The Added Point and Its Neighborhoods
Adding one point to an infinite discrete space changes its behavior at infinity without changing the topology around any of its original points. The key is to make a neighborhood of the new point contain all but finitely many original points. This gives a concrete setting in which compactness, convergence, and isolation can each be checked directly from definitions.
Let \(X\) be an infinite set with the discrete topology \(\mathcal{T}=\mathcal{P}(X)\), so every subset of \(X\) is open. Write \(X^*=X\cup\{\infty\}\), where \(\infty\notin X\). A subset of a topological space is compact if every open cover of it has a finite subcover.
In the general construction of a one-point compactification, neighborhoods of the added point are complements of compact subsets of the original space. Here the compact subsets of an infinite discrete space are exactly its finite subsets: a finite set is compact, while the cover by its open singletons has no finite subcover when the set is infinite. Thus the displayed definition specializes that construction to this case.
These sets do form a topology. The empty set and \(X^*\) are included. Any union that contains no \(\infty\) is a subset of \(X\), hence open. If a union contains \(\infty\), at least one set in it is \(X^*\setminus F\) for a finite \(F\subseteq X\). The complement of the union is a subset of \(F\), so is finite, and the union is open. For finite intersections, an intersection of subsets of \(X\) is a subset of \(X\); intersecting a subset of \(X\) with a set containing \(\infty\) again gives a subset of \(X\); and \( (X^*\setminus F)\cap(X^*\setminus G)=X^*\setminus(F\cup G)\), where \(F\cup G\) is finite. These cases verify the topology axioms.
Worked Example: Neighborhoods in \(\mathbb{N}^*\)
Take \(X=\mathbb{N}=\{1,2,3,\dots\}\) with the discrete topology. The set \(U=\mathbb{N}^*\setminus\{2,5\}\) is an open neighborhood of \(\infty\); it contains \(\infty\) and every positive integer other than \(2\) and \(5\). The set \(V=\{\infty\}\cup\{n\in\mathbb{N}:n\geq 4\}\) is also open, because its complement in \(\mathbb{N}^*\) is \(\{1,2,3\}\), a finite set.
Not every neighborhood of \(\infty\) is a tail of the natural numbers: \(U\) omits \(2\) and \(5\) but includes \(3\) and every integer greater than \(5\). Conversely, every neighborhood \(W\) of \(\infty\) omits only finitely many natural numbers. If \(F=\mathbb{N}\setminus W\), then \(F\) is finite and \(W=\mathbb{N}^*\setminus F\). This verifies the neighborhood description in both directions.
Compactness, Separation, and Isolated Points
Proof. Let an open cover of \(X^*\) be given. Some member \(U\) of the cover contains \(\infty\). By the definition of the topology, \(X^*\setminus U\) is a finite subset of \(X\). For each point of this finite remainder, choose a member of the cover that contains it. Together with \(U\), these finitely many members cover \(X^*\). Therefore \(X^*\) is compact.
For Hausdorffness, take distinct points \(x,y\in X\). The disjoint open sets \(\{x\}\) and \(\{y\}\) contain them. For a point \(x\in X\) and \(\infty\), the sets \(\{x\}\) and \(X^*\setminus\{x\}\) are disjoint open neighborhoods of those points. These cases separate every pair of distinct points, so \(X^*\) is Hausdorff.
Every subset of \(X\) is open in \(X^*\), so the subspace topology on \(X\) is discrete. To see that \(X\) is dense, use the neighborhood test for closure: every open neighborhood of a point of \(X\) meets \(X\), since it contains that point; and every open neighborhood \(X^*\setminus F\) of \(\infty\) meets \(X\), since \(X\setminus F\ne\varnothing\) when \(X\) is infinite and \(F\) is finite. Hence every point of \(X^*\) lies in \(\overline{X}\).
For each \(x\in X\), the open singleton \(\{x\}\) shows that \(x\) is isolated. The point \(\infty\) is not isolated: every open neighborhood of it omits only finitely many points of the infinite set \(X\), and therefore contains points of \(X\) as well as \(\infty\). \(\square\)
Worked Example: Isolated Points and the Point at Infinity
In \(\mathbb{N}^*\), consider \(x=4\). The set \(\{4\}\) is open because every subset of \(\mathbb{N}\) is open in \(\mathbb{N}^*\), so \(4\) is isolated. The set \(\{\infty\}\) is not open: if it were a neighborhood of \(\infty\), its complement \(\mathbb{N}\) would have to be finite, but \(\mathbb{N}\) is infinite.
More generally, if \(F\subseteq\mathbb{N}\) is finite, the neighborhood \(\mathbb{N}^*\setminus F\) contains infinitely many natural numbers. For example, when \(F=\{1,2,3\}\), this neighborhood contains \(4,5,6,\dots\) as well as \(\infty\). Thus no choice of finite exclusions produces the singleton \(\{\infty\}\).
Convergent Sequences
A sequence \((x_n)\) in a topological space converges to \(x\) if, for every open neighborhood \(U\) of \(x\), there is an index \(N\) such that \(x_n\in U\) for every \(n\geq N\). At a point of \(X\), the open singleton makes this test especially restrictive. At \(\infty\), the test is governed by finite sets.
Proof. Suppose \(x_n\to x\in X\). The singleton \(\{x\}\) is an open neighborhood of \(x\), so the definition of convergence requires \(x_n\in\{x\}\), and hence \(x_n=x\), for all sufficiently large \(n\). Conversely, if \(x_n=x\) for all sufficiently large \(n\), every open neighborhood of \(x\) contains \(x\), so it contains all sufficiently late terms. Thus \(x_n\to x\).
