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How to prove the following proposition?

Let $(X, d)$ be a metric space, if for any metric space $(Y,\tilde{d})$, any continuous map $f:(X,d)\rightarrow (Y,\tilde{d})$ is uniformly continuous then $(X,d)$ satisfies the following property: For any open cover of $X$, there exist a real number $\delta >0$, such that every subset of X having diameter less than $\delta$ is contained in some member of the cover.

The property stated above comes from the well known Lebesgue number lemma(See James R. Munkres Topology second edition Lemma 7.5): If the metric space $(X,d)$is compact and $\mathcal{A}$is an open cover of $X$, then there exists a number $\delta >0$ such that every subset of $X$having diameter less than $\delta$ is contained in some member of $\mathcal{A}$.

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  • $\begingroup$ @JoséCarlosSantos Thank you for your advice $\endgroup$ – user658532 Mar 28 at 9:04
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    $\begingroup$ What I know: If $X$ is connected and every continuous function is uniformly continuous, then $X$ is compact (which implies it satisfies Lebesgue number lemma). See here. $\endgroup$ – YuiTo Cheng Mar 28 at 11:55
  • $\begingroup$ @YuiToCheng Thank you $\endgroup$ – user658532 Mar 28 at 12:05
  • $\begingroup$ I'm curious where you found the problem. Can you state the source? (I assume this is not come up on your own; otherwise, how can you assert the proposition is true?) $\endgroup$ – YuiTo Cheng Mar 28 at 12:05
  • $\begingroup$ I can prove that X is complete, but I don’t how to continue. $\endgroup$ – user658532 Mar 28 at 12:06
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We are going to prove it by contradiction. Suppose $X$ doesn't satisfy the Lebesgue Number Lemma, i.e.

There exists an open covering $\{U_{\alpha}\}_{\alpha\in J}$ such that for all $\delta>0$, there exists $A_{\delta}$, diam $A_{\delta}<\delta$ and $A_{\delta}\subsetneq U_{\alpha}$ for all $\alpha\in J$.

It suffices to prove that there exists $E= \{x_{1},y_{1},x_{2},y_{2},\dots\}$ such that $E$ has no limit point and $d(x_n,y_n) \to 0$ when $n\to \infty$.

Because $E$ has no limit points, it follows that $E $ is closed in $X.$ In its subspace topology, $E$ is discrete. Hence any function defined on $E$ is continuous. Let's take $g = 0$ on $\{x_{n}\},$ $g = 1$ on $\{y_{n}\}$. By the Tietze extension theorem (since $E$ is closed), $g$ extends to a function $G:X\to \mathbb {R}$ that is continuous on $X$. But $G$ is not uniformly continuous (As $d(x_{n},y_{n}) \to 0,|G(x_{n})-G(y_{n})|=1$). Contradictory to the starting assumption that every continuous function on $X$ is uniformly continuous.

How to obtain such $E$? Simply choose $x_n,y_n\in A_{1/n}$ with $x_n\neq y_n$. I claim that no subsequence $x_{n_k}$ can converge in $X$. If $x_{n_k}\to x$, then $x\in U_\alpha$ for some $\alpha$. Because $U_\alpha$ is open, there exists $\epsilon>0$ such that the open ball $B(x,\epsilon)$ of radius $\epsilon$ centered at $x$ is contained in $U_\alpha$. Since $x_{n_k}\to x$, every neighboorhood of $x$ contains infinitely many $x_{n_k}$. Choose $n_k$ large enough so that $1/{n_k}<\epsilon$. Then $A_{n_k}\subset B(x,\epsilon) \subset U_{\alpha}$. Contradiction. Similarly, no subsequence $y_{n_k}$ can converge in $X.$

Now if $E$ has a limit point, a subsequence $x_{n_k}$ or $y_{n_k}$ must converge in $X.$ Hence it's our desired $E$.

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  • $\begingroup$ @PaulFrost Still thanks for pointing out the error that diam$A_\delta<\delta$, not = $\endgroup$ – YuiTo Cheng Mar 29 at 10:43
  • $\begingroup$ I guess the "=" was the reason for my misunderstanding. If $diam(A) > 0$, you automatically find distinct points in it even without the assumption that it is not contained in any $U_\alpha$. $\endgroup$ – Paul Frost Mar 29 at 10:48
  • $\begingroup$ By the way, +1 for your nice answer. $\endgroup$ – Paul Frost Mar 29 at 16:40

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