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While doing some practices, I've come across an interesting question... the 'converse' of the Lebesgue Number Lemma.

The Lebesgue Number Lemma: Any open covering of a sequentially compact subset of a metric space has a Lebesgue Number $\lambda>0$

Now the question is: Suppose that every open covering of $M$ (metric space) has a positive Lebesgue number. Give an example of such an $M$ that is not compact.

My personal thoughts: Although I find the concept of Lebesgue Number to be easy to understand, however I'm usually clueless when it comes to open coverings.

Anyone can kindly provide insights on this question?

ireallydonknow
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1 Answers1

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An example of such $M$ is the set of integers $\mathbb Z$ with the standard metric $d(x,y)=|x-y|$. You can take $\lambda=1/2$; it works as a Lebesgue number for any open cover of $\mathbb Z$. It works, for the simple reason that every nonempty subset with diameter at most $1/2$ consists of exactly one point, which is contained in some member of the cover.

Aside. The above is closely related to the fact that every continuous function on $\mathbb Z$ is uniformly continuous. Along these lines, you may find the following fact interesting:

Fact. If a metric space $X$ has the property that every open cover of $X$ has a Lebesgue number, then every continuous function on $X$ is uniformly continuous.

After proving the above (which isn't hard), consider whether the converse of the fact holds. (I don't know; haven't given it much thought).