Proof of Tietze Extension Theorem in Munkres

**Bingk** · September 12th 2011, 05:21 PM

Hi! I don't understand a part of the proof.

The theorem states:
Let X be a normal space; let A be a closed subspace of X.
(a) Any continuous map of A into the closed interval [a,b] of R may be extended to a continuous map of all of X into [a,b].
(b) Any continuous map of A into R may be extended to a continuous map of all of X into R.

My question is about part (b).

Let f be a continuous map from A into (-1,1).
It then states that "The half of the Tietze theorem already proved shows that we can extend f to a continuous map g:X --> [-1,1] mapping X into the closed interval".

How does part (a) show this?
(a) tells us that f should be a continuous map from A into the closed interval [-1,1] so that we can extend it to a continuous map g from the entire space X into [-1,1]. It doesn't say anything about A being mapped into open intervals. What's the connection I'm missing?

Thanks!

**Drexel28** · September 12th 2011, 05:35 PM

Originally Posted by Bingk

Hi! I don't understand a part of the proof.

The theorem states:
Let X be a normal space; let A be a closed subspace of X.
(a) Any continuous map of A into the closed interval [a,b] of R may be extended to a continuous map of all of X into [a,b].
(b) Any continuous map of A into R may be extended to a continuous map of all of X into R.

My question is about part (b).

Let f be a continuous map from A into (-1,1).
It then states that "The half of the Tietze theorem already proved shows that we can extend f to a continuous map g:X --> [-1,1] mapping X into the closed interval".

How does part (a) show this?
(a) tells us that f should be a continuous map from A into the closed interval [-1,1] so that we can extend it to a continuous map g from the entire space X into [-1,1]. It doesn't say anything about A being mapped into open intervals. What's the connection I'm missing?

Thanks!

I don't quote know what book you are using, but here is a way to think about it. It clearly suffices to prove this for mappings $A\to(-1,1)$ since $\mathbb{R}\approx(-1,1)$ . So, let $f:A\to (-1,1)$ be continuous, we know that we can extend $f$ to $\widetilde{f}:X\to[-1,1]$ . So, let $Y=\widetilde{f}^{-1}(\{1\})\cup\widetilde{f}^{-1}(\{-1\})$ . By continuity we know that $Y$ is closed in $X$ . That said, since $\widetilde{f}$ is an extension of $f$ and we know $\widetilde{f}(A)=f(A)\subseteq(-1,1)$ and so $A$ is disjoint from $Y$ . Now, by Urysohn's lemma there exists some continuous map $h:X\to[0,1]$ with $h(Y)=\{0\}$ and $h(A)=\{1\}$ . So, define $k(x)=h(x)\wildetile{f}(x)$ . From basic topology we know that $k$ is continuous and we know it's an extension for $f$ since $k(a)=h(a)\widetilde{f}(a)=1\cdot \widetilde{f}(a)=f(a)$ for all $a\in A$ . I'm pretty sure then you can checking where elements of $Y$ and $X-Y$ go under $k$ separately that $h(A)\subseteq(-1,1)$ . Make sense?

**Bingk** · September 12th 2011, 06:20 PM

Thanks!

I got the rest of the proof, but it's that particular part that I don't get ... how do we know that we can extend $f$ to $\widetilde{f}:X\to[-1,1]$ so that $\widetilde{f}$ is continuous?

Sorry, it seems like it should be something obvious, but I guess I'm not thinking along the right lines

Oh, and your proof is very similar to that of the book, same ideas

**Drexel28** · September 12th 2011, 06:36 PM

Originally Posted by Bingk

Thanks!

I got the rest of the proof, but it's that particular part that I don't get ... how do we know that we can extend $f$ to $\widetilde{f}:X\to[-1,1]$ so that $\widetilde{f}$ is continuous?

Sorry, it seems like it should be something obvious, but I guess I'm not thinking along the right lines

Oh, and your proof is very similar to that of the book, same ideas

Because Tietze's extension theorem, the first part, says that you can extend things into a closed interval to the whole space, no?

**Bingk** · September 13th 2011, 06:52 AM

I guess ... I thought Tietze's theorem tells us that if the domain is a closed subspace that maps into a closed interval, then we can extend the function to map the entire space into the same closed interval. Wherein was my problem, because the function we are talking about for the second part maps into an open interval, yet we still extended the domain to the entire space mapping into the closed interval. It was the open/closed combination that was bugging me, and I'm assuming that first the range was extended into the closed interval, then we used Tietze to extend the domain. Is that right?

Or is it because in the proof of the first part, where we divided the closed interval [a,b] (actually the interval [-1,1] to make things simpler), we could replace the closed interval with an open interval and the proof would still work?

Thanks!

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