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A A pasting lemma counterexample (of sorts)

  1. Dec 12, 2016 #1
    From the pasting lemma we have that if ## X = A \cup B ## and ##f: A \rightarrow Y ## and ## g: A \rightarrow Y## are continuous functions that coincide on ## A \cup B ##, they combine to give a continuous function ## h: X \rightarrow Y ## s.t. ## h(x) = f(x) ## for ## x \in A ## and ## h(x) = f(g) ## for ## x \in B ##, so long as both ## A ## and ## B## are both open or closed.

    But what if ##A## is open and ##B## is closed? This should not work. My reasoning is that, if it worked, then the pasting lemma would already say so! But this obviously does not constitute a proof or counterexample. I think it shouldn't be difficult, but I am missing something. The "obvious" choice was to take ##A## and ##B## as complements, but I am not sure how to show a lack of continuity. The intersection is empty, so the functions agree on that. Another example?

    I feel like this should be very easy or obvious. A nudge in the right direction would be appreciated.

    -Dave K
     
  2. jcsd
  3. Dec 12, 2016 #2

    fresh_42

    Staff: Mentor

    With complementary sets ##A,B## one could define ##f: (-\infty,0) \rightarrow \mathbb{R}## by ##1/x## and ##g: [0,+\infty) \rightarrow \mathbb{R}## by ##g(x)=0##. It would be more interesting / challenging to find an example with ##A \cap B \neq \emptyset##.
     
  4. Dec 12, 2016 #3
    Indeed it would. This works for the time being or at least gets me started. I think my mental block was that the functions I was coming up with were all onto their domain even though there is no good reason for this. Thanks.
     
  5. Dec 12, 2016 #4

    fresh_42

    Staff: Mentor

    In case you find an example with overlapping domains, please post it. I'm curious now. (And confident, that there are some - I mean it's topology, isn't it?)
     
  6. Dec 12, 2016 #5
    Will do.
     
  7. Dec 17, 2016 #6
    You can overlap the domains away from the problem points. Let [itex]X=S^1[/itex]. The usual angle function [itex]S^1\to [0,2\pi)[/itex] is of course discontinuous. But it continuous when restricted to the arc consisting of points with angles in [itex][0,\pi][/itex] and also to the arc of points with angles in [itex](\pi/2,2\pi)[/itex] (notice that the first arc is a closed subset of [itex]S^1[/itex] and the second is a open).

    Edit: One could also have just enlarged the domain of your functions, so that, for example, [itex]f[/itex] is also defined on [itex](0,1)[/itex] and is zero here. But my example still holds if you want [itex]A[/itex] and [itex]B[/itex] to be connected.
     
    Last edited: Dec 17, 2016
  8. Feb 8, 2017 #7
    Consider the functions ##f : (-\infty, 0]\cup\{1\} \rightarrow \mathbb{R} ##, defined by ##f(x)=1## and ##g: (0,\infty) \rightarrow \mathbb{R}##, defined by ##g(x)=x##.
     
  9. Feb 10, 2017 #8
    Hey, I hadn't realized there were new posts on this thread. Thank you @Infrared and @Mayaka Ibara . I will check them out later.
     
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