What Is the One-Point Compactification of a Discrete Space?

For an infinite discrete space $X$, the one-point compactification adds a point, usually written $\infty$, that represents going off to infinity in $X$. The original points remain isolated. The new point’s neighborhoods are the sets containing all but finitely many points of $X$.

This construction works because compact subsets of a discrete space are finite. Here is the topology and why it gives a compact Hausdorff space.

The topology on the added point

Let $X^*=X\cup\{\infty\}$, where $\infty$ is not in $X$. Define the open sets as follows:

• Any subset of $X$ is open in $X^*$.
• A set containing $\infty$ is open exactly when its complement in $X$ is finite.

Equivalently, the neighborhoods of $\infty$ have the form

$$\{\infty\}\cup (X\setminus F),$$

where $F$ is a finite subset of $X$. Since $X$ is discrete, every subset of $X$ is already open. The topology induced on $X$ by $X^*$ is therefore its original discrete topology.

The word “cofinite” describes a subset whose complement is finite. Thus the neighborhoods of the added point are cofinite in $X^*$, with the point $\infty$ itself included.

Why compact subsets of a discrete space are finite

If $K\subseteq X$ is infinite, the collection of singleton sets

$$\bigl\{\{x\}:x\in K\bigr\}$$

is an open cover of $K$. No finite subcollection covers $K$, so $K$ is not compact. Conversely, every finite subset is compact: from any open cover, choose one member covering each of its finitely many points.

Therefore, in the one-point compactification of a discrete space, the neighborhoods of $\infty$ are exactly the sets whose complement in $X$ is compact—here, finite.

Why the result is compact and Hausdorff

To check compactness, take any open cover of $X^*$. Some member of the cover contains $\infty$. Its complement is a finite subset of $X$, so finitely many more cover members suffice to cover the points left over. The whole space thus has a finite subcover.

To check the Hausdorff property, consider two distinct points. If both are in $X$, their singleton sets are disjoint open neighborhoods. If one is $x\in X$ and the other is $\infty$, use $\{x\}$ and $X^*\setminus\{x\}$. Both are open and disjoint: the second contains $\infty$ and omits only one point of $X$.

The original space $X$ is dense in $X^*$ because every neighborhood of $\infty$ contains points of $X$ when $X$ is infinite. Thus $X^*$ is a compact Hausdorff space in which $X$ sits densely as an open subspace.

Example: the integers

Give $\mathbb{Z}$ the discrete topology and add $\infty$. A typical neighborhood of $\infty$ is

$$\{\infty\}\cup\bigl(\mathbb{Z}\setminus\{-2,0,5\}\bigr).$$

It contains every integer except a finite set. The resulting space is compact: any open cover member containing $\infty$ leaves only finitely many integers to cover. The integers are still isolated, but $\infty$ is not: every open neighborhood of it contains infinitely many integers.

More generally, if $X$ is countably infinite, enumerate it as $x_1,x_2,\ldots$. The neighborhoods of $\infty$ include tails such as $\{\infty,x_n,x_{n+1},\ldots\}$. In fact, this compactification is homeomorphic to a convergent sequence together with its limit point. For an uncountable discrete $X$, the same cofinite-neighborhood description applies, but the compactification is not metrizable: in a metric space, a countable neighborhood base at $\infty$ would imply that $X$ is countable.

Common mistakes

Treating every subset as compact. Discrete means every subset is open, not that every subset is compact. An infinite subset is not compact, as its cover by singletons shows.

Using complements of arbitrary compact sets without checking what they are. In a general locally compact Hausdorff space, open neighborhoods of the added point have compact complement in the original space. For a discrete space, those compact sets are precisely the finite ones.

Applying the construction to a finite discrete space without qualification. The usual one-point compactification is defined for a noncompact locally compact Hausdorff space. A finite discrete space is already compact, so it has no one-point compactification under that convention. If you simply adjoin a point and use the cofinite rule, the new point is isolated and the original space is not dense; that is not a compactification of the original space in the usual sense.