Uniformly continuous function from a subspace of $\mathbb{R}$ maps bounded sets to bounded sets

Question

I know this question has been asked on this site before (here, here and a few more), but here I have less assumptions. $f: X \to Y$ is uniformly continuous such that $X \subseteq \mathbb{R}$, but I am not assuming anything on $Y$, i.e. It is a general metric space, not necessarily complete. I am also not assuming that $X$ is closed.

Is this, generally, true that $f$ maps bounded sets in $X$ to bounded sets in $Y$? My professor seems to suggest so, but it seems impossible without compactness, either assuming $X$ is closed or that $Y$ is complete such that there is a uniformly continuous extension of $f$ to $\mathrm{Cl}(A)$.

If it is not possible, can you provide me with a counterexample?

(From wikipedia): "A continuous function {\displaystyle f:S\rightarrow R} f:S\rightarrow R whose restriction to every bounded subset of S is uniformly continuous is extendable to X, and the converse holds if X is locally compact." Here $X$ and $R$ are general metric spaces, and $S \subseteq X$. In your case, $\Bbb{R}$ is certainly locally compact, though it may be better to use the closure of your subspace as the extension domain. This is not a full answer, but may help guide one. — B. Mehta, Dec 03 '17 at 18:12

DanielWainfleet · Answer 1 · 2019-11-21T03:40:31.903

1

Let $\rho$ be the metric on $Y.$ By contradiction, suppose $D$ is a bounded subset of $X$ but $f(D)$ is unbounded in $Y.$ Then there exists a sequence $(d_n)_{n\in \Bbb N}$ in $D$ such that $\rho ((f(d_m),(f(d_n))>1$ whenever $m\ne n.$

$D$ is bounded in $\Bbb R$ so $(d_n)_{n\in \Bbb N}$ has a sub-sequence $(d_{n_j})_{j\in \Bbb N}$ that is Cauchy. So for $r>0$ let $m_r \in \Bbb N$ be such that $m_r<n<n'\implies |d_n-d_{n'}|<r.$

There exists $n_j>m_r$ so $|d_{n_j}-d_{n_{j+1}}|<r$ for such $j,$ but $\rho (f(d_{n_j}),f(d_{n_{j+1}}))>1.$

So for every $r>0$ there exist $d,d'\in D\subset X$ with $|d-d'|<r$ and $\rho (f(d),f(d'))>1,$ contradicting the uniform continuity of $f$ on $X$.

edited Nov 21 '19 at 03:40

answered Dec 03 '17 at 20:32

DanielWainfleet

59,529

Some of your parentheses are messed up. particularly $\rho (d(f(d_m),d(f(d_n))$. Also, what is $d$ here? – Bary12 Dec 03 '17 at 21:34
I had typos in the 1st paragraph. Thanks. – DanielWainfleet Dec 04 '17 at 01:51

score 0 · Answer 2 · answered Dec 03 '17 at 18:42

0

Take $\widetilde{Y}$ the completion of $Y$, and consider the extension $$\widetilde{f}:\overline{X} \to \widetilde{Y}.$$

If $A \subset X$ is bounded, then so is $\mathrm{cl}_{\overline{X}}A=\mathrm{cl}_{\mathbb{R}}A$, and thus $\mathrm{cl}_{\overline{X}}A$ is compact. Hence, $\widetilde{f}(\mathrm{cl}_{\overline{X}}A)$ is bounded, since it is compact. In particular, $\widetilde{f}(A)=f(A)$ is bounded.

As a sidenote, $X$ being a subspace (metric subspace) is important. $\mathbb{Z}$ with the discrete metric (which, "as a topological space", is a subspace of $\mathbb{R}$) is such that $f:\mathbb{Z} \to \mathbb{R};$ $x \mapsto x$ is uniformly continuous but does not map bounded sets to bounded sets.

answered Dec 03 '17 at 18:42

Aloizio Macedo

35,361
6
74
155

What is the completion of $Y$ and how is it defined? – Bary12 Dec 03 '17 at 18:44
@Bary12 See here. – Aloizio Macedo Dec 03 '17 at 18:46
I am afraid completion is outside of the course curriculum, is there a different approach? – Bary12 Dec 03 '17 at 19:02
This is correct but incomplete in that we must show that $f$ can be extended to a continuous $\tilde f$. – DanielWainfleet Dec 03 '17 at 19:44
@DanielWainfleet This is a standard fact about uniformly continuous functions with complete codomain. – Aloizio Macedo Dec 03 '17 at 19:45
@Yes. It is. But beyond the proposer's background, so I attempted to give an "elementary" solution' – DanielWainfleet Dec 03 '17 at 20:42

Uniformly continuous function from a subspace of $\mathbb{R}$ maps bounded sets to bounded sets

2 Answers2