We give a proof of Kuratowski theorem. This is the Kuratowki Theorem involving the projection map from a product space onto one of the factors. All spaces in consideration are at least Hausdorff.

A map from the space into the space is said to be a closed map if is closed in for every closed subset of . Here’s the statement of the theorem.

**Kuratowski Theorem**

The following conditions are equivalent.

- The space is compact.
- For every space , the projection is a closed map.
- For every normal space , the projection is a closed map.

Before going to the proof, note that the theorem is not true when none of the factors is compact. Consider the product space where is the real number line. The graph of the curve with ranges over the non-zero real numbers is a closed subset of . The projection of graph of into the x-axis is the set , which is not closed in .

**The Proof**

The direction is clear. We prove and .

Let be compact. Let be a closed subset of . We show that is open in . To this end, let and . Note that . By the Tube Lemma (see here), there exists an open set with such that . It follows that . This means that is open and is closed in .

Suppose condition 3 holds. We show that is compact. Suppose not and let be a collection of closed subsets of with the finite intersection property such that (see here for the characterization of compactness using the finite intersection property). Choose a point . Let . Define a topology on using the following base:

where sets in are of the form where is finite and . In words, we can describe the topology on in this way. The set is a discrete subspace of and each basic open set containing the point is the union of three sets – the singleton set . the intersection of finitely many sets in and a subset of . It follows that the space is Hausdorff. To show this, we only need to separate and any point in by disjoint open sets. Let . Note that . Thus, for some . As a result, and are disjoint open sets separating and , respectively. It also follows that the space is normal. To see this, let and be disjoint closed subsets of such that . Then and are open sets separating and , respectively.

Let where . By assumption, is closed in . Note that . Since , we have . This means that . Furthermore, there exists such that . We claim that for all . This contradicts that the intersection of is empty. Let . Let be open such that . Consider the open set , which contains the point . Since , and since , the open set contains points of . Thus for some . This means that and . What we have just shown is that for every open set containing , for all . It follows that for all , producing that contradiction indicated earlier. Thus, every collection of closed subsets of with the finite intersection property must have non-empty intersection. This implies that is compact. With this observation, the proof of Kuratowski Theorem is complete.

**Perfect Mapping**

The function is a perfect map if is continuous and a closed map such that each point inverse is compact. When each point inverse under the mapping is compact, the map is said to have compact fibers. A corollary of Kuratowski Theorem is that the projection map in Kuratowski Theorem is also a perfect map.

**Theorem 1**

Let be a space and be a compact space. Then the projection is a perfect map.

The projection map is always continuous. According to Kuratowski Theorem, it is a closed map. For each , , which is compact. Thus, each point inverse of the projection map is compact.

Dan Ma topology

Daniel Ma topology

Dan Ma Kuratowski Theorem

Daniel Ma Kuratowski Theorem

Dan Ma perfect map

Daniel Ma perfect map

2023 – Dan Ma