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A Tychonoff space is Čech-complete if it is a G-delta set in a Hausdorff compactification. The G-delta compactification invariance makes this independent of the chosen compactification, and equivalent to being a G-delta set in its Stone-Čech compactification.

If $\pi :\beta X\to K$ is the canonical map from the Stone-Čech compactification onto a Hausdorff compactification, its fibers over the original points are singletons. If $\pi (z)=x\in X$ and $z\ne x$, choose an open neighborhood $U$ of $x$ in $\beta X$ whose closure excludes $z$. Choose open $V\subseteq K$ with $x\in V$ and $V\cap X\subseteq U$. The open neighborhood ${\pi }^{-1}V$ of $z$ has dense trace $V\cap X$, placing $z$ in $\overline{U}$, a contradiction. This controls when images of closed remainder sets stay in the remainder.

For the comparison map $\pi :\beta X\to K$, the fibers of compactification comparison maps identity ${\pi }^{-1}(X)=X$ pulls a G-delta set presentation of $X$ back from $K$. Conversely if $\beta X\setminus X={\bigcup }_n{F}_n$ with $F_n$ closed compact, then $K\setminus X={\bigcup }_n\pi (F_n)$. Each image is closed and misses $X$. Thus $X$ is a G-delta set in one Hausdorff compactification exactly when it is in every one.

A Hausdorff compactification of $X$ is a pair $(h,Y)$ in which $Y$ is compact Hausdorff and $h:X\to Y$ is a topological embedding with dense image.

For a locally compact Hausdorff noncompact topological space, form $X^{\ast }=X\cup \{\infty \}$ with one new point. Keep all open subsets of $X$, and take the neighborhoods of infinity to contain setsCompact subsets of a Hausdorff space are closed. Finite unions of compact sets are compact, so finite intersections of these basic neighborhoods are again of the same form; their intersections with ordinary open sets are open in $X$. This specifies a topology restricting to the original one on $X$.

It is compact: from an open cover choose a member containing infinity. Its complement is contained in some compact $K$, and finitely many more members cover that compact remainder. It is Hausdorff: two points of $X$ are separated as before; to separate $x\in X$ from infinity, choose an open neighborhood $U$ of $x$ lying inside a compact neighborhood $K$. The open set $U$ and the infinity neighborhood $X^{\ast }\setminus K$ are disjoint.

The inclusion of $X$ is an embedding. Every basic neighborhood of infinity meets $X$, because otherwise its compact complement $K$ would equal all of $X$, contrary to noncompactness. Thus $X$ is dense in $X^{\ast }$. This proves that $X^{\ast }$ is the one-point compactification.

First prove the suggested density fact. If $Z$ is dense in $Y$, $V$ is open and $y\in V$, then every open neighborhood $W$ of $y$ has $W\cap V$ nonempty open. It meets $Z$, so $W$ meets $V\cap Z$. HenceIdentify $X$ with its dense embedded image in the hausdorff compactification. For $x\in X$, local compactness supplies an open neighborhood $U$ in $X$ and a compact set $K\subseteq X$ with $x\in U\subseteq K$. Choose an open set $V$ in $Y$ with $V\cap X=U$. The inclusion into $Y$ is continuous, so $K$ is compact in $Y$ and therefore closed, since $Y$ is Hausdorff. The density fact now yieldsThus every point of the embedded $X$ has a $Y$-open neighborhood contained in $X$. Consequently

A Hausdorff compactification of $X$ is an embedding $h:X\to Y$ into a compact Hausdorff space with dense image. The complement $Y\setminus h[X]$ is the remainder; a one-point or two-point compactification has a remainder of that cardinality.

For one point, take the unit circle andThe angle runs over $(-\pi ,\pi )$, so this is a homeomorphism onto the circle with $(-1,0)$ removed. That image is dense, and the circle is compact Hausdorff. For two points, take $Y=[-\pi /2,\pi /2]$ and $h_2(x)=\arctan x$. This is a homeomorphism onto its dense open interval, leaving exactly the two endpoints. Therefore both required compactifications exist; in the first the positive and negative ends meet, whereas in the second they remain distinct.

Write $T_n=h(\{x\in \mathbb{R}:|x|>n\})$, making explicit that these are subsets of $Y$. If $p\notin h[\mathbb{R}]$ and $U$ is an open neighbourhood of $p$, then $h([-n,n])$ is compact and hence closed in $Y$. The open set $U\setminus h([-n,n])$ still contains $p$ and meets the dense image of $h$. Such an intersection point belongs to $T_n$, so $p\in \overline{T_n}$ for every $n$.

