One-point compactification · Topology

Intuition

A space that is not compact can often be made compact by adding a single point. The new point is thought of as infinity, and a set around it is declared open when what it leaves out is compact. The line becomes a circle, the plane becomes a sphere, and the construction is the standard way of putting a boundary on a space that had none.

Draw the whole plane on a sheet and then wrap the sheet round a ball, pulling every direction of escape towards one point at the top. Everything that ran off to infinity now arrives somewhere.

sets diagram: one-point-added-at-the-edge

The new space is $X$ with one extra point. A set around that point is open exactly when what it omits is a compact closed subset of $X$ — so the open sets round infinity are the complements of the compact parts, and a cover of the whole space always has a member catching everything outside one compact set.

Adding a point at infinity

Let $X$ be a space and let $X^+=X\cup\{\infty\}$ with one new point. Declare $U\subseteq X^+$ open when either is an open subset of , or and is a compact closed subset of . This is a topology, is a subspace of it, and is compact.

U\ \text{open}\iff U\in\tau_X\ \text{or}\ X^+\setminus U\ \text{compact closed}

What it builds

$\mathbb{R}^+\cong S^1$ and $(\mathbb{R}^2)^+\cong S^2$ : the second is stereographic projection, which sends the sphere without its north pole onto the plane and sends the pole to infinity.

The one-point compactification is compact

X^+=X\cup\{\infty\}\ \text{is compact}

Take an open cover of the new space. Some member of it contains the added point, and by the definition of the topology what that member leaves out is a compact subset of the original space. The rest of the cover covers that compact subset, so finitely many of them already do. Put those finitely many together with the one member containing the added point: every point of the new space is either in that member or in the compact part, so the finite family covers everything. A cover has been thinned, and the cover was arbitrary.

Proof steps

X^+=\textstyle\bigcup_i U_i,\ U_i\ \text{open}

Take any open cover of the new space.

\infty\in U_{i_0}

Some member catches the added point.

K=X^+\setminus U_{i_0}\ \text{compact}

By the definition of the topology, what that member omits is compact.

K\subseteq U_{i_1}\cup\cdots\cup U_{i_n}

The rest of the cover covers that compact set, so finitely many suffice.

X^+=U_{i_0}\cup U_{i_1}\cup\cdots\cup U_{i_n}

Those finitely many together with the first cover everything.

Applications

Practice

Open Around Infinity

A set containing the new point is open when what it leaves out is compact and closed.

\infty\in U,\ X^+\setminus U\ \text{compact closed}\implies U\ \text{open}

Try it

Which sets containing $\infty$ are open in $X^+$ ?