## Compactness, Fiber Products, and Proper Maps

Most references on topological spaces seem to define a compact space in terms of open coverings of a space having finite subcoverings. Formally, we say that a collection of open sets $\left\{U_{i}\right\}_{i \in I}$, indexed by some set $I$, is an open covering of a topological space $K$ if $K \subset \bigcup_{i \in I} U_{i}$. We define $K$ to be compact if there exist finitely many open sets $U_{i_{1}},\cdots,U_{i_{n}}$ in the open covering such that $K \subset \bigcup_{j=1}^{n}U_{i_{j}}$.

There exist a number of equivalent definitions of a compact space, which we briefly mention here. A topological space $K$ is compact if and only if $K$ has the finite intersection property: if $\left\{E_{i}\right\}_{i \in I}$ is a collection of a closed sets indexed by a set $I$ such that, for any finite subset $I_{0} \subset I$, $\bigcap_{i \in I_{0}} E_{i} \neq \emptyset$, then $\bigcap_{i \in I}E_{i} \neq \emptyset$.

$\displaystyle p(x) := (x-z)(x-\overline{z}) = x^{2} - \overline{z}x - zx + \left|z\right|^{2} = x^{2} - 2\text{Re}(z)x + \left|z\right|^{2}$

Another equivalent definition is in terms of nets, also known as Moore-Smith sequences. A topological space $K$ is compact if and only if every net $(x_{\alpha})_{A}$ in $K$ has a convergent subnet.

We now prove the following theorem:

A topological space $K$ is compact if and only if, for every topological space $Z$, the projection map $\pi_{Z}: K \times Z \rightarrow Z$ is closed, where $K \times Z$ is endowed with the product topology.

Here is the proof.