Compactifications as Dense Copies
A compactification places a space inside a compact Hausdorff space without changing the topology of the original space. The copy must be dense: every point added to make the larger space compact is approached by points of the original space. Compactifications can be compared in two useful ways. Their remainders record which points were added, while maps that agree on the embedded space record whether the surrounding spaces fit together.
Throughout, a compact space means a space in which every open cover has a finite subcover. An embedding \(e:X\to K\) is a continuous injective map that is a homeomorphism from \(X\) onto \(e(X)\) with its subspace topology. The closure \(\overline{A}\) of a subset \(A\) is the smallest closed set containing \(A\); a subset is dense when its closure is the whole space.
This definition does not assert that every topological space has a compactification. It says what a compactification is when one exists. Requiring an embedding, rather than merely an injective map, ensures that the copy of \(X\) has exactly its original topology.
When Two Compactifications Are Equivalent
Suppose \((K,e)\) and \((L,f)\) are compactifications of the same space \(X\). They are equivalent if there is a homeomorphism \(h:K\to L\) satisfying \(h\circ e=f\). Thus the homeomorphism must match the two copies of every point of \(X\); a homeomorphism that sends the embedded copy somewhere else does not establish this equivalence.
Proof. Let \(h:K\to L\) be a homeomorphism with \(h\circ e=f\). If \(z\in K\setminus e(X)\) and \(h(z)=f(x)\) for some \(x\in X\), then \(h(z)=f(x)=h(e(x))\). Injectivity of \(h\) gives \(z=e(x)\), a contradiction. Hence \(h\) sends \(K\setminus e(X)\) into \(L\setminus f(X)\). Applying the same argument to \(h^{-1}\) shows that it maps the first remainder onto the second. The restriction of a homeomorphism to these subspaces is a homeomorphism. \(\square\)
Remainders are therefore an invariant of equivalence, but the remainder alone need not determine how the embedded points approach it. The next example makes this distinction explicit.
Worked Example: The Rationals and Their Remainder
Let \(X=\mathbb{Q}\cap[0,1]\) with the subspace topology inherited from \(\mathbb{R}\), and let \(K=[0,1]\) with its usual topology. The inclusion \(e:X\to K\) is an embedding: the topology on \(X\) is, by definition, the subspace topology from \(K\). The rational numbers are dense in \(\mathbb{R}\), so every relative neighborhood in \([0,1]\) of each point of \([0,1]\) meets \(X\). Therefore \(\overline{e(X)}=K\). Since \([0,1]\) is compact and Hausdorff, this is a compactification.
The remainder is \(K\setminus e(X)=[0,1]\setminus\mathbb{Q}\), the irrational numbers in \([0,1]\). To verify that it is dense, take any nonempty relatively open set \(U\subseteq[0,1]\). It contains a nonempty interval of positive length lying in \([0,1]\), possibly with one endpoint equal to \(0\) or \(1\). Choose a rational number \(r\) in the interior of that interval. Since the interval is open around \(r\), there is a positive distance from \(r\) to its endpoints. Choose \(n\in\mathbb{N}\) large enough that \(\sqrt{2}/n\) is smaller than that distance. Then \(r+\sqrt{2}/n\) lies in the interval and is irrational: if it were rational, subtracting the rational \(r\) and multiplying by \(n\) would make \(\sqrt{2}\) rational. Thus every such \(U\) meets the remainder.
It follows that the closure of the remainder is all of \([0,1]\), so the remainder is not closed. In particular, density of the rationals does not by itself prove that the irrationals are dense; the argument above verifies the density needed for this conclusion.
Maps That Preserve the Embedded Space
For compactifications written as \((K,e)\) and \((L,f)\), a continuous map \(F:K\to L\) preserves the embedded space if \(F\circ e=f\). If the copies of \(X\) are already identified as the same subset, this condition is written \(F(x)=x\) for every \(x\in X\). Such a map is unique if it exists, but it need not be a homeomorphism.
Proof. We first show that \(E=\{z\in K:F(z)=G(z)\}\) is closed. If \(z\notin E\), then \(F(z)\ne G(z)\). The Hausdorff property gives disjoint open neighborhoods \(V\) of \(F(z)\) and \(W\) of \(G(z)\). The set \(F^{-1}(V)\cap G^{-1}(W)\) is an open neighborhood of \(z\). For every \(u\) in this neighborhood, \(F(u)\in V\) and \(G(u)\in W\), so \(F(u)\ne G(u)\). Thus \(K\setminus E\) is open and \(E\) is closed. Since \(F\) and \(G\) agree on the dense set \(A\), \(E\) contains \(A\). A closed set containing a dense set is all of \(K\), so \(E=K\) and \(F=G\). \(\square\)
Proof. The image \(F(K)\) is compact: for any open cover of \(F(K)\), its inverse images under \(F\) cover \(K\); a finite subcover of \(K\) gives a finite subcover of \(F(K)\). A compact subset of a Hausdorff space is closed. To recall why, if \(y\) lies outside a compact set \(C\), separate \(y\) from each \(c\in C\) by disjoint open neighborhoods. The neighborhoods of points \(c\) cover \(C\), so finitely many suffice. Intersect the corresponding finitely many neighborhoods of \(y\); this gives a neighborhood of \(y\) disjoint from \(C\). Hence the complement of \(C\) is open. Applying this fact to \(F(K)\subseteq L\), the image is closed. It contains \(B=F(A)\), which is dense in \(L\), so \(F(K)=L\). \(\square\)
Consequently, a continuous map between compactifications that preserves the embedded space is surjective: its image contains the dense embedded copy in the target. The uniqueness proposition shows there is at most one such continuous map. Neither statement says that this map is injective.
