Let (X,d) be the discrete metric space and let S be a infinite subset of X. Create a proof that S is non-compact in X.
Thanks.
To prove that S is non-compact in the discrete metric space X, we can show that S does not satisfy the definition of compactness.
The definition of compactness states that a set S is compact if and only if every open cover of S has a finite subcover. In other words, for any collection of open sets that cover S, there exists a finite subcollection that still covers S.
In the case of the discrete metric space X, the open sets are the subsets of X. Since X is discrete, every subset of X is open. Therefore, each point in X is an open set.
Now, for the infinite subset S of X, we can choose an open cover for S by considering every individual point in S as an open set. In other words, for each element s in S, we take the open set {s}.
This open cover for S consists of one open set for each element of S. However, since S is an infinite set, this open cover has infinitely many open sets.
Now, if S were compact, there would exist a finite subcover that covers S. However, in this case, it is impossible to find a finite subcover since the open cover we chose contains infinitely many open sets.
Therefore, we have shown that S does not satisfy the definition of compactness, and hence, S is non-compact in the discrete metric space X.