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Let $X$ be a metric space such that for every metric space $Y$ and any continuous function $f : X \to Y$ , $f(X)$ is closed in $Y$ , then is $X$ compact ?

Compare with this $A \subseteq \mathbb R^n $ s.t. for every continuous function $f : A \to \mathbb R$ , $f(A)$ is closed in $\mathbb R$ , is $A$ closed $\mathbb R^n$?

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  • You could apply a much more difficult result : If $X$ is metrizable and no-compact then $X$ admits an incomplete metric $d$. Then the identity function on $X$ into the metric completion of $(X,d)$ is not closed in the range. But the answers given are good . – DanielWainfleet Jul 31 '16 at 17:08

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Any real-valued function $f$ on $X$ is bounded since the image $\tilde{f}\colon X \xrightarrow{f} \Bbb{R} \xrightarrow{\cong} (0,1) \hookrightarrow \Bbb{R}$ is closed.

Using this you can go to the question: If every real-valued continuous function is bounded on $X$ (metric space), then $X$ is compact.

Edit: A metric space $X$ is compact if and only if it is complete and totally bounded.

There is a inclusion of $X$ into one of its completions and the image of $X$ is dense. By the assumption the image of $X$ is closed we know that $X$ is complete.

Assume $X$ is not totally bounded. So there is an $\varepsilon > 0$, such that we can not cover $X$ by a finite number of balls with radius $2\varepsilon$. Now we construct pairwise disjoint open balls $B_i= B_{\varepsilon_i}(x_i)$ with $i \in \Bbb{N}$. Let $x_0$ be any point in $X$ and $\varepsilon_0 = \varepsilon$. Now assume we have given $(x_i, \varepsilon_i )$ for $0\le i \le n$ such that the $B_i$ are pairwise disjoint. Then $\overline{\bigcup_{i=0}^n B_i} \subset \bigcup_{i=0}^n B_{2\varepsilon}(x_i) \subsetneq X $. So we find $(x_{n+1} , \varepsilon_{n+1})$ such that $B_i$ are paarwise disjoint for $0\le i \le n+1$.

By Urysohn's lemma we find functions $f_i$ with $f_i(x_i)=i$ and $f_i(x)=0$ for $x \notin B_{1/2 \cdot \varepsilon_i} (x_i)$. For any $x\in X$ there exists a neighborhood of $X$ such that only finitely many $f_i$ are nonzero. Thus $f = \sum f_i$ is a well defined unbounded function on $X$. This is a contradiction to the fact that any function on $X$ is bounded. Thus $X$ is totally bounded.

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user60589
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  • Any real valued function is bounded ? How can you conclude that !? –  Jul 13 '16 at 13:06
  • Sorry I mixed up $X$ and $A$. Thanks to Professor Scott for pointing out my argument. I added way to show that $X$ is compact using that all functions are bounded. – user60589 Jul 15 '16 at 14:05
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The answer given by user60589 is correct but so compressed that it was downvoted; here is an expanded version.

Let $f:X\to\Bbb R$ be any continuous function. There is a homeomorphism $h:\Bbb R\to(0,1)$, and since $(0,1)\subseteq\Bbb R$, $h\circ f$ is a continuous function from $X$ to $\Bbb R$. Let $g=h\circ f$. By hypothesis $g[X]$ is closed in $\Bbb R$, and $g[X]$ is bounded (since it’s a subset of $(0,1)$), so $g[X]$ is compact. Finally, $h$ is a homeomorphism, so $h^{-1}$ is also a homeomorphism, and therefore

$$f[X]=h^{-1}\big[h\big[f[X]\big]\big]=h^{-1}\big[g[X]\big]$$

is also compact and hence bounded. Thus, every real-valued function on $X$ is bounded, and by this question (cited by user60589) $X$ is compact.

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Brian M. Scott
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  • M.Scott : Isn't $h \circ f$ a continuous function from $X$ to $(0,1)$ ? In that case , by my hypothesis , $g(X)$ is closed in $(0,1)$ ( not necessarily closed in $ \mathbb R$ right ? .. ) –  Jul 14 '16 at 12:08
  • Or are we arguing here like this : Since $h : \mathbb R \to (0,1)$ is continuous , so $h: \mathbb R \to \mathbb R$ is continuous , so $(h \circ f)(X)$ is closed in $\mathbb R$ ... –  Jul 14 '16 at 12:13
  • Under these observations , I think that the statement of the problem can be made stronger by reducing the hypothesis : "Let $X$ be a metric space such that for any continuous function $f : X \to \mathbb R$ , $f(X)$ is closed in $\mathbb R$ , then $X $ is compact " ; am I correct ? If I am correct then it also gives a simple answer to this question which I think is a special case of my question http://math.stackexchange.com/questions/1232788/does-there-exist-non-compact-metric-space-x-such-that-any-continuous-functio?rq=1 right ? –  Jul 14 '16 at 12:26
  • @user228169: Yes, we can regard $g$ as a continuous function from $X$ to $\Bbb R$, not just to $(0,1)$. I'm not sure that this argument is simpler than the one that Martin found for his question: this one does require the additional result that a pseudocompact metric space is compact. – Brian M. Scott Jul 14 '16 at 14:02
  • yes , I think this is much more simpler and stronger too as we can fix the co-domain space to be $\mathbb R$ here .. –  Jul 14 '16 at 15:20
  • There have been two downvotes tonight of both this answer and that of which it is an expansion. Apparently neither downvoter is interested in performing a genuine service by pointing out an actual error. Until such an error is pointed out, I will assume that they simply failed to understand what they read. – Brian M. Scott Oct 17 '16 at 09:34