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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 $f:Xtomathbb{R}$ 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 $vin V$ and $r>0$ such that $V=B_r^V(v)$. Since $f$ is uniformly continuous, for $varepsilon=1$ we find $delta>0$ such that $d(x,y)leqdeltaimplies d(f(x),f(y))leqvarepsilon=1$. Then for any $xin V$ we can move from $v$ to $x$ in steps of size $leqdelta$. This can be done in $lceil d(v,x)/deltarceilleq r/delta+1$ 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 $f:Xtomathbb{R}$ is compact. In particular, the image of every uniformly continuous function $f:Xtomathbb{R}$ 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 $$f(x):=supleft{1-frac1n-d(x,x_n):ninmathbb{N}right}.$$ We want to show $f$ is uniformly continuous. Let $varepsilon>0$ and choose $delta=varepsilon>0$. Let $x,yin X$ 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 $ninmathbb{N}$. We get $$1-frac1n-d(y,x_n)leq1-frac1n-d(x,x_n)+d(x,y)leq f(x)+delta$$ for all $ninmathbb{N}$, so $f(y)leq f(x)+delta$, so $d(f(x),f(y))leqdelta=varepsilon$.

By hypothesis, because $f$ is uniformly continuous, the image $f(X)$ is closed. Since $f(x_n)=1-frac1nto1$ as $ntoinfty$ we find $1inoverline{f(X)}$. Because $f(X)$ is closed, there exists $xin X$ 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 $ninmathbb{N}$ 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.