Question

Let S be a subset of R. Prove that S is compact iff every infinite subset of Shas an accumulation point in S. The last 3 sentences in the first paragraph. Start from (Thus by Considering the neighborhood N(x; 1/n), where n ∈N, we may have a count-able set of distinct points, sayT={xn:n∈N},contained in S and there exists xn ∈N(x; 1/n). Therefore x ∈T′ and by assumption some accumulation point of T is in S. We claim that x is the only accumulation point of T). This is a true statement. Just need justification.

Answer 1

To prove that a subset S of R is compact, we need to show that S is closed and bounded. On the other hand, to prove that every infinite subset of S has an accumulation point in S, we need to show that every sequence in S has a limit point in S.

Step-by-step explanation:

To prove that a subset S of R is compact, we need to show that S is closed and bounded.

On the other hand, to prove that every infinite subset of S has an accumulation point in S, we need to show that every sequence in S has a limit point in S.

Firstly, if S is compact, then it is closed and bounded. To prove this, we can use the fact that every open cover of S has a finite subcover, and that S is a closed subset of itself. Therefore, any infinite subset of S must have an accumulation point in S.

Conversely, if every infinite subset of S has an accumulation point in S, then S is compact. This can be proven by showing that S is closed and bounded. By assuming the opposite, we can construct an open cover of S that does not have a finite subcover. This implies the existence of an infinite subset of S without an accumulation point, which contradicts the given assumption.