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

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 $$\tau$$?

• For what it's worth I always heard/interpreted the definition for the topology of $X$ but as the three answers point out it would be an equivalent definition either way. – fleablood Jun 18 at 23:40

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