Conversely, if $p=h(x)$ choose an integer $n>|x|$. Since $h$ is an embedding, there is an open neighbourhood $U$ of $p$ in $Y$ whose intersection with $h[\mathbb{R}]$ is $h((-n,n))$. It misses $T_n$, so $p\notin \overline{T_n}$. HenceFor the two ends of a real-line compactification, let $A_n^{+}=\overline{h((n,\infty ))}$ and $A_n^{-}=\overline{h((-\infty ,-n))}$. Each is nonempty compact and connected, being the closure of a connected image. The two nested families therefore have nonempty connected intersections $C^{+}$ and $C^{-}$. Since closure commutes with finite unions, $\overline{T_n}=A_n^{+}\cup A_n^{-}$. MoreoverFor if a point fails both intersections, it fails one positive member and one negative member, hence both members at the larger index by nesting.

If the remainder were a finite three-point set, its Hausdorff subspace topology would be discrete. Each nonempty connected subset $C^{\pm }$ would then be a singleton, so their union would have at most two points. This contradiction proves there is no three-point Hausdorff compactification of the real line. The argument does not assert that every possible remainder is finite.

For $y\in Y$, every neighborhood of $y$ in the subspace is of the form $V\cap Y$ with $V$ an ambient open neighborhood. Since $A\subseteq Y$, this meets $A$ exactly when $V$ meets $A$. Thus

A Hausdorff compactification is a topological embedding $h:X\to K$ with dense image in a compact Hausdorff space $K$. Define the domination preorder byIt is reflexive by the identity map and transitive by composition. Any comparison map is onto: its image is compact and hence closed in $K$, and contains the dense set $h(X)$. Two compactifications are equivalent if a homeomorphism $u:K\to K^{\prime }$ satisfies $u\circ h=h^{\prime }$.

Such a homeomorphism and its inverse give domination both ways. Conversely, if comparison maps exist in both directions, each composite agrees with the identity on the dense embedded copy of $X$. Two continuous maps to a Hausdorff space agreeing on a dense set agree everywhere. The comparison maps are therefore inverse homeomorphisms. Thus equivalence is exactly mutual domination.

The Stone-Čech extension property says that every continuous map from $X$ into any compact Hausdorff space extends continuously to the compactification. The extension is unique by density. For two compactifications with this property, extend each embedding through the other compactification. This gives comparison maps both ways, hence equivalence by the preceding argument.

For the perfect-map characterization by Stone-Čech remainders, the universal property first gives a unique continuous $F:\beta X\to \beta Y$ extending the composite $X\stackrel{f}{\to }Y\hookrightarrow \beta Y$. The required remainder condition is equivalent toOne inclusion is already forced by the extension property; the other forbids any point of the remainder mapping into $Y$.

Suppose first that $f$ is a perfect map. Fix $y\in Y$ and let $K=f^{-1}(y)$, a compact subset of $X$ and hence a closed subset of $\beta X$. If $p\in F^{-1}(y)\setminus K$, normality of $\beta X$ supplies an open $U$ with $K\subseteq U$ and $p\notin \overline{U}$. Because $f$ is closed,is an open neighborhood of $y$ and $f^{-1}(V)\subseteq U\cap X$. Choose an open $V^{\prime }$ of $\beta Y$ with $V^{\prime }\cap Y=V$. The open set $F^{-1}(V^{\prime })$ contains $p$, and its intersection with the dense subset $X$ lies in $U$. Every point of that open set is consequently in $\overline{U}$, contradicting the choice of $p$. Therefore $F^{-1}(y)=K\subseteq X$ for every $y\in Y$, proving the remainder condition. This argument also handles an empty fiber.

Conversely suppose the remainder condition holds. Every fiber $f^{-1}(y)=F^{-1}(y)$ is a closed subset of $\beta X$, hence compact, and lies in $X$. To show that $f$ is closed, let $A$ be closed in $X$ and set $D={\overline{A}}^{\beta X}$. The initial closure identity gives $D\cap X=A$. Compactness makes $F(D)$ closed in $\beta Y$, and the remainder condition givesIndeed any preimage in $D$ of a point of $Y$ must belong to $X$ and hence to $A$. Thus $f(A)$ is closed in $Y$. Together with compact fibers this proves

Compactifications. A Hausdorff compactification of $X$ is a compact Hausdorff space $K$ together with a topological embedding $j:X\to K$ having dense image. We identify $X$ with that image.

For noncompact locally compact Hausdorff $X$, adjoin $\infty$ and give $X^{+}=X\cup \{\infty \}$ its one-point compactification topology: retain the open sets of $X$, and take $\{\infty \}\cup (X\setminus C)$, with $C$ compact, as a neighborhood base at $\infty$. Compact subsets are closed, making this a topology with the original subspace topology on $X$.

