I Gluing lemma / infinite corollary

[PsychonautQQ]

Gluing Lemma: Let X be a topological space, and suppose X = A_1 U A_2 U ... U A_k, which each A_i is closed in X. For each i, let f_i: A_i ---> Y be a continious map s.t f_i = f_j on the intersection of A_i and A_j.

Then the book goes on to give an example of where this is not true for an arbitrary union rather than a finite union:
By consider the subspace X=[0,1] of R, and the sets A_0 = {0} and A_i = [1/(i+1) , 1/i] for i=1,2,3... the gluing lemma is false.

The proof is left as an exercise. This is not homework so i feel no guilty posting it here, where it's more likely to be read by somebody that has some insight into the problem. Can anybody show me why the gluing lemma no longer applies?

 

[fresh_42]

I assume the statement of the lemma is, that the $f_i$ can be extended to a continuous map $f\, : \,X\longrightarrow Y$ ? In the example, how should continuity at $x=0$ be achieved?

 

[PsychonautQQ]

So given a neighborhood V of f(0) in Y, we need a neighborhood U of 0 in X s.t. f(U) is contained in f(V).

That being said, we know that f_0: {0} ---> Y is a continuous map. Therefore given a neighborhood M of f_0({0}) in Y, there exists a neighborhood N of x in X s.t. f_0(N) is contained in M. But this is obviously, because there in the subspace {0} of [0,1], the neighborhood N can be chosen to be the open set that contains only 0. This no longer works when we take the union of the singleton set {0] with all the closed intervals because there is no longer an open set containing only 0. I feel that I'm missing some details and possibly some notation errors.

 

[WWGD]

I think it has to see with the fact that an arbitrary union of closed sets is not necessarily closed. Consider a closed subset $W \subset Y$ , so that $f^{-1}(W)|{A_i}:=V_i$ is closed in each $A_i$, by continuity. If we consider $f^{-1}(W)$ over $X=A_1 \cup A_2 \cup...\cup A_j \cup ...$, we will get an arbitrary union of closed sets, which may not be closed. Let me see if I can think of an example of a function.

 

[Infrared]

If your question is to give a counterexample, here's a hint:

$A_0$ is disjoint from each $A_i$ (if $i\geq 1$) so changing the value of a function $f$ on $A_0$ doesn't affect whether the hypotheses are met. But clearly at most a single value of $f$ at $0$ can give continuity there.

 

[WWGD]

It may also be the case that in this case $0$ is a limit point of your set, so you need to use sequential continuity.

 

[fresh_42]

So given a neighborhood V of f(0) in Y, we need a neighborhood U of 0 in X s.t. f(U) is contained in f(V).

That being said, we know that f_0: {0} ---> Y is a continuous map. Therefore given a neighborhood M of f_0({0}) in Y, there exists a neighborhood N of x in X s.t. f_0(N) is contained in M. But this is obviously, because there in the subspace {0} of [0,1], the neighborhood N can be chosen to be the open set that contains only 0. This no longer works when we take the union of the singleton set {0] with all the closed intervals because there is no longer an open set containing only 0. I feel that I'm missing some details and possibly some notation errors.

What is the intersection of $A_0$ with the $A_i \; (i>0)$ which forces you to clue them there?