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?