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The following problem is from Munkres's Topology (exercise 13 in section 18 "Continuous Functions"; page 112, 2nd edition).

Let $A \subset X$; let $f:A \to Y$ be continuous; let $Y$ be a Hausdorff. Show that if $f$ may be extended to a continuous function $g:cl(A)\to Y$, then $g$ is uniquely determined by $f$.

hengxin
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Jack Wislly
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1 Answers1

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HINT: Suppose that $g_0:\operatorname{cl}A\to Y$ and $g_1:\operatorname{cl}A\to Y$ both extend $f$. We know that $g_0(x)=g_1(x)$ for each $x\in A$, so suppose that $x\in(\operatorname{cl}A)\setminus A$. For convenience let $y_0=g_0(x)$ and $y_1=g_1(x)$. Suppose that $y_0\ne y_1$; $Y$ is Hausdorff, so there are open sets $U_0$ and $U_1$ in $Y$ such that $y_0\in U_0$, $y_1\in U_1$, and $U_0\cap U_1=\varnothing$.

Note: Everything that I’ve done up to here is just ‘following my nose’, i.e., doing the most natural, straightforward thing that might work; it’s the sort of preliminary reasoning that should quickly become nearly automatic. For instance, if you want to prove that something is unique, one obvious approach is to assume that it isn’t unique and try to get a contradiction; that’s what I’m doing here. That gives me the distinct points $y_0$ and $y_1$; now what is there in the statement of the problem that naturally goes with having a pair of distinct points in $Y$? Here that’s easy. The hypothesis that $Y$ is Hausdorff immediately tells me something about those two points: they can be separated by disjoint open nbhds, and I might as well do it and see whether it’s useful.

And here’s the rest of the hint: let $V_0=g_0^{-1}[U_0]$ and $V_1=g_1^{-1}[U_1]$, and consider the sets $V_0\cap A$ and $V_1\cap A$ to get a contradiction.

Brian M. Scott
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  • Thx. However, that is exactly where I failed to continue. What is the contradiction then? I would expect to show that $x$ is not a limit point of $A$ (thus, contrary to $x \in \bar{A} \setminus A$), but I failed to get that. Could you offer me some more hints? – hengxin Jan 23 '14 at 05:08
  • I think I have found the solution. Following your arguments, we can consider the set $V_0 \cap V_1$. First, $V_0 \cap V_1$ is an open set of $X$. Second, $x \in V_0 \cap V_1$. Third, $V_0 \cap V_1 \subset \bar{A} \setminus A$. In all, we obtain a neighbor (i.e., $V_0 \cap V_1$) of $x$ does not intersect $A$. Therefore, $x$ is not a limit point of $A$. Is this argument correct? – hengxin Jan 23 '14 at 05:29
  • @hengxin How did you conclude that $V_0 \cap V_1 \subset \bar{A} \setminus A$. – Muno Jun 04 '17 at 07:28
  • @Muno Sorry, I have forgotten it. I taught myself a little bit about topology about 4 years ago. I find it not easy for me to pick them up. Sorry for that. – hengxin Jun 04 '17 at 07:40
  • @Muno I think we can't conclude $V_0 \cap V_1 \subset \bar{A} \setminus A$. It might be better to consider $x \in V_0 \cap V_1$ where $x\in(\operatorname{cl}A)\setminus A$. Then by definition of limit point, any neighborhood of $x$ ... – user398843 Feb 25 '18 at 18:27
  • I see that $V_0$ and $V_1$ are disjoint sets that both include x. Is that correct? – Dannyu NDos May 07 '19 at 23:53
  • @DannyuNDos: Almost: $V_0\cap A=f^{-1}[U_0]$ and $V_1\cap A=f^{-1}[U_1]$ are disjoint sets that both contain $x$. $V_0\cap V_1$ can be non-empty, but it must be contained in $(\operatorname{cl}A)\setminus A$. – Brian M. Scott Aug 02 '20 at 01:11