# A Lemma In Proving Sequentially Compact Implies Every Open Cover Has a Finite Subcover

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!

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?

• I'm not sure how to use $B_{1/k_\ell}(x^{k_\ell})$ to my advantage. I understand the $x^{k_\ell}$s are very close to $x$ inside some $B_R(x)$ and I take very small radii, but that doesn't gurantee $B_{1/k_\ell}(x^{k_\ell})\subset U_i$! In other words, how do I make sure I take $\frac{1}{k_\ell}$ small enough to ensure that $B_{1/k_\ell}(x^{k_\ell})\subseteq B_R(x)\subset U_i$? Oct 31, 2019 at 17:29
• There is an $L$ such that $x^{k_l}\in B_R(x)\subseteq U_i$ for all $l\ge L$, $1/k_l\rightarrow0$ and $B_R(x)$ is open, so it does. Oct 31, 2019 at 17:35