A compact locally connected metric space is "uniformly locally connected"\ That is, for any $\epsilon > 0$, there is some $\delta > 0$ such that whenever $\rho(x, y) < \delta$, then $x$ and $y$ both lie in some connected subset of $X$ of diameter $<\epsilon$.


Since $X$ is locally connected metric space then each $x\in X$ has a nhood base of open connected sets\ Given $\epsilon > 0$, let $x \in X $ and $U_x=\rho(x, \epsilon) $ be a nhood of $x$\ There exist an open connected basic nhood $V_x$ with diameter $<\epsilon$, Now $$X=\bigcup_{x\in X }{V_x}$$, hence cover $X$ by open connected nhoods of diameter $<\epsilon$.\ Since $X$ is compact, reduce this to a finite subcover $\{V_{x1},. . . , V_{xn}\}$ and let $\delta$ be a Lebesgue number (22.5) for this cover.\ Then if $\rho(x, y) < \delta$, both $x$ and $у$ belong to some $V_{xi}$. \

{Theorem 22.5} (Lebesgue covering lemma). If $\{U_1..., U_n\}$ is a finite open cover of a compact metric space X, there is some $\delta > 0$ such that if A is any subset of $X$ of diameter $< \delta$, then $A \subset U_i$ for some i.

I try to write the proof better than this.

I would like to confirm this proof

If acceptable, I would like to clarify and improve it (Language and Mathematical)as much as possible

  • $\begingroup$ Do not use \ or , or .\ at the end of a sentence. Just use . $\quad$ And if a sentence ends in a displayed line, include the . in that line. Write in the same style as for an an essay about something else. What you have is not so bad. I've seen far worse. And your proof is correct . $\endgroup$ – DanielWainfleet Dec 12 '17 at 9:25

The proof is correct.

As for writing it up. That depends on your audience. For a general audience, the first rule is - avoid abbreviations ("neighborhood", not "nhood"). And I assume that the "/" scattered through the text was some sort of misguided attempt to show where you put new lines? Just let Latex have it's way here. If a new line is needed, let it form a paragraph break.

Also "and $U_x=\rho(x,\epsilon)$ be a nhood of $x$" doesn't make sense. $\rho$ is a function that takes points in your space in both arguments, and produces real numbers. You have it with a real number as the second argument, and producing a set. Now it is evident that you mean $U_x$ to be a ball of radius $\epsilon$ about $x$. But if someone has told you this is a good way to denote the ball, do NOT trust them about notations in the future! Commonly used ways of denoting balls include $B(x, \epsilon)$ and $B_x(\epsilon)$. If the metric to be used is not obvious, then usually $B_\rho(x, \epsilon)$ is preferred. (Of course, it you had intended $B(x, \epsilon)$ and only accidently used $\rho$, just be careful in your final write-up.)

But you don't do anything with "$U_x$", which makes introducing a notation for it pointless. You don't actually need to name the balls, since you don't plan on using that name.

There are also some phrasing differences I would suggest. Your phrasing is clear enough, but sounds awkward sometimes to a native English speaker (at least - to this native English speaker). There is nothing particularly bad, though.

I would write it up like this:

Since $X$ is locally connected, for $\epsilon > 0$, each point $x \in X$ has a open connected neighborhood $V_x$ contained within the ball of radius $\epsilon$ about $x$. The collection $\{V_x \mid x \in X\}$ forms an open cover of $X$ by sets of diameter $< \epsilon$. Since $X$ is compact, it has a finite subcover $\{V_1, V_2, ..., V_m\}$. By the Lesbegue covering lemma (Theorem 22.5), there is a $\delta > 0$ such that if $\rho(x, y) < \delta$, then there is some $V_i$ with $\{x, y\} \subset V_i$, which completes the proof.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.