While proving that if a set $A$ is sequentially compact in $\mathbb R^n$ (Every sequence has a converging subsequence in $A$), then every open cover has a finite subcover, I was asked to prove the following:
Let $\left \{ U_i \right \}_{i \in I}$ an open cover of $A$. Prove by contradiction that there exists $r>0$ such that for all $x\in A$ there exists $i \in I$ such that $B_r(x)\subset U_i$.
My attempt: Assuming this is not true, I'm trying to construct a sequence. for all $k\in \mathbb N$ there exists $x^k\in A$ such that for all $i\in I$, $B_{1/k}(x^k)\nsubseteq U_i$. I then thought of using the convergent subsequence $x^{k_\ell}\rightarrow x\in A$ and perhaps denoting by $i_\ell$ an index such that $x^{k_\ell} \in U_{i_\ell}$. But I couldn't get any further.
Any help is appreciated!
Best Answer
That is certainly a good approach. Consider: there is an $i\in I$ such that $x\in U_i$; now $x^{k_l}\rightarrow x$ means nothing but that any ball around $x$ contains all but finitely many of the $x^{k_l},l\in\mathbb{N}$ and there is some ball around $x$ that is contained in $U_i$, because that is an open set. Can you finish?