Now suppose each point of \(X\) occurs only finitely often. Let \(U\) be any open neighborhood of \(\infty\), so \(U=X^*\setminus F\) for some finite \(F\subseteq X\). Each point of \(F\) occurs at only finitely many indices. The union of these finitely many finite sets of indices is finite, so there is an index after which no term lies in \(F\). All terms from then on lie in \(U\), proving \(x_n\to\infty\).
Conversely, suppose \(x\in X\) occurs infinitely often. The open neighborhood \(X^*\setminus\{x\}\) of \(\infty\) excludes \(x\). Since infinitely many terms equal \(x\), the sequence is not eventually in that neighborhood, so it does not converge to \(\infty\). This proves both directions of the criterion. \(\square\)
Worked Example: Two Sequences in \(\mathbb{N}^*\)
Let \(a_n=n\). Each fixed natural number \(m\) occurs in this sequence at exactly one index, namely \(n=m\). Thus every \(m\in\mathbb{N}\) occurs only finitely often, and the theorem gives \(a_n\to\infty\). Directly, any neighborhood of \(\infty\) omits a finite set \(F\subseteq\mathbb{N}\); after the largest element of \(F\), all terms \(a_n=n\) lie outside \(F\). If \(F\) is empty, all terms already lie in the neighborhood.
Now let \(b_n\) alternate between \(1\) and \(2\): \(b_n=1\) for odd \(n\), and \(b_n=2\) for even \(n\). The point \(1\) occurs at infinitely many odd indices. The neighborhood \(\mathbb{N}^*\setminus\{1\}\) therefore fails to contain all sufficiently late terms, so \(b_n\) does not converge to \(\infty\). The sequence also does not converge to a point \(m\in\mathbb{N}\), since convergence to \(m\) would require it to be eventually equal to \(m\), whereas it continues to alternate between two distinct values.
Check: \(a_n\) converges to no natural number because for any fixed \(m\), the terms \(a_n\) with \(n>m\) are not \(m\). The criterion above says \(a_n\to\infty\), matching the direct neighborhood check.
Exactly Which Subsets Are Compact?
Proof. First suppose \(C\subseteq X\) is finite. Given any open cover of \(C\), choose one cover member for each point of \(C\). This gives a finite subcover; if \(C=\varnothing\), the empty subcollection is already a finite subcover. Next suppose \(\infty\in C\), and take any open cover of \(C\). Choose a cover member \(U\) containing \(\infty\). The complement \(X^*\setminus U\) is finite, so only finitely many points of \(C\) remain to be covered. Choose one cover member for each of those points. Together with \(U\), these form a finite subcover. Thus both types of set in the statement are compact.
For the converse, suppose \(C\subseteq X\) is infinite. The collection \(\{\{x\}:x\in C\}\) is an open cover of \(C\), since each singleton is open in \(X^*\). A finite subcollection covers only finitely many points, so it cannot cover the infinite set \(C\). Therefore \(C\) is not compact. The cases are exhaustive: a set either contains \(\infty\), or is a subset of \(X\); in the latter case it is compact exactly when it is finite. \(\square\)
Worked Example: Compact and Noncompact Subsets of \(\mathbb{N}^*\)
Let \(E=\{2,4,6,\dots\}\) and \(C=\{\infty\}\cup E\). The set \(C\) contains \(\infty\), so the theorem says it is compact. To check the open-cover argument directly, take any open cover of \(C\), and choose a member \(U\) containing \(\infty\). Its complement in \(\mathbb{N}^*\) is finite, so \(E\setminus U\) is finite. Choose cover members for those remaining even integers. Along with \(U\), these finitely many sets cover \(C\).
In contrast, let \(O=\{1,3,5,\dots\}\), which does not contain \(\infty\) and is infinite. Its cover by open singletons \(\{\{m\}:m\in O\}\) has no finite subcover: any finite selection covers only finitely many odd integers, while \(O\) contains infinitely many. Therefore \(O\) is not compact. A finite subset such as \(\{2,5\}\) is compact, since it has only two points and any open cover has a finite subcover.
A Common Pitfall: Repeated Terms and Compactness
A sequence converging to \(\infty\) need not eventually have distinct terms. For instance, define \(c_{2k-1}=c_{2k}=k\) for \(k\in\mathbb{N}\). Every fixed \(m\in\mathbb{N}\) occurs exactly twice, at indices \(2m-1\) and \(2m\), so the sequence converges to \(\infty\) by the criterion. Requiring all sufficiently late terms to be distinct would incorrectly reject this convergent sequence. The correct condition is that each fixed point occurs only finitely many times.
There is a related compactness error: an infinite subset of the original discrete space is not compact, but an infinite subset of \(X^*\) that includes \(\infty\) is compact. The singleton-cover argument proves noncompactness only for subsets contained in \(X\): a cover by singletons of \(C\subseteq X\) need not be an open cover of a set containing \(\infty\), because it does not cover \(\infty\). For sets containing \(\infty\), the neighborhood of \(\infty\) in any open cover leaves only finitely many points uncovered. This is precisely why the two cases in the compact-subset theorem differ.
Check Your Understanding
Use the neighborhood and compactness criteria proved above to answer these questions.
- In \(\mathbb{N}^*\), describe an open neighborhood of \(\infty\) that omits \(3\) and \(7\), and state why it is open.
- Does the sequence \(1,1,2,2,3,3,\dots\) converge to \(\infty\)? Justify your answer using the occurrence condition.
- Which subsets of \(X^*\) are compact if \(X\) is infinite and discrete? Include the empty set and explain its status.
- Why is an infinite subset \(A\subseteq X\) not compact, even though \(X^*\) itself is compact?
- Is \(\infty\) isolated in \(X^*\)? Give the neighborhood argument that decides the question.