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Prove A~B=>f(A)~f(B) for a continuous f:X->Y

  1. Nov 17, 2015 #1

    BiGyElLoWhAt

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    So proofs are a weak point of mine.
    The hint is that a composite of a continuous function is continuous. I'm not really sure how to use that. What I was thinking was something to the effect of an epsilon delta proof, is that applicable?

    Something to the effect of:
    ##A \sim B\text{ and let } f \text{ be a continuous function X} \to \text{Y:}##
    ##\text{by definition, if } f \text{ is continuous, then there exists an } \epsilon \text{ for every } \delta \text{ such that ... blah...so }##
    ##f(A) \sim f(A+\epsilon) \sim f(A+n\epsilon) \text{ and by induction } f(A) \sim f(B) \text{ for large enough n}##
    is that strong enough? Since A+n*epsilon = B it goes to f(B). Would that be necessary in the proof?
    How can I do a proof with composite continuity?
     
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  3. Nov 17, 2015 #2

    micromass

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    You will have to say what ##X## and ##Y## are, and what ##\sim## is.
     
  4. Nov 17, 2015 #3

    Krylov

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    At first I thought I was going to make an effort to understand the OP, but I just gave up.
     
  5. Nov 17, 2015 #4

    micromass

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    My bet is that the OP means that if two numbers ##A## and ##B## are infinitesimally close, then ##f(A)## and ##f(B)## are also infinitesimally close. This is indeed a characterization of continuity in the hyperreal number system.
     
  6. Nov 17, 2015 #5

    Krylov

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    Both your bet and such characterisation sound plausible. I just experienced a parsing error when I encountered the "... blah...so".
     
  7. Nov 17, 2015 #6

    micromass

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    My parsing error came a bit earlier with "there is an epsilon for each delta"
     
  8. Nov 17, 2015 #7

    Krylov

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    I remember a mischievous "true/false" question on an introductory analysis exam, where the definition of continuity was stated as usual, but with the roles of the symbols ##\varepsilon## and ##\delta## reversed, to the dismay of the audience.
     
  9. Nov 17, 2015 #8

    micromass

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    You mean the following?
    [tex]\forall \delta >0: \exists \varepsilon>0: \forall x: |x-a|<\varepsilon~\Rightarrow~|f(x) - f(a)| < \delta[/tex]
    That's a really good question :D
     
  10. Nov 17, 2015 #9

    Krylov

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    Yes, precisely. :smile:
     
  11. Nov 17, 2015 #10

    micromass

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    Could be worse:
    A function ##x:\mathbb{R}\rightarrow \mathbb{R}## is continuous at ##\delta## if and only if
    [tex]\forall a>0:~\exists \varepsilon>0: \forall f\in \mathbb{R}:~|f-\delta|<\varepsilon~\Rightarrow~|x(f)-x(\delta)|<a[/tex]
     
  12. Nov 17, 2015 #11

    fresh_42

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  13. Nov 17, 2015 #12

    BiGyElLoWhAt

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    Terribly sorry, all.
    http://inperc.com/wiki/index.php?title=Homology_classes
    There's where it comes from.
    It isn't defined anywhere, but I'm assuming ~ means connected, no cuts/holes/ etc.
    X is a subspace of R^n, and I'm assuming Y is as well, X is the only one that has actually been defined, but thus far, as has been given, we're only working in subspaces of R^n.

    Does that clear things up? The Blah... is the usual definition for continuity in calculus, I was just being lazy.

    *The exercise is about 1/3 of the way down, just before section 3 about counting features.
     
  14. Nov 17, 2015 #13

    micromass

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    So ##A\sim B ## means that there is a continuous path ##q:[0,1]\rightarrow X## such that ##q(0)=A## and ##q(1)=B##? Can you find a continuous path between ##f(A)## and ##f(B)## then?
     
  15. Nov 17, 2015 #14

    BiGyElLoWhAt

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    Yes, I believe.
    I mean, intuitively, yes. The problem lies in the proof part, I think.
    Am I supposed to use ##f(q): [0,1] \to Y## and then just the fact that q is continuous and f is continuous therefore f(q) is continuous?
     
  16. Nov 17, 2015 #15

    micromass

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    What is ##f(q)## supposed to mean? Do you mean ##f\circ q##?
     
  17. Nov 17, 2015 #16

    BiGyElLoWhAt

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    Yes, sorry. The composite.
     
  18. Nov 17, 2015 #17

    BiGyElLoWhAt

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    I guess it would also be : [q(0), q(1)], would it not?
     
  19. Nov 17, 2015 #18

    micromass

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    That's an interval, that has no meaning in general topological spaces.
     
  20. Nov 17, 2015 #19

    micromass

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    Yes. And yes ##q## is continuous (by definition of being a path) and ##f## is continuous (given), so there composition ##f\circ q## is too.
     
  21. Nov 17, 2015 #20

    BiGyElLoWhAt

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    Is that literally it?
     
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