Closing an Embedded Image
A space can sit densely inside a compact Hausdorff space in many ways. Once such an embedding is given, the closure of its image is the smallest closed part of the ambient space that still contains the image. It provides a compactification even when the original ambient compact space contains points that are not needed. A further question is whether the embedded space occupies an open part of this compactification. The answer is governed precisely by local compactness.
An embedding \(e:X\to K\) is an injective continuous map that is a homeomorphism from \(X\) onto the subspace \(e(X)\) of \(K\). A subset is dense if its closure is the whole space. In this tutorial, a compactification of \(X\) means a compact Hausdorff space \(Y\) together with an embedding \(j:X\to Y\) whose image is dense in \(Y\). The topology on every subset mentioned below is its subspace topology unless otherwise stated.
The closure is taken in \(K\). Since \(Y\) is a subspace of \(K\), it is Hausdorff. It is also compact: \(Y\) is closed in the compact space \(K\), and a closed subset of a compact space is compact. Indeed, an open cover of \(Y\), together with the open set \(K\setminus Y\), covers \(K\); a finite subcover of \(K\) then yields a finite subcover of \(Y\). By the definition of closure, \(e(X)\) is dense in \(Y\). The map \(e\) remains an embedding when its codomain is restricted to \(Y\), because the subspace topology on \(e(X)\) is unchanged. These facts prove the construction works.
Proof. The set \(Y\) is closed in \(K\), so it is compact by the closed-subset property of compact spaces. Since \(K\) is Hausdorff, its subspace \(Y\) is Hausdorff: disjoint open neighborhoods in \(K\) of two distinct points restrict to disjoint open neighborhoods in \(Y\). The map \(e:X\to Y\) is injective and continuous, and it is a homeomorphism onto \(e(X)\) with the subspace topology, because that topology is the same whether computed in \(K\) or in \(Y\). Finally, \(Y=\overline{e(X)}\) means exactly that \(e(X)\) is dense in \(Y\). Therefore \(Y\) is a compactification of \(X\). \(\square\)
The construction needs no separate assumption that \(X\) is Hausdorff: an embedding into \(K\) already identifies \(X\) with a subspace of a Hausdorff space, so \(X\) is Hausdorff. Different embeddings can give different compactifications; taking the closure specifies the compactification associated with the embedding at hand.
When Is the Image Open?
For the criterion, use the following local-compactness convention: a space is locally compact if every point has an open neighborhood whose closure in that space is compact. The theorem below concerns openness in \(Y=\overline{e(X)}\), not necessarily in the original ambient space \(K\).
Proof. First suppose \(D\) is locally compact, and fix \(x\in D\). Choose an open set \(U\) in \(D\) with \(x\in U\) and with compact closure \(C=\overline{U}^{\,D}\) contained in \(D\). Since compact subsets of Hausdorff spaces are closed, \(C\) is closed in \(Y\). By the subspace topology there is an open set \(O\) in \(Y\) such that \(O\cap D=U\). We claim \(O\subseteq C\). For any \(y\in O\), every open neighborhood of \(y\) in \(Y\), intersected with \(O\), is an open neighborhood of \(y\). Since \(D\) is dense, that intersection meets \(D\), and the resulting point lies in \(O\cap D=U\). Thus \(y\in\overline{U}^{\,Y}\subseteq C\), where the last inclusion holds because \(C\) is closed in \(Y\) and contains \(U\). Consequently \(x\in O\subseteq C\subseteq D\), with \(O\) open in \(Y\). Each \(x\in D\) therefore has an open neighborhood in \(Y\) contained in \(D\), proving \(D\) is open.
Now suppose \(D\) is open in \(Y\), and fix \(x\in D\). We use the following shrinking fact for compact Hausdorff spaces: if \(V\) is open in \(Y\) and \(x\in V\), there is an open set \(W\) with \(x\in W\) and \(\overline{W}^{\,Y}\subseteq V\). To verify it, let \(F=Y\setminus V\), a closed and hence compact set. For each \(y\in F\), the Hausdorff property gives disjoint open sets \(U_y\) and \(W_y\) containing \(x\) and \(y\), respectively. Finitely many \(W_y\)'s cover \(F\). Intersect the corresponding finitely many \(U_y\)'s to obtain an open neighborhood \(W\) of \(x\), and let \(G\) be the union of those \(W_y\)'s. Then \(W\cap G=\varnothing\), and since \(G\) is open, no point of \(G\) belongs to \(\overline{W}^{\,Y}\). Thus \(\overline{W}^{\,Y}\subseteq Y\setminus G\subseteq V\). (If \(F=\varnothing\), take \(W=Y\).) Apply this fact with \(V=D\). The set \(W\) is an open neighborhood of \(x\) in \(D\), and its closure in \(D\) equals \(\overline{W}^{\,Y}\), because that closure is already contained in \(D\). This closure is closed in compact \(Y\), hence compact. So \(D\) is locally compact. \(\square\)
For an embedding \(e:X\to K\), take \(D=e(X)\) and \(Y=\overline{e(X)}\). Since \(e\) is a homeomorphism from \(X\) to \(D\), local compactness transfers between them. The theorem therefore gives the useful form of the criterion.
