If $X$ is a metrizable space such that every metric generating the topology is bounded, then $X$ is compact.

I found a proof by Mr.@Henno Brandsma

**proof **

suppose to the contrary that $X$ is not compact

Then there is a closed subset $Y$ of $X$ homeomorphic to the integers (a closed and discrete subset, we get this from any sequence without a convergent subsequence.)

Then define an unbounded function $f$ on $Y$ (map the n-th point in the homeomorphism with $N$ to $n$, and extend by Tietze theorem.) $$F: X \to R $$ Then if $d$ is a compatible metric for X, then so is $$d'(x,y) = d(x,y) + |F(x) - F(y)|$$ (a sequence convergent in $d'$ converges in $d$, because $d \le d' $ and one in $d$ converges in $d'$ by continuity of $F$, essentially. Same convergent sequences means same closed sets, so same open sets).

But $d'$ is unbounded. Contradiction !

Therefore $X$ is compact.

My questions :- 1) Why $X$ is not compact gives that there is a closed subset $Y$ of $X$ homeomorphic to the integers ??

2) The Tietze theorem is for continuous function , but I do not know how the unbounded function $f$ will be continuous in this case, could any one give an example for such function ?? Thanks alot.

  • $\begingroup$ Sure. Take $X = \mathbb{R}$ and define $Y = \mathbb{Z}$. Then, $f : Y\to \mathbb{R}$, $x\mapsto x$ is continuous and unbounded and extends to the identity map on $X$. $\endgroup$ – Michael Lee Dec 2 '17 at 17:57
  • $\begingroup$ Assume $X$ is not compact, so that there exists a sequence $a_n$ in $X$ that has no converging subsequence, and denote $A$ to be the underlying set of this sequence. Define $f:A\to R$ via $f(a_n)=n$. It is continuous on A, and $A\subset X$ is a closed set, so by the Tietze extension theorem $f$ extends to a (non-bounded) function on all of $X$. $\endgroup$ – Hamada Al Dec 2 '17 at 17:58
  • $\begingroup$ How $f(a_n)=n$ is continuous ? $\endgroup$ – Hamada Al Dec 2 '17 at 18:00
  • 1
    $\begingroup$ Well, if you've chosen $A = \{a_n : n\in \mathbb{N}\}$ to be discrete, then every function from $A$ to $\mathbb{R}$ is continuous, as every subset of $A$ is open in $A$. $\endgroup$ – Michael Lee Dec 2 '17 at 18:02

1) In the question you wrote about it:

we get this -i.e. $Y$ - from any sequence without a convergent subsequence

So, if $X$ is not compact, it contains a sequence $(y_n)$ which has no convergent subsequence. Define then $Y:=\{y_n:n\in\Bbb N\}\subseteq X$. This is closed and inherits the discrete topology -- we use here that $y_n$ has no convergent subsequence --, so $Y$ is a countably infite discrete subspace of $X$. As such, $Y$ is homeomorphic to either $\Bbb N$ or $\Bbb Z$.

2) Since $Y$ is discrete, every function from $Y$ is continuous. In particular, we can take $f:=\ Y\ni y_n\,\mapsto\, n\in\Bbb R$.
Then, it extends to $X$ and allows to define the unbounded metric $d'$.


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