# Possible characterization of compact metric spaces via real-valued uniformly continuous functions?

1) If $$X$$ is a metric space such that the image of every uniformly continuous function $$f: X \to \mathbb R$$ is bounded, then is it necessarily true that $$X$$ is compact ?

2) If $$X$$ is a metric space such that the image of every uniformly continuous function $$f: X \to \mathbb R$$ is compact, then is it necessarily true that $$X$$ is compact ?

If either of (1) or (2) is not true, what happens if we also assume $$X$$ is complete ?

(Note that (1), hence (2), is true if we require all real valued "continuous" image to be bounded ... )

I'm not sure if you're familiar with this, but for (1) you can take $$X$$ to be the closed unit ball of any Banach space. Because $$X$$ is bounded, any uniformly continuous $$f:X\to\mathbb{R}$$ is bounded. However, $$X$$ is not compact. Note that $$X$$ is complete.

Proof that any uniformly continuous function $$f$$ from a bounded convex linear set $$V$$ to a metric space $$W$$ must be bounded: Since $$V$$ is bounded, there exists $$v\in V$$ and $$r>0$$ such that $$V=B_r^V(v)$$. Since $$f$$ is uniformly continuous, for $$\varepsilon=1$$ we find $$\delta>0$$ such that $$d(x,y)\leq\delta\implies d(f(x),f(y))\leq\varepsilon=1$$. Then for any $$x\in V$$ we can move from $$v$$ to $$x$$ in steps of size $$\leq\delta$$. This can be done in $$\lceil d(v,x)/\delta\rceil\leq r/\delta+1$$ steps. By the triangle inequality, we get $$d(f(v),f(x))\leq r/\delta+1$$.

Statement (2) is true, actually. Let $$X$$ be a metric space such that the image of every uniformly continuous function $$f:X\to\mathbb{R}$$ is compact. In particular, the image of every uniformly continuous function $$f:X\to\mathbb{R}$$ is closed, which is all we really need. The goal is to show that $$X$$ is compact, which means that every sequence has a convergent subsequence. So let $$\{x_n\}$$ be a sequence in $$X$$.

Consider the function $$f(x):=\sup\left\{1-\frac1n-d(x,x_n):n\in\mathbb{N}\right\}.$$ We want to show $$f$$ is uniformly continuous. Let $$\varepsilon>0$$ and choose $$\delta=\varepsilon>0$$. Let $$x,y\in X$$ with $$d(x,y)<\delta$$. Assume without loss of generality that $$f(y)\geq f(x)$$, so $$d(f(x),f(y))=f(y)-f(x)$$. Then $$f(x)\geq1-\frac1n-d(x,x_n)$$ for all $$n\in\mathbb{N}$$. We get $$1-\frac1n-d(y,x_n)\leq1-\frac1n-d(x,x_n)+d(x,y)\leq f(x)+\delta$$ for all $$n\in\mathbb{N}$$, so $$f(y)\leq f(x)+\delta$$, so $$d(f(x),f(y))\leq\delta=\varepsilon$$.

By hypothesis, because $$f$$ is uniformly continuous, the image $$f(X)$$ is closed. Since $$f(x_n)=1-\frac1n\to1$$ as $$n\to\infty$$ we find $$1\in\overline{f(X)}$$. Because $$f(X)$$ is closed, there exists $$x\in X$$ such that $$f(x)=1$$. We can use this to find a subsequence of $$\{x_n\}$$ converging to $$x$$ as follows.

Take $$x_{n_1}=x_1$$. Then if $$x_{n_k}$$ is defined, by definition of the supremum there exists $$n\in\mathbb{N}$$ such that $$1-\frac1n-d(x,x_n)>1-\frac1{n_k}$$. We define $$x_{n_{k+1}}=x_n$$. This way, $$\{n_k\}$$ is increasing, and $$1-\frac1{n_k}-d(x,x_{n_k})\to1$$, so $$d(x,x_{n_k})\to0$$, so $$x_{n_k}\to x$$. Therefore, $$X$$ is compact.

• why is the image of $f$ compact ? – user521337 Dec 29 '18 at 2:47
• I thought a compact function is a function that maps compact sets to compact sets. If you want a function which has a compact image, just take $X$ as the closed unit ball. – SmileyCraft Dec 29 '18 at 2:49
• I specifically said that I want the image of $f$ to be compact .... but again ... why is the image of every $f$ in your example compact ? – user521337 Dec 29 '18 at 2:50
• Sorry misread that, and I see your point, indeed the image need not be compact. It is bounded though, but I'll continue thinking about this. – SmileyCraft Dec 29 '18 at 2:54
• why is it even bounded ? – user521337 Dec 29 '18 at 3:08