Worked Examples
Worked Example: The Open Interval Inside Its Closure
Embed \(X=(0,1)\), with its usual topology as a subspace of \(\mathbb{R}\), into \(K=[0,1]\) by inclusion. The space \(K\) has the subspace topology inherited from \(\mathbb{R}\); it is compact and Hausdorff. The image is \(D=(0,1)\). Every open neighborhood in \(K\) of \(0\) meets \(D\), since it contains points of \((0,1)\); every neighborhood of \(1\) does too. Points of \(D\) already belong to its closure. There are no other points of \(K\), so \(\overline{D}^{\,K}=[0,1]\). Hence the compactification is \([0,1]\), and its remainder consists of \(0\) and \(1\).
The image \(D\) is open in \([0,1]\): it is the intersection of \([0,1]\) with the open interval \((0,1)\) in \(\mathbb{R}\). The criterion predicts that \(X\) is locally compact. To check this directly, fix \(x\in(0,1)\) and set \(r=\min(x,1-x)/3>0\). The open interval \((x-r,x+r)\) is a neighborhood of \(x\) in \(X\), and its closure in \(X\) is \([x-r,x+r]\), a compact closed interval contained in \(X\). This verifies local compactness at every point.
Check: the closure calculation gives a compact Hausdorff space containing \(D\) densely, and \(D\) is open in that closure, in agreement with the criterion.
Worked Example: The Rationals Are Dense but Not Open
Let \(K=[0,1]\) with its usual topology and \(D=\mathbb{Q}\cap[0,1]\), with the subspace topology. The set \(D\) is dense in \(K\): every nonempty open interval in \([0,1]\) contains a rational number. Therefore \(\overline{D}^{\,K}=K\), and \(K\) is the compactification obtained from this inclusion.
The set \(D\) is not open in \(K\). For example, \(1/2\in D\), but every open neighborhood of \(1/2\) in \(K\) contains an interval around \(1/2\) and hence contains irrational points, so it is not contained in \(D\).
We can also verify directly that \(D\) is not locally compact at \(1/2\). Suppose \(C\subseteq D\) were a compact neighborhood of \(1/2\) in \(D\); that means \(C\) is compact and contains an open neighborhood \(U\) of \(1/2\) in \(D\). Some \(\delta>0\) satisfies \((1/2-\delta,1/2+\delta)\cap D\subseteq U\). As a compact subset of the Hausdorff space \([0,1]\), \(C\) must be closed in \([0,1]\). But \(C\) contains all rationals in that interval, whose closure in \([0,1]\) contains the whole interval, including irrational points. This contradicts \(C\subseteq D\). Thus no compact neighborhood exists at \(1/2\), as the criterion requires.
Check: \(D\) is dense, so the closure is all of \(K\); it is not open there, and the explicit failure of local compactness at \(1/2\) matches the theorem.
Worked Example: A Compact Image That Is Not Ambient-Open
Let \(X=[0,1]\) and \(K=[0,1]^2\), both with their usual Euclidean subspace topologies. Define \(e:X\to K\) by \(e(t)=(t,t)\). This is an embedding: it is continuous and injective, and its inverse on the diagonal image is the continuous first-coordinate projection. The image \(D=\{(t,t):0\le t\le1\}\) is compact, so it is closed in the Hausdorff space \(K\). Thus \(\overline{D}^{\,K}=D\), and the closure compactification adds no points.
The diagonal \(D\) is not open in \(K\). Every open neighborhood in the square of, for example, \((1/2,1/2)\) contains points off the diagonal. But \(D\) is open in its closure \(Y=D\): every space is open in itself. Correspondingly \(X\) is locally compact; it is compact, and \(X\) itself is an open neighborhood of each of its points with compact closure.
Check: “open in the closure” means open in the subspace \(Y=\overline{D}^{\,K}\). It does not mean open in the original ambient space \(K\).
Why the Distinction Matters
The closure construction gives a compactification directly from an embedding, while the openness criterion tells us whether the original space sits inside that compactification as an open subspace. This is useful when comparing compactifications: a locally compact space has an open image in every compact Hausdorff closure arising from an embedding, whereas a failure of local compactness rules out openness in any such closure.
Common pitfall. A tempting but incorrect claim is that an embedded image is open in its compactification only if it is open in the original ambient compact space. In the diagonal example, the image is not open in \(K=[0,1]^2\), but its closure is the diagonal itself, and the image is open in that closure. The criterion concerns the compactification \(Y=\overline{e(X)}\); replacing \(Y\) by a larger ambient \(K\) changes the question.
There is a second useful check on the construction: density does not mean that the image is the whole closure. In the interval example the two endpoints are in the closure but not in the image; in the diagonal example the image is already closed and no points are added. In both cases the closure itself, rather than the original ambient space automatically, is the compact Hausdorff space used.
Check Your Understanding
Use the closure construction and the openness criterion to organize your reasoning.
- For an embedding \(e:X\to K\) with \(K\) compact Hausdorff, explain why \(\overline{e(X)}\) is both compact and Hausdorff.
- Let \(D=(0,1]\subseteq[0,1]\). Find its closure in \([0,1]\), its remainder, and decide whether it is open in that closure.
- Why does the density of \(\mathbb{Q}\cap[0,1]\) in \([0,1]\) not imply that it is open there?
- State the local-compactness criterion for a dense subspace of a compact Hausdorff space.
- In the diagonal example, distinguish whether \(D\) is open in \(K\) from whether it is open in \(\overline{D}^{\,K}\).