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I am seeking help on intuition behind the idea of the proof of this theorem. Part of my confusion stems from the fact that I don't quite understand what it means for an open cover to "have a Lebesgue number".

The definition I have for Lebesgue number for an open cover $\mathbb{U}$ of $A\subset X$ where $(X,d)$ is an arbitrary metric space is: a fixed real number $\epsilon>0$ is a Lebesgue number for $\mathbb{U}$ if for every $x\in A$ $\exists$ set $U(x)$ in $\mathbb{U}$ such that $B_\epsilon(x)\subset U(x)$.

So an open cover having a Lebesgue # means there is a certain radius of open ball for each element of A, which is contained in an element of the open cover? But I don't understand the significance behind this, and how this is related to the theorem I stated above.

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You have the correct definition and interpretation of a Lebesgue number, but seem to be struggling with the intuition. Personally, I tend to find it more instructive to see what goes wrong (rather than what goes right if we're following along with a theorem). As such, consider the following "non-example":

Let $(X, \mathcal{T})$ be the (not sequentially compact) interval $(0,1)$ with the standard topology. Let $\mathbb{U}$ be the open cover $\{U_n\}_{n=1}^\infty$ where $U_n = (1/(n+2) , ~1/n)$, e.g, $U_1 = (1/3, 1)$, $U_2 = (1/4, 1/2)$, et cetera. Note that this is indeed an open cover: given $x \in (0,1)$, there exists $n \in \mathbb{N}$ such that $1/(n+2) < x < 1/n$.

Since any $x$ will have a $U_n(x)$, we can stick a ball of radius $\epsilon(x)$ about $x$ inside $U_n(x)$, that is $x \in B_{\epsilon(x)} (x) \subset U_n(x)$. However, the issue is that the radius of the ball depends on the particular $x$ value. As $x \to 0$, we have that $\epsilon(x) \to 0$ as well, so there is no way to pick one $\epsilon$ that works for every $x \in X$.

In some way, having a Lebesgue number for an open cover is sort of like a function being uniformly continuous, instead of just continuous. For a function to be continuous, we can pick $\delta$ that depends upon $x$ and $\epsilon$, whereas uniform continuity requires that we choose $\delta$ depending only on $\epsilon$--the same $\delta$ has to work for every $x$.

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  • $\begingroup$ Excellent answer i think. Thanks. $\endgroup$ – Frank Swanton Nov 8 '16 at 3:08
  • $\begingroup$ =) I remember being confused upon first introduction as well--the key for me was the analogy with uniform continuity, the idea that this one Lebesgue number has to work for every single $x$, i.e., we don't get to pick a "different Lebesgue number for each $x$" or something like that. $\endgroup$ – erfink Nov 8 '16 at 3:10

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