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.

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