
     
  20. Sep 13, 2007 #19

    matt grime

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    Then i suspect you do not understand why the other set in the question is closed either. A function is continuous if and only if the inverse image of closed sets is closed (equivalently open sets is open). This makes the question trivial. If you wish to do it with your definition, what is the definition of a boundary point?



    This is just the English meaning of the phrases.

    If I say

    (I go to work) if (it is Monday)

    you agree that the meaning of that sentence is, well, if it is monday, then I go to work, but I might also work on every other day.

    If I say

    (I go to work) only if (it is Monday)

    you agree that I don't go to work on Tuesdays, or any other day than Monday. However, you can't say for sure that I go to work on a Monday.

    (I go to work) if and only if (it is monday)

    means that if it is monday I go to work, and if I'm at work then it is a Monday.


    It's Wednesday, and I have to go to work now, by the way.
     
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