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Continuous functions

  1. Jan 15, 2008 #1
    I'm having trouble with the third part of a three part problem (part of the problem is that I don't even see how what I'm trying to prove can be true).

    The problem is:

    Let X and Y be topological spaces with X=E u F. We have two functions: f: from E to Y, and g: from F to Y, with f=g on the intersection of E and F. f and g are continuous with respect to the relative topologies. We are interested in the function h=f u g, from X into Y.

    There were three questions and the one I can't get is to prove that if E and F are both closed then h is continuous.

    The problem is this doesn't seem to be true to me. Nothing is said about which topologies are used, so we can assume pretty much anything for Y. Now the intersection of E and F is closed, since they are both closed. We can then assume Y takes on a single value in this intersection (for example, if graphing on the real line we could have E=[0,2] and F=[1,3] then the intersection = [1,2], and then we could have h be a trapezoidal function ... f(X)=x if x is in [0,1] f(x)=g(x)=1 if x is in [1,2], and g(x)=3-x if x is in [2,3] ... if we assume that {1} is an open set in Y, then the inverse image is the closed set [1,2] in X, so this isn't continuous under these topologies).

    Clearly I am missing something and my reasoning must be off. Any help would be appeciated.
  2. jcsd
  3. Jan 15, 2008 #2


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    Something is said; f and g are continuous....
    Last edited: Jan 15, 2008
  4. Jan 15, 2008 #3
    That doesn't help me.
  5. Jan 15, 2008 #4
    Obviously the continuity is the limitation on the topologies here and the problem must turn on that point. But I don't see how to work out a proof from that (even if I could maybe see how it applies to my specific bad example).
  6. Jan 15, 2008 #5
    if {1} is an open set in Y then your functions f and g are not continous either because the inverse image [1,2] is not an open set (i asume that you are using the standard topologies on [0,2] and [1,3])
  7. Jan 15, 2008 #6
    Okay, we can forget my example. It was a bad example. But still don't see how to start on a positive proof. I was able to do it for E and F open, but I seem to run into problems for them both closed.
  8. Jan 16, 2008 #7
    dear gonzo
    in your problem there are 4 tolpological spaces to consider:
    [tex] (X, \mathbf{X}) , (Y,\mathbf{Y}),(E,\mathbf{E}) [/tex]and[tex] (F,\mathbf {F})[/tex].
    you know that [tex]E [/tex] and [tex]F[/tex] are closed in [tex] (X, \mathbf{X}) [/tex].
    the proof will consist of the following steps:
    1.prove a map [tex] h : X \rightarrow Y[/tex] is continuous if the preimage of any closed subset of [tex] Y [/tex] is a closed subset of [tex] X [/tex].
    2. Prove if a subset [tex]A[/tex] of [tex] E[/tex] is closed in [tex] (E,\mathbf{E})[/tex] than [tex]A[/tex] is closed in [tex] (X,\mathbf{X})[/tex]. the same holds for closed subsets of [tex] (F,\mathbf{F}) [/tex]
    3. use 1 and 2 to prove that h is continuous
  9. Jan 16, 2008 #8
    Thanks, dalle. That seemed to be helpful, but I ran into a problem.

    1. I have as a theorem I can use directly

    I thought I could prove 2. easily, but the problem was it didn't seem to depend on E or F being closed.

    Then I seem to be making some leap again to the final proof that doesn't depend on E and F being closed ... which is wrong, since the first part of the problem is an example where they are not both open nor both closed and h is not continuous.

    I'm not sure where I need to bring in the fact the E and F are closed.
  10. Jan 16, 2008 #9
    let us consider the following example first:
    let [tex] X = \mathbf{R} [/tex] with the usual topology and [tex] E[/tex] =(-1,1). [tex] E [/tex] is not closed in [tex] (X,\mathbf{X}) [/tex]. The set [tex] A [/tex]= (-1,0] is closed in [tex] (E,\mathbf{E}) [/tex] but it is not closed in [tex] \mathbf{R} [/tex] . [tex] A \subset E [/tex] is a closed subset with respect to [tex] (E,\mathbf{E}) [/tex] does not imply that [tex] A [/tex] is closed in [tex] (X,\mathbf{X}) [/tex].

    Let [tex] C [/tex] be a closed set in [tex] (Y,\mathbf{Y}) [/tex]. you know that [tex] C_f =f^{-1}(C)[/tex] is closed in [tex] (E,\mathbf{E} )[/tex] and [tex] C_g =g^{-1}(C)[/tex] is closed in [tex] (F,\mathbf{F}) [/tex].

    For h to be continous you have to show that [tex] C_f \cup C_g = h^{-1}(C) [/tex] is a closed set in [tex] (X,\mathbf{X}) [/tex].
  11. Jan 16, 2008 #10
    Okay, I think I'm making progress now. If [tex]G_E \in (E,\mathbf{E})[/tex] is open, then we have [tex]G_E = G \cap E[/tex] for some open set G in X.

    This means that [tex]G_E^c=G^c \cap E[/tex] where [tex]G_E^c[/tex] is the complement of [tex]G_E[/tex] in E and thus an arbitrary closed set in E. Since E is closed and the intersection of two closed sets is closed, we have shown that [tex]G_E^c[/tex] is also closed in X, and the same applies for F. So that's step one.

    Now the union of two closed sets is closed, so that should be the final step, right? Am I still missing something?

    Thanks for all your help by the way.
  12. Jan 16, 2008 #11


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    It doesn't look like you're using the continuity of f or g at all, and this is pretty vital here.
  13. Jan 16, 2008 #12
    the continuity is what says that the inverse image of any closed set is closed, and dalle wrote in his last post. My comment on the union of two closed sets was responding to that.
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