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I have been reading about Tietze extension theorem and i wondered if we could do it for any subset of $\mathbb{R}$, i.e ,

Let $A$ be a closed subset of a normal space $X$ and we have a continuous function $f: A \rightarrow C$ where $C$ is any subset of $\mathbb{R}$, is there any way we can make a continuous extension?

My intuition says that we cant because in the proof of the theorem we use the fact that we are working with intervals, but im not quite seeing how to create a counterexample, so any enlightenment would be appreciated.

Thanks.

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  • $\begingroup$ Isnt the function supposed to be defined on a closed subset so we could in theory apply the Tietze extension theory? $\endgroup$ – Pedro Santos Oct 19 at 18:23
  • $\begingroup$ Yes but i meant that subset being the image of the function. $\endgroup$ – Pedro Santos Oct 19 at 18:25
  • $\begingroup$ Could you edit your question to precisely state what you were asking. Which version of Tietze's theorem are you familiar with, please state it, and state what generalization you wish to consider. There is a standard trick, since $\Bbb R$ is homeomorphic to $(0,1)$ which is a subset of $[0,1]$, so you may assume your function is into $[0,1]$ if that is relevant to the answer. $\endgroup$ – Mirko Oct 19 at 18:26
  • $\begingroup$ Yes i think its clear now , sorry. $\endgroup$ – Pedro Santos Oct 19 at 18:29
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    $\begingroup$ You already got an answer to your question, but if you want to know more about the topic the keywords to google are "absolute (neighbourhood) extensor" $\endgroup$ – Alessandro Codenotti Oct 19 at 18:56
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The answer is no, in general. Let $A=\{0,1\}=C$, and $f: A \rightarrow C$ be the identity function, where $X=\Bbb R$. If $g$ were an extension of $f$, then the image of $g$ must be connected (since $\Bbb R$ is). On the other hand the image must be $C$ (since the image of $A$ is $C$, and the extension is only supposed to take values in $C$, according to my understanding of your question), which is not connected.

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  • $\begingroup$ Yeah that was my question , Thank you !!! $\endgroup$ – Pedro Santos Oct 19 at 19:07
  • $\begingroup$ thank you, you are welcome! $\endgroup$ – Mirko Oct 19 at 19:10
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In the statement of the Tietze extension theorem you can in general replace $\Bbb R$ by any so-called AR (sometimes also called AE) space, absolute retracts. These have a quite general (but technical) characterisation, but among the main examples are all convex subsets of separable metric locally convex linear spaces, and this thus includes all intervals in $\Bbb R$ or disks in the plane and solid disks in higher dimensions. See Hu's book theory of retracts, among others, or books like van Mill's "infinite-dimensional topology (prerequisites and an introduction)", or Borsuk's original texts, also see Wikipedia for some basic info and references.

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If $A$ is a non-closed subset of $\Bbb R$, let $a$ be an element of the closure of $A$ in $\Bbb R$ that is not an element of $A$. Then $x\mapsto 1/(x-a)$ is a continuous function from $A$ to $\Bbb R$ which does not extend to a continuous function from $\Bbb R$ to $\Bbb R$.

The same trick works if $\Bbb R$ is replaced by a metric space $X$. This time take $x\mapsto d(x,a)^{-1}$ to be the function.

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  • $\begingroup$ I had made a similar comment, but it turned out the question was asking something else. The set $A$ is assumed to be closed in $X$, $f: A \rightarrow C$ and extension $g: X \rightarrow C$. $\endgroup$ – Mirko Oct 19 at 19:13

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