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At the $27$-th paragraph of his topology text Munkres prove the following result, well know as Lebesgue number lemma.

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So by the proof (not by the statement!) the diameter must be strictly less than $\delta$ and not simply less but I think it is possible to require that it is just such really: indeed, if we pick any $\epsilon\in(0,\delta)$ then any set with diameter less (or equal) to $\epsilon$ it has surely diameter strictly less then $\delta$ and so that it is contained in an element of $\cal A$; moreover, into the proof we can pick $\delta$ strictly less the the minimum value of $f$ so that the proof holds for the above described case. So I ask if what I observed is actually true or false because I think that if it is then actually the Lebesgue number lemma holds with more general condition. So could someone help me, please?

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  • $\begingroup$ We can infer 'strictly less' from the word 'less', many authors would explicitly say 'less than or equal to' if necessary. It's not clear what the question is, aside from that $\endgroup$
    – FShrike
    Dec 4, 2022 at 11:54
  • $\begingroup$ @FShrike Well, the question is: can the diameter be less (or equal) to δ instead of strictly less? Does still the lemma holds with this more general hypothesis? $\endgroup$ Dec 4, 2022 at 13:08
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    $\begingroup$ This is not a hypothesis, though. The existence of $\delta$ is a conclusion, not an assumption. $\endgroup$
    – FShrike
    Dec 4, 2022 at 13:37
  • $\begingroup$ @FShrike Oh, right! So can I conclude that $\delta$ is such that if $Y$ is a subset of $X$ with diameter less or equal to $\delta$ then it is contained in an elemento of $\cal A$? Forgive my confusion. $\endgroup$ Dec 4, 2022 at 14:03

2 Answers 2

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You are right, Munkres says that if $\operatorname{diam}(B) < \delta$, then $B$ is contained in some element of $\mathcal A$. But of course it is also true that there exists $\delta' > 0$ such that if $\operatorname{diam}(B) \le \delta'$, then $B$ is contained in some element of $\mathcal A$. You do not need to go into Munkres proof, just take any $\delta' \in (0,\delta)$.

You can also define $$\lambda = \sup \{ \delta > 0 \mid \delta \text{ is a Lebesgue number for } \mathcal A\}. $$ Clearly if $\operatorname{diam}(B) < \lambda$, then $B$ is contained in some element of $\mathcal A$. But you cannot expect that that this is true under the assumption $\operatorname{diam}(B) \le \lambda$.

The above $\lambda$ is the biggest Lebesgue number for $\mathcal A$. We may have $\lambda = \infty$, depending on $\mathcal A$.

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  • $\begingroup$ Okay, thanks for your check!!! Answer upvoted and approved: thanks yet... ;-) $\endgroup$ Dec 4, 2022 at 15:43
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$\newcommand{\eps}{\varepsilon}$ $\DeclareMathOperator{\diam}{diam}$ The proof provides a $\delta$ which satisfies the requirements of the theorem. However, the theorem says that there is one such $\delta$ and as you pointed out, by no means this $\delta$ is unique since you can always find a $\delta'$ such that $0<\delta'<\delta$. So I think your question is the following: Is the $\delta$ found in the proof the largest $\delta$ we can have so that the diameter of a set has to be strictly less than $\delta$? I would say no. To illustrate this, I need to provide an example. Fix $\eps \in \mathbb{R}$ with $0<\eps$ and consider the compact set $K=[-\eps,\eps]$. Further, consider the collection of open sets $\mathcal{A}=\{A_1,A_2\}$, where $A_1=(-\infty,\eps)$ and $A_2=(-\eps,\infty)$. Clearly, $K\subset \bigcup_i A_i$. Note that $C_1=[\eps,\infty)$ and $C_2 = (-\infty,-\eps]$. We will compute $f(x)$ for $x$ in different regions.

  1. When $x\in (0,\eps)$: $$ d(x,C_i) = \left\{ \begin{array}{rl} \eps-x,& i=1 \\ \eps+x,& i=2 \end{array} \right. $$ Hence, $f(x) = \eps$.
  2. When $x\in (-\eps,0)$: $$ d(x,C_i) = \left\{ \begin{array}{rl} \eps-x,& i=1 \\ \eps+x,& i=2 \end{array} \right. $$ Hence, $f(x) = \eps$.
  3. When $x=0$: $$ d(x,C_i) = \left\{ \begin{array}{rl} \eps,& i=1 \\ \eps,& i=2 \end{array} \right. $$ Hence, $f(x) = \eps$.
  4. When $x\in C_1$: $$ d(x,C_i) = \left\{ \begin{array}{rl} 0,& i=1 \\ \eps+x,& i=2 \end{array} \right. $$ Hence, $f(x) = (\eps+x)/2\geq (\eps+\eps)/2=\eps$.
  5. When $x\in C_2$: $$ d(x,C_i) = \left\{ \begin{array}{rl} \eps-x,& i=1 \\ 0,& i=2 \end{array} \right. $$ Hence, $f(x) = (\eps-x)/2\geq (\eps+\eps)/2=\eps$. We conclude that $$ \min_{x\in K}\left\{f(x)\right\} = \eps. $$ However, note that any subset of $K$ whose diameter is less than $2\eps$ is completely covered by one $A_i$. Here's a proof. Let $W \subset K$ with $\diam(W) < 2\eps$. First observe that it cannot be the case that $\inf(W)=-\eps$ and $\sup(W)=\eps$ simultaneously since this would imply that $\diam(W)=2\eps$ contradicting the assumption. If the $\sup(W) < \eps$,
    then, $W \subset (-\infty, \eps) = A_1$. If $-\eps < \inf(W)$, then $W \subset (-\eps, \infty) = A_2$. In any case we were able to show that $W$ is fully contained in one of the $A_i$. In summary, though the $\delta$ provided by the proof works, in certain cases it is possible to find a Lebesgue number that is larger than that $\delta$.
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