Theorem. Let $D \subset Y$. Suppose that D is non-empty and sequentially compact. Let C be an open cover of D. Then there is a real number $\epsilon > 0$ such that if $E \subset D$ with $d(E)<\epsilon$ then $E \subset A_{\alpha}$ for some $A_{\alpha}\in C$.
Note that d(E) is the diameter of E.

This seems so intuitively obvious to me... or am I not reading it correctly? Can't I just choose $\epsilon$ small enough that E is approximately a point?

2. ## Re: Theorem about covers

Originally Posted by paupsers
Theorem. Let $D \subset Y$. Suppose that D is non-empty and sequentially compact. Let C be an open cover of D. Then there is a real number $\epsilon > 0$ such that if $E \subset D$ with $d(E)<\epsilon$ then $E \subset A_{\alpha}$ for some $A_{\alpha}\in C$. Note that d(E) is the diameter of E.
This seems so intuitively obvious to me... or am I not reading it correctly? Can't I just choose $\epsilon$ small enough that E is approximately a point?
This is far from a trivial result. It is known as proving that sequentially compact metric space has Lebesgue number.
Suppose not. If $n\in\mathbb{Z}^+$ there must be a set $B_n$ such that $0 but $B_n$ is a subset of no open set in the cover.
See what you can do with that setup.