$X$ is metric space s.t. for every metric space $Y$ and any continuous function $f : X \to Y$ , $f(X)$ is closed in $Y$ ; is $X$ compact?

Question

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 $\mathbb R^n$?

Answer 1

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.

Answer 2

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.