An open cover contains a neighborhood of $\infty$, leaving a compact subset of $X$ to cover by finitely many more members. Hence $X^{+}$ is compact. Points of $X$ are separated using its Hausdorff property; to separate $x\in X$ from $\infty$, use local compactness to choose an open $U\ni x$ with compact closure and take $X^{+}\setminus \overline{U}$ at $\infty$. Finally every neighborhood of $\infty$ meets $X$, because $X$ itself is not compact. Thus $X$ is dense and $X^{+}$ is a Hausdorff compactification with one-point remainder.

Comparison maps. The Stone-Čech compactification of a Tychonoff space has the universal property that every continuous map from $X$ to a compact Hausdorff space extends continuously to $\beta X$. Applying it to any Hausdorff compactification $K$ gives a continuous surjection $\pi :\beta X\to K$ fixing $X$. Surjectivity follows because its compact image is closed and contains the dense set $X$.

We need the fibers of compactification comparison maps propertySuppose $\pi (z)=x\in X$ with $z\ne x$. By regularity of compact Hausdorff $\beta X$, choose an open neighborhood $U$ of $x$ with $z\notin \overline{U}$. The embedding $X\subseteq K$ supplies open $V\subseteq K$ containing $x$ and satisfying $V\cap X\subseteq U\cap X$. The open set ${\pi }^{-1}(V)$ contains $z$. Density of $X$ implies $z\in \overline{{\pi }^{-1}(V)\cap X}=\overline{V\cap X}\subseteq \overline{U}$, a contradiction. Thus the displayed identity holds.

Equivalence of (a), (b), (c). Clearly (b) implies (c). If (c) holds in $K$, inverse images under $\pi$ and ${\pi }^{-1}(X)=X$ show that $X$ is a G-delta set in $\beta X$, giving (a). Conversely assume (a) and write $X={\bigcap }_n{U}_n$ in $\beta X$. The sets $F_n=\beta X\setminus U_n$ are compact and miss $X$. Their images $\pi (F_n)$ are closed in $K$ and miss $X$ by the fiber identity. Since $\pi$ is onto, they exhaust $K\setminus X$. HenceThis proves (b), for every $K$. This G-delta compactification invariance defines a Čech-complete space.

The Baire property. Suppose $X={\bigcap }_n{G}_n$ in a compact Hausdorff $K$, with $X$ dense and all $G_n$ open. Each $G_n$ is dense. For dense open subsets $U_n$ of $X$, choose open $O_n\subseteq K$ with $U_n=X\cap O_n$. Every nonempty open subset of $K$ meets $X$ in a nonempty relatively open set and hence meets $U_n$, so each $O_n$ is dense in $K$.

The assumed Baire category theorem in $K$ makes ${\bigcap }_n(G_n\cap O_n)$ dense in $K$. This set lies in $X$ and equals ${\bigcap }_n{U}_n$, so it is also dense in $X$. Therefore Čech-complete spaces are Baire spaces.

Enumerate $\mathbb{Q}=\{q_n:n\in \omega \}$. Each $\mathbb{Q}\setminus \{q_n\}$ is open and dense in $\mathbb{Q}$, while their intersection is empty. Thus $\mathbb{Q}$ is not a Baire space, and consequentlyIf $\mathbb{Q}$ were a G-delta set in $\mathbb{R}$, it would also be one in ${\mathbb{R}}^{+}$: the subspace $\mathbb{R}$ is open in its one-point compactification, and adjoining that open condition to a countable intersection preserves the G-delta set property. Since $\mathbb{Q}$ is dense in $\mathbb{R}$ and hence in ${\mathbb{R}}^{+}$, this would be a Hausdorff compactification of $\mathbb{Q}$ contradicting the preceding conclusion. Therefore $\boxed{\mathbb{Q}\text{ is not }{G}_{\delta }\text{ in }\mathbb{R}}$.

Every continuous map from a space to a compact Hausdorff space extends continuously to its Stone-Čech compactification. Uniqueness follows from density of the original space and the Hausdorff property of the target. Two Hausdorff compactifications with this property have comparison maps both ways; their composites fix the dense original space and therefore are identity maps, proving equivalence of the compactifications.

For a Hausdorff compactification $h:\mathbb{R}\to Y$, let $C^{\pm }$ be the intersections of the closures of the positive and negative tails, respectively. Each is nonempty and connected by the closed-filter-base theorem. Their union equals the remainder: an image point has a bounded-interval neighbourhood excluding sufficiently long tails, whereas every neighbourhood of a remainder point meets tails by density and compactness of bounded intervals. If the remainder is finite, each $C^{\pm }$ must be a singleton, so there can be only one or two remainder points. In particular there is no three-point compactification of the real line.