A zero-dimensional space has a base of clopen sets. A totally disconnected space has no connected subset with more than one point, equivalently all its connected components are singletons. We use these definitions in the stated Hausdorff setting.

If a Hausdorff space has a clopen base, any two different points can be separated by a clopen set. A connected subset containing both would be split into two nonempty disjoint relatively open sets. Thus zero-dimensional implies totally disconnected here.

The compact converse, including the component lemma. For a point $x$ of a compact Hausdorff $K$, putThis is its quasicomponent, a closed compact set. We prove the Šura-Bura component lemma that $Q_x$ equals the connected component of $x$.

Suppose $Q_x=A\cup B$ were a separation into nonempty disjoint relatively closed sets, with $x\in A$. Both are compact, hence closed in $K$. Normality of compact Hausdorff spaces gives disjoint open neighborhoods $U\supseteq A$, $V\supseteq B$. The compact set $H=K\setminus (U\cup V)$ misses $Q_x$. For each $y\in H$, choose a clopen neighborhood $N_y$ of $x$ missing $y$; these exist by the definition of $Q_x$. The open complements cover $H$, so finitely many suffice. Their intersection $N$ is clopen, contains $Q_x$, and lies in $U\cup V$. If $H$ is empty take $N=K$.

Now $N\cap U$ is clopen in $K$: it is open, and its complement is $(K\setminus N)\cup (N\cap V)$, also open. It contains $x$ and misses $B$, contradicting that $B\subseteq Q_x