# Every continuous function is uniformly continuous $\Rightarrow$ Lebesgue number exists for every open cover

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 $$\mathcal{A}$$.

• @JoséCarlosSantos Thank you for your advice – user658532 Mar 28 at 9:04
• 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. – YuiTo Cheng Mar 28 at 11:55
• @YuiToCheng Thank you – user658532 Mar 28 at 12:05
• 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?) – YuiTo Cheng Mar 28 at 12:05
• I can prove that X is complete, but I don’t how to continue. – user658532 Mar 28 at 12:06

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 $$E$$.

• @PaulFrost Still thanks for pointing out the error that diam$A_\delta<\delta$, not = – YuiTo Cheng Mar 29 at 10:43
• 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 $U_\alpha$. – Paul Frost Mar 29 at 10:48
• By the way, +1 for your nice answer. – Paul Frost Mar 29 at 16:40