Let $S$ be a nonempty subset of $\mathbb{R}$. If ܵS is a finite union of disjoint bounded intervals, then which one of the following is true?

Question

Let $S$ be a nonempty subset of $\mathbb{R}$. If ܵS is a finite union of disjoint bounded intervals, then which one of the following is true?

(A) If ܵS is not compact, then $\sup S \notin S$ and $\inf S ܵ\notin S$

(B) Even if sup S ∈ ܵS and $\inf S\in S$, S need not be compact

(C) If $\sup S \in S$ and $\inf S\in S$, then S is compact

(D) Even if ܵS is compact, it is not necessary that $\sup S \in S$ and $\inf S \in S$

My attempt : I have discarded the option a) and c) by taking $ S = [1,2) \cup (3,4]$ here $\inf S = 1 \in S$ and $\ sup S = 4 \in S$

I'm comfusion about option B and option C

thanks in advance

Answer 1

Your example $[1,2)\cup(3,4]$ is fine and proves that (A) and (C) don't hold in general. It also shows that (B) is true.

The statement (D) is false. If $S$ is compact, then $\sup S\in S$ because:

1. $\sup S$ is the limit of a sequence of elements of $S$;
2. since $S$ is compact, the limit of any convergent sequence of elements of $S$ also belongs to $S$.

The same argument applies to $\inf S$.