Worked Example: A Continuous Map That Is Not an Equivalence
Take \(X=(0,1)\) with its usual topology. One compactification is \(K=[0,1]\), embedded by inclusion \(i(t)=t\). Another is \(L=S^1=\{(u,v)\in\mathbb{R}^2:u^2+v^2=1\}\), with its subspace topology, embedded by \(j(t)=(\cos(2\pi t),\sin(2\pi t))\). The map \(j\) is injective on \((0,1)\) and is a homeomorphism onto \(S^1\setminus\{(1,0)\}\); this image is dense in \(S^1\). The interval and the circle are compact Hausdorff, so both pairs are compactifications.
Define \(F:K\to L\) by \(F(t)=(\cos(2\pi t),\sin(2\pi t))\). This map is continuous and satisfies \(F(i(t))=j(t)\) for every \(t\in X\). It is surjective, since the full interval of angles from \(0\) to \(2\pi\) traces the circle. However, \(F(0)=F(1)=(1,0)\), so it is not injective and cannot be a homeomorphism. The remainders also differ: \(K\setminus i(X)=\{0,1\}\), while \(L\setminus j(X)=\{(1,0)\}\). This example shows why a continuous map preserving the embedded space does not in general establish equivalence.
Check: the endpoints are distinct in \(K\), while they have the same image in \(L\). Thus the failure of injectivity is exactly at the two points added to the interval.
The Remainder Does Not Record Every Attachment
Worked Example: Two Compactifications with the Same Remainder
Let \(X=\mathbb{N}\) with the discrete topology. Partition \(\mathbb{N}\) into two infinite sets \(P\) and \(Q\), and form a space \(K_{P,Q}\) by adjoining two points \(p,q\). Give it the topology of the disjoint union of \(\{p\}\cup P\) and \(\{q\}\cup Q\), where each component is a convergent sequence: its natural-number points are isolated, and every neighborhood of its added point contains all but finitely many points of its corresponding set. For example, a neighborhood base at \(p\) consists of \(\{p\}\cup(P\setminus E)\) for finite \(E\subseteq P\).
Each component is compact. Given an open cover of \(\{p\}\cup P\), choose a member containing \(p\); it covers all but finitely many points of \(P\), and finitely many additional cover members suffice for the remaining points. The same argument applies to \(\{q\}\cup Q\), so their finite disjoint union is compact. It is Hausdorff: distinct isolated points have disjoint singleton neighborhoods; an isolated point can be separated from a component’s added point by a neighborhood of that point omitting the isolated point; and \(p,q\) lie in disjoint components. The subspace \(X=P\cup Q\) is discrete, and it is dense because each neighborhood of \(p\) or \(q\) contains points of \(X\). Thus \(K_{P,Q}\) is a compactification of \(X\), with remainder \(\{p,q\}\).
Now take \(K_1=K_{P,Q}\) with \(P\) the even positive integers and \(Q\) the odd positive integers. Take \(K_2=K_{P',Q'}\), where \(P'\) is the set of positive multiples of \(3\) and \(Q'=\mathbb{N}\setminus P'\). Both remainders are discrete two-point spaces, hence homeomorphic. But there is no homeomorphism \(H:K_1\to K_2\) fixing every \(n\in\mathbb{N}\). In \(K_1\), the sequence \(2,4,6,\ldots\) converges to \(p\), since all its terms lie in \(P\) and every neighborhood of \(p\) omits only finitely many points of \(P\). A homeomorphism fixing \(\mathbb{N}\) cannot send \(p\) to a point of \(\mathbb{N}\), because it already fixes every natural number and is injective. It must send \(p\) to one of the two remainder points of \(K_2\).
But the even sequence does not converge to either remainder point of \(K_2\). Infinitely many even terms are multiples of \(3\), and infinitely many are not. Every neighborhood of the added point corresponding to \(P'\) can exclude all points in \(Q'\), so it misses infinitely many even terms. Every neighborhood of the added point corresponding to \(Q'\) can exclude all points in \(P'\), so it also misses infinitely many even terms. This contradicts preservation of convergence by a homeomorphism. The two remainders are homeomorphic, but the compactifications are not equivalent by a homeomorphism fixing \(X\).
Common pitfall. Knowing that two compactifications have homeomorphic remainders does not prove that they are equivalent. Equivalence must respect how the original space sits in the compact space. Conversely, a continuous map that respects the embedded space is unique when the target is Hausdorff and is surjective when both spaces are compact Hausdorff, but it can still identify distinct remainder points. The relevant tests are therefore different: compare remainders for a necessary invariant, and require a homeomorphism preserving the embedding for equivalence.
Check Your Understanding
Use the definitions and comparison results above to answer the following.
- Why does an embedding, rather than an arbitrary injective map, appear in the definition of compactification?
- Prove that if \(X\) is compact Hausdorff, then any compactification \((K,e)\) of \(X\) has \(e(X)=K\).
- Can two equivalent compactifications have remainders of different cardinalities? Justify your answer.
- Why does a continuous map preserving the embedded space have to be surjective when the target is compact Hausdorff and the embedded copy is dense?
- In the two sequence compactifications, identify an explicit subsequence that prevents a homeomorphism fixing \(\mathbb{N}\).