# Subsets of a set with trivial topology

Question: Let $X$ be any set with at least two elements. Assume that the only open subsets of $X$ are the empty set $\emptyset$ and $X$ itself. - Which subsets of $X$ are closed? - Which subsets of $X$ are compact?

My thoughts: Thus also $\emptyset$ and $X$ have to be also closed subsets. As their complements are both open and by definition the set is closed if the complement is open.

The open set is compact as it is a finite set and also $X$ is compact as it has a finite amount of closed subsets, thus is bounded and closed. Am I in any way correct with these thoughts?

## 2 Answers

For a set to be $not$ compact, it needs to have an open cover without any finite subcovers. Are there any sets which have open covers which can't be reduced to finite ones? Think about what open covers must look like in this topology.

The equivalence between compactness and bounded+closed is only true in metric spaces. In fact not even that, just in a particular class of metric spaces, not even in all of them. For general metric spaces we need to strengthen bounded to "totally bounded".

And in general for topological spaces we don't have a notion of boundedness, so compactness is instead characterized using open sets:

$A$ is compact if and only if whenever $\{U_i\mid i\in I\}$ is a collection of open sets such that $A\subseteq\bigcup_{i\in I}U_i$, there are some $i_1,\ldots,i_n\in I$ such that $A\subseteq\bigcup_{j=1}^n U_{i_j}$.

Meaning that $A$ is compact if whenever it is covered by open sets, we can find just a finite subset of this cover which already covers $A$.

If $A$ is a subset of a finite set, how many subsets does that finite set has? What sort of open covers (regardless to the topology!) could it even have?