Answer:
It has been proved that T is compact
Step-by-step explanation:
To prove this using the definition of compactness, let's assume that T is
not compact. Now, if that be the case, an open cover of T will exist. Let's call this open cover "A". Now, this open cover will have no finite subcover.
Now, from the question, since T is closed, it’s complement R\T will be open.
Therefore, if we add the set R\T to the collection of sets A, then we'll have an open cover of R and also of S.
Due to the fact that S is compact, this
cover will have a finite sub - cover which we will call B.
Finally, either B itself or B\{R\T} would be a finite sub - cover of A. This is a contradiction.
Thus, it proves that T has to be compact if S is to be a compact subset of R and T is to be a closed subset of S.
