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Suppose the countable subspace $D$ is dense in the separable Tychonoff space $X$ and $f$ is a continous function from $D$ to the closed unit interval. What are some conditions on $X$ or $D$, which make $f$ continuously extendable over $X$？

edited Oct 20 2011 at 3:25

Ricky Demer
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asked Oct 20 2011 at 3:21

Paul
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Are you aware of Tietze's Extension Theorem? It's related, but with a $T_4$ space $X$ rather than $T_{3.5}$. See: en.wikipedia.org/wiki/Tietze_extension_theorem – David White Oct 20 2011 at 4:12

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But here $D$ is dense, so Tietze applies only if $D=X$... – Pietro Majer Oct 20 2011 at 8:30

Pietro Majer · Answer 1 · 2011-10-20 08:27:35Z UTC

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The situation is analogous to the particular case of $X$ a metric space, for any Tychonoff space $X$ is uniformisable, and a real valued function $f$ on a dense subset $D$ of a uniform space $X$ is certainly continuously extendable to $X$ provided it is uniformly continuous. This is also a necessary condition if $X$ is compact, for any continuous function on a compact uniform space is always uniformly continuous.

answered Oct 20 2011 at 8:27

Pietro Majer
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Is "Any Tychonoff space X is uniformisable" means that every Tychonoff space is a uniform space? – Paul Oct 20 2011 at 9:34

Pietro provided a link to Wikipedia in his answer. But the answer is essentially yes; the topology on $X$ will be compatible with a construction that makes $X$ into a uniform space. – Christopher A. Wong Oct 20 2011 at 9:47

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Yes, it admits a uniform structure (not unique in general). So the complete answer is: $f$ is uniformly continuous on $D$ wrto one such uniform structures. – Pietro Majer Oct 20 2011 at 10:31

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To be specific, $f$ is extendable if and only if it is uniformly continuous wrt (the restriction to $D$ of) the fine uniformity on $X$. – Emil Jeřábek Oct 20 2011 at 13:35

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With respect to – Richard Rast Oct 21 2011 at 1:38

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Christopher A. Wong · Answer 2 · 2011-10-20 08:51:01Z UTC

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At least in the case of a metric space $X$, such a function $f$ extends from $D$ to all of $X$ if and only if $f$ maps Cauchy sequences to Cauchy sequences (note that this is a weaker condition than uniform continuity).

As mentioned by Pietro, for your general Tychonoff space $X$, you make it a uniform space, so I think you can generalize my statement above to the following: $f$ extends if and only if $f$ maps Cauchy nets to Cauchy nets.

edited Oct 20 2011 at 8:51

answered Oct 20 2011 at 8:39

Christopher A. Wong
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Also note that a function maps Cauchy nets to Cauchy nets if and only if it is uniformly continuous. – Pietro Majer Oct 20 2011 at 10:32

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@Pietro: Are you sure? For example, every continuous function from $\mathbb R$ to any uniform space also maps Cauchy nets to Cauchy nets. (Let $\{x_a\}_{a\in D}$ be a Cauchy net. There is $a_0$ such that $|x_a-x_{a_0}|\le1$ for every $a\ge a_0$, hence all such $x_a$ are confined to the compact interval $I=[x_{a_0}-1,x_{a_0}+1]$. Then $f$ is uniformly continuous on $I$, which implies that $\{f(x_a)\}_{a\in D}$ is Cauchy.) – Emil Jeřábek Oct 20 2011 at 13:24

OTOH, Christopher’s condition is clearly not necessary, even in the metric case. For instance, if $D=X$, then every continuous $f$ trivially extends, but in general does not map Cauchy sequences to Cauchy sequences (e.g., take $X=\mathbb Q$, $f(x)=0$ for $x<\pi$, $f(x)=1$ for $x>\pi$). – Emil Jeřábek Oct 20 2011 at 13:31

@Emil: oh yes, my distraction! – Pietro Majer Oct 20 2011 at 14:46

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@ Emil: In the metric case, I suppose I was thinking of the process of completion as uniquely determined by Cauchy sequences. So perhaps what I really want is that $f$ maps Cauchy sequences to Cauchy sequences, wherever these Cauchy sequences converge to a point $p \notin D$, $p \in X$. – Christopher A. Wong Oct 20 2011 at 16:27

Todd Eisworth · Answer 3 · 2011-10-20 14:00:35Z UTC

The criterion for "EVERY continuous map from $D$ to $[0, 1]$ has a continuous extension to $X$" is that any two disjoint zerosets in $D$ have disjoint closures in $X$. You can find this in Chapter 6 of Gillman and Jerison's classic "Rings of Continuous Functions". They also consider the "local problem" of continuously extending a single map at length in some of the exercises, e.g. given $f:D\rightarrow Y$ (not necessarily $Y=[0, 1]$) Exercise 6G characterizes the largest subspace of $X$ to which $f$ can be continuously extended in terms of $z$-filters.

Bill Johnson · Answer 4 · 2011-10-20 20:11:08Z UTC

A relevant paper is

Taĭmanov, A. D. On extension of continuous mappings of topological spaces. (Russian) Mat. Sbornik N.S. 31(73), (1952). 459–463. 56.0X

The MR of this paper is:

Let $S$ be a $T_1$-space, $A$ a dense subspace of $S$, and $R$ a compact Hausdorff space. Let $f$ be a continuous mapping of $A$ into $R$. Then $f$ admits a continuous extension over $S$ if and only if for all disjoint closed subsets $A_1,A_2$ of $R$, the relation $(f^{-1}(A_1))^-\cap(f^{-1}(A_2))^-=0$ obtains (closure in $S$). From this result, a theorem of Yu. M. Smirnov [Uspehi Matem. Nauk 6, no. 4(44), 204--206 (1951)] is easily proved, as well as a theorem of Vulih [Mat. Sbornik N.S. 30(72), 167--170 (1952); MR0048790 (14,70c)]. A final corollary is a special case of a theorem widely known and recently published by Katětov [Fund. Math. 38, 85--91 (1951); MR0050264 (14,304a)].

I remembered this because as a graduate student I used it to give a (I think new at the time) proof that every compact Hausdorff space is a continuous image of a compact totally disconnected compact Hausdorff space (which, in turn, I use these days to reduce proving the Riesz representation theorem for $C(K)$ to the case where the compact space $K$ is totally disconnected).

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