# Implications of the Lebesgue covering lemma

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 $X$ ?

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 $S$.
• 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. – José Carlos Santos Aug 17 '18 at 19:36