7

I'm dealing with the following problem:

Let $X$ a topological space, $Y$ a metric space and $A$ a subspace of $X$. If $f$ is a continuous mapping of $A$ into $Y$, show that $f$ can be extended in at most one way to a continuous mapping of $\bar{A}$ into $Y$.

Uniqueness is not a problem, existence is the difficult part for me. I did the proof for the case in that $X$ is a metric space one year ago, but it seems I can't use the same idea here. I hope someone could give me a hand. Tips will suffice, of course.

Daniel
  • 7,185
  • 2
    The key is that $Y$ is a Hausdorff space. Hence if two continuous functions $f,g\colon Z\to Y$ differ at a point $z_0\in Z$, they differ ... – Daniel Fischer Nov 08 '14 at 14:42
  • Thanks for your answer but I've forgot to say that uniqueness is actually not a problem, the problem is in the existence part. – Daniel Nov 08 '14 at 14:46
  • 8
    For the existence, you need stronger assumptions. A map that is only continuous can in general not be continuously extended to the closure. Consider the case where $Y$ is an incomplete metric space, $X$ its completion, $A = Y$ and $f = \operatorname{id}$. But the problem doesn't ask you to prove existence, it asks you to show that it can be extended in at most one way, that is, it asks you to show uniqueness. – Daniel Fischer Nov 08 '14 at 14:49
  • 1
    it says "at most" one way. So you don't have to show existence. This is good because I guess this is not possible. For example $D\supset int D \to int D$. – Daniel Valenzuela Nov 08 '14 at 14:50
  • Ohh, that's right! I'm sorry, I didn't understand the problem completely. The problem is straightforward now. A good question is under what hypothesis existence holds, Is it sufficient with $Y$ being complete? If we take $a\in \bar{A}$, is there a sequence $a_n\in A$ such that $f(a_n)$ is a Cauchy sequence in $Y$? – Daniel Nov 08 '14 at 14:54
  • 1
    The standard calc 3 counterexamples where you get different values of the function as you approach the origin are the intuition for why existence won't work in general. – Neal Nov 08 '14 at 14:55
  • 2
    A useful sufficient condition for existence of an extension is that $X$ is a metric space, $Y$ is a complete metric space, and $f$ is uniformly continuous. – Nate Eldredge Nov 08 '14 at 15:07
  • FYI: Here you find existence results: https://math.stackexchange.com/questions/989205/extending-a-bounded-linear-operator – user3810316 Jul 29 '23 at 09:47

0 Answers0