A Topological space $X$ is compact if every Filter base $\mathcal{B}$ on $X$ has an accumulation point. That is, there exists $x \in X$ such that

for all $N \in \mathcal{N}(x)$, for all $A \in \mathcal{B}$, $N \cap A \neq \emptyset$.

An alternative, well-known (equivalent?) definition involves properties of ‘coverings’ of X by families of open sets. The definition used to define a Compact set

Proposition. A closed subset of a compact space is compact.

Propostion. Every compact subset of a Hausdorff space is closed.

Lemma. If $f: X \rightarrow Y$ is continuous and $X$ is compact then $f(X)$ is compact ("the image of a compact set is compact")

Theorem. Let $f: X \rightarrow Y$ be bijective and continuous. If $X$ is comapct and $Y$ is Hausdorff, then $f$ is a Homeomorphism.

Tychonoff's theorem. Arbitrary products of compact spaces are compact.

Theorem. Let $X$ be an ordered set with the Least upper bound property. Then, in the Order topology, its Closed interval is compact. Corollary. Each closed real interval $[a,b] \subset \mathbb{R}$ is compact. The Ordered square is compact.

Heine-Borel theorem