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The Lebesgue covering lemma states :

If $\{G_{\alpha} \subset X : \alpha \in I\}$ is an open cover of sequentially compact set , then $\exists \delta >0 , s.t. \forall x \in X$ there is some $\alpha \in I$ with $B_{\delta}(x)\subset G_{\alpha}$.

My Question :

Consider a metric space $(X,d)$. Say we know that a set is not open then we do not have the case that around every point in $X \exists x, s.t. B_{\delta}(x) \subset X$ for $\delta >0$. So does this mean that the set is not sequentially compact or does the implication of the lemma require that the set be sequentially compact first and only then we can say that there exists an open ball around each point x in $X$,contained in $X$ ?

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The concept of open ball depends upon the space that we're working with. If you have a subset $S$ of $X$ and you talk about an open ball $B_\delta(x)$ in that subset, then what that means is$$\{y\in X\,|\,d(x,y)<\delta\}\cap S.$$So, such a ball is always a subset of $S$.

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  • $\begingroup$ I'm afraid this isn't quite what I meant. I must not have phrased my question very clearly, I apologise. what I meant was If we know a set is not open can we use the above lemma to decide if it is sequentially compact ( and hence compact in a metric space) $\endgroup$ – excalibirr Aug 17 '18 at 19:32
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    $\begingroup$ Think about $\[0,1]$, with the usual distance. It is compacy. Now think about $\{0\}$. It is not an open subset. However, it is still compact. $\endgroup$ – José Carlos Santos Aug 17 '18 at 19:36
  • $\begingroup$ ah okay , thank you :) I see now the implications of the lemma only go one way. May I ask is this lemma much use for introductory metric spaces where I only want to show if a set is open\closed, complete, compact... or does it really only become useful in more advanced topology ? $\endgroup$ – excalibirr Aug 17 '18 at 19:48
  • $\begingroup$ @exodius the lemma is used to show that a continuous function on a compact metric space is uniformly continuous, e.g. $\endgroup$ – Henno Brandsma Aug 18 '18 at 16:13

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