Does compactness of a subset use open sets from subspace topology for its finite subcover?

Question

I read that a subset $S$ of a topological space $(X,\tau)$ is compact if for every open cover of $S$, there exists a finite subcover, i.e.:

$$ S \subseteq \bigcup_{i \in J}O_i $$

for a finite $J$, and where $O_i$ are open sets.. Now, are these open sets $O_i$ taken from the subspace topology of $S$? , i.e. $(S,\tau_1)$ s.t. $O_i \in \tau_1$ or are they from $\tau$?

Answer 1

It's a distinction without a difference. $O_i \in \tau_1 \iff \exists U_i \in \tau (O_i=U_i \cap S)$, so if any open cover in the subspace topology must have a finite subcover, the same must be true in $\tau$, and vice versa.

Answer 2

Since open sets in the subspace topology are intersections of open sets in the ambient space with $S$, the two conditions are equivalent.

Answer 3

Either one is fine. If you cover $S$ by open sets $O_i$ in $\tau$ then $S$ is also covered by $O_i\cap S$ and these are open in $S$. On the other hand if you cover $S$ by open sets $O_i$ which are ope in $S$ then there exist open sets $O_i'$ that are in $\tau$ such that $O_i=O_i'\cap S$ and $S$ is covered by $(O_i')$ also.