Find the supremeum and infimum, and is the set compact?

Question

A) $S=(1,2) \cup (2,3]$

Here the Supremum would be 3, since it's the least upper bound of the set. Which 3 would be in the set.

The infimum would be 1, since it's the greatest lower bound, but it's not included in the set.

For the compact I know it's supposed to be closed and bounded so looking at this set, there is a finite subcover covering [3], but when looking at the open set (1,2), there doesn't exist a finite subcover covering (1). Would this be a non compact case?

Answer 1

The set is bounded but not closed. For example, the collection $(1,2) \cup (2+{1 \over n},4)$ is an open cover that has no finite subcover.

Answer 2

The set is not compact because the sequence $x_n=\frac{1}{n+1}+1 $ which lies in $S,\forall n \in \Bbb{N}$, does not have a convergent subsequence in $S$

$x_n \to 1 $ thus every subsequence of $x_n$ converges to $1$

Answer 3

The logic about the supremum and infimum is correct.

This is non-compact, more precisely because compact sets are to be closed, and this set is not closed : $2$ is a limit point of the set(why?), and $2$ is not part of the set. Note that you need not consider the logic of open covers if you are looking at closedness and boundedness.