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I am reasking a year-old math.stackexchange.com question asked by someone else.

(For my needs every space $X$ and $Y$ will be Polish---that is a completely separably metrizable space.)

The Tietze extension theorem says that if $X$ is a Polish space (even a normal space) and $Y=\mathbb{R^n}$, then a continuous function $f:C \rightarrow Y$ on a closed set $C \subseteq X$ can be extended to a continuous function $g:X \rightarrow Y$.

It seems important to the theorem in general that $Y = R^n$, however there are some examples of pairs $X,Y$ where the theorem also holds. For example, it is true if $X, Y \in \{2^\mathbb{N},\mathbb{N}^\mathbb{N}\}$. (Although, other pairs like $X=\mathbb{R}, Y = 2^\mathbb{N}$ do not hold.)

Is there a characterization of the pairs $X,Y$ (or $X,Y,C$) for which the Tietze extension theorem holds?

If not, what extensions are known, especially those that include my $2^\mathbb{N}$ example above?


One motivation for asking this question is Lusin's theorem: If $(X,\mathcal{B},P)$ is a Borel probability measure on a Polish space $X$ (even a Radon measure on a finite measure space) and $f:X \rightarrow Y$ is a measurable map (again $Y$ is Polish, or even second countable), then for all $\varepsilon > 0$, there is a closed set $C$ of measure $1-\varepsilon$ such that $f$ is continuous on $C$.

If $Y$ is $\mathbb{R}^n$, we can apply the Tietze extension theorem to find some continuous $g:X \rightarrow Y$ such that $g = f$ on $C$. Wikipedia currently (12 July 2013) has a false statement that for any locally compact $X$ we can find such a continuous $g:X \rightarrow Y$. For a counterexample, take $X=[0,1]$ and $Y=2^\mathbb{N}$ and $f$ to be the bit representation of the reals.

I am interested in which cases this stronger version of Lusin's thoerem (with the continuous $g:X \rightarrow Y$) holds.

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Tietze's extension theorem also holds for mappings into locally convex spaces, see msp.org/pjm/1951/1-3/pjm-v1-n3-p04-p.pdf . –  alexod Jul 12 '13 at 21:21
    
The spaces $Y$ where Tietze's extension theorem holds are called absolute retracts. –  Joseph Van Name Jul 12 '13 at 23:12
    
(Tietze is not characteristic, so to speak). –  Wlodzimierz Holsztynski Jul 12 '13 at 23:25
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@Wlodzimierz Holsztynski. Can you please comment using complete sentences? Not using complete sentences is very annoying. –  Joseph Van Name Jul 12 '13 at 23:57
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@Wlodzimierz Holsztynski, I agree with Joseph Van Name, I imagine your comments are insightful. However, as they are, I can't understand them. I would value (and upvote) an answer from you that elaborates on your comments. –  Jason Rute Jul 13 '13 at 22:37

1 Answer 1

There is a nice characterization of the spaces $X$ where the Tietze extension theorem holds for all complete separable metric spaces $Y$. We say that a Hausdorff space $X$ is ultranormal if whenever $R,S$ are disjoint closed subsets of $X$, then there is some clopen set $C$ with $R\subseteq C$ and $S\subseteq C^{c}$. A Hausdorff space $X$ is ultraparacompact if every open cover of $X$ is refinable by a partition of $X$ into clopen sets. Every ultraparacompact space is ultranormal, every ultranormal space is zero-dimensional, and every ultranormal metric space is ultraparacompact. Furthermore, a space is ultraparacompact if and only if it is paracompact and ultranormal. Every zero-dimensional separable metric space is ultraparacompact and hence ultranormal. Therefore for separable metric spaces, the notions of ultranormality, ultraparacompactness, and zero-dimensionality coincide. See my paper for more information on ultranormality and ultraparacompactness. This paper is an expanded version of a long answer I gave to this question here on MO.

$\mathbf{Theorem}$ (Ellis)

  1. Let $X$ be a Hausdorff space. Then $X$ is ultranormal if and only if whenever $C\subseteq X$ is a closed subspace, $Y$ is a complete separable metric space, $C\subseteq X$ is a closed subspace, then every map from $C$ to $Y$ can be extended to a map from $X$ to $Y$.

  2. Let $X$ be an ultraparacompact space and let $C$ be a closed subspace of $X$. Let $Y$ be a complete metric space. Then every continuous map from $C$ to $Y$ can be extended to a map from $X$ to $Y$.///

For a proof of the above result, see the paper Extending Continuous Functions on Zero-Dimensional Spaces by Robert Ellis.

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This is a nice answer. Although, I am still holding out hope for something more general---say where $X$ is not zero-dimensional and $Y$ is not an absolute retract. –  Jason Rute Jul 13 '13 at 14:35
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The problem whether a map $f:C\rightarrow Y$ extends to a map $F:X\rightarrow Y$ is undecidable in even the simplest cases. If $X=I\times I$ and $C=\partial X$ (here we take the boundary of $X$ as a subset of $\mathbb{R}^{2}$) and $\ell:I\rightarrow Y$ is a loop with basepoint $r$, then let $f(x,0)=\ell(x)$ and $f(x,y)=r$ otherwise. The extensions of $f$ are the homotopies of $\ell$ to a point. However, it is well known that the word problem in group theory can be formulated in terms of whether a loop is equivalent to the identity in the fundamental group, and the word problem is undecidable. –  Joseph Van Name Jul 13 '13 at 16:51
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I would therefore be a bit surprised if there is a nice general characterization of when we can extend a map from a subset to the whole space. –  Joseph Van Name Jul 13 '13 at 16:58

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