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!


1 Answer 1


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?

  • $\begingroup$ 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 $B_{1/k_\ell}(x^{k_\ell})\subseteq B_R(x)\subset U_i$? $\endgroup$
    – Theorem
    Oct 31, 2019 at 17:29
  • 1
    $\begingroup$ 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. $\endgroup$
    – Thorgott
    Oct 31, 2019 at 17:35

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