The whole point of this article is the following seemingly trivial observation.

Theorem. A topological space X is compact if and only if it satisfies thefinite intersection property(F.I.P.):

- if is a collection of closed subsets of X such that every finite intersection then

**Proof**.

Suppose *X* is compact. Then given a collection {*C _{i}*} of closed subsets of

*X*with empty intersection, we have so the open cover {

*X*–

*C*} has a finite subcover. This gives a finite collection of

_{i}*C*with empty intersection.

_{i}The converse is just as easy: if F.I.P. holds and is an open cover of *X*, then By F.I.P., there’s a finite collection so is a finite subcover. ♦

Though this looks like a simple rephrasing of the definition of compact spaces, its underlying philosophy embodies an entire paradigm shift. [ Ok, that’s probably stretching it a little … ] The idea is that a closed subset of *X* typically represents a subset which satisfies certain nice conditions. Hence F.I.P. means that if every finite set of conditions can be satisfied by some object, then there’s some object that satisfies all conditions.

To make it a little more explicit, let’s prove the following.

Proposition. Suppose are non-empty compact Hausdorff spaces and is a collection of continuous maps:Then there exists a sequence where and for each n≥1.

Compactness is essential here. For example, if and each *f _{n}* is the inclusion map, then no such sequence exists since the intersection of all

*X>n*is empty.

**Proof**.

Let which is compact by Tychonoff’s theorem. For each *n*≥1, let:

Note that Also is homeomorphic to via the following mutually inverse maps:

Thus is non-empty and compact. Since *X* is Hausdorff, is closed in *X*. By F.I.P., ♦

In particular, if *X _{n}* is finite with the discrete topology, then it’s compact Hausdorff and any map

*f*is continuous.

_{n}

Corollary. If is a sequence of non-empty finite sets and is a collection of functions, then we can find such that for each n≥1.

Once again, a deceptively simple result. It looks obvious until one really attempts to prove it. [ Try proving it directly; bear in mind *f _{n}* is not surjective in general. ]

## Application: Infinite Four-Colour Theorem

We know the four-colour theorem for maps with finitely many countries, namely, any such map can be coloured by at most four colours such that no two countries which share a border have the same colour.

What about a map with countably infinite number of regions? It’s not entirely clear that four colours suffice. One may be inclined to reason as follows: start with a region with finitely many countries – we know this can be coloured by four colours – then extend the colouring to other countries one at a time. The problem with this approach is that if we start with a bad colouring, then we may get stuck and be forced to retrace the steps, and how do we prove that retracing the steps must eventually lead to a valid colouring?

Compactness to the rescue.

Label each country as 1, 2, 3, … . Let *X _{n}* be the set of colourings of the map restricted to countries 1, …,

*n*, with 4 colours such that two countries which share a border do not have the same colour. We have a map by ignoring the colour of the

*n*-th country.

Now, classical four-colour theorem tells us each *X _{n}* is non-empty. So by the above corollary, there’s a sequence with such that for each

*n*. This corresponds to a colouring for the entire infinite map. ♦

## Application: Infinite Tiling

Suppose we wish to fit a countably infinite set of shapes in a bounded region.

Now we assume that every finite subset of shapes can be fitted within the bounded region without overlaps. [ We say that two shapes overlap if their interiors share a common point. Also a shape is within the bounded region as long as its interior is inside. ]

Claim.Under this assumption, the entire infinite class of shapes can be placed in the region without overlaps.

**Proof**.

Label the shapes by 1, 2, 3, … . For each *n*, let *X _{n}* be the set of ways to fit shapes 1 to

*n*within the region. The map is given by removing shape

*n*.

The placement of each shape *i* can be parametrised by coordinates and a point on the unit circle representing its orientation. Thus *X _{n}* can be parametrised by 2

*n*coordinates and

*n*points on the unit circle, which forms a subset of

**R**

^{4n}. This subset is clearly bounded; it’s also closed because its complement is open: e. g. if two shapes have overlapping interiors, then by perturbing all shapes a little, those two shapes still overlap. Thus, by the Heine-Borel theorem,

*X*is compact. The above proposition tells us there’s a way to fit all shapes in. ♦

_{n}**Note**

Suppose we change the condition of overlap just a little: *now two shapes are said to overlap if there’s a common point (even at the boundary)*. Then consider rectangles 1 × 2^{–n}, for *n* = 1, 2, 3, … , to be packed in the square 1 × 1. Since there’s a way to pack every finite set of rectangles in the square, but if we attempt to pack the entire infinite set, some of them must share an edge and thus overlap.