# A question about Lebesgue number lemma: can be the diameter not strictly less than the Lebesgue number?

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

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?

• 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 Dec 4, 2022 at 11:54
• @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? Dec 4, 2022 at 13:08
• This is not a hypothesis, though. The existence of $\delta$ is a conclusion, not an assumption. Dec 4, 2022 at 13:37
• @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. Dec 4, 2022 at 14:03

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 $$\mathcal A$$.

• Okay, thanks for your check!!! Answer upvoted and approved: thanks yet... ;-) Dec 4, 2022 at 15:43

$$\newcommand{\eps}{\varepsilon}$$ $$\DeclareMathOperator{\diam}{diam}$$ 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 $$\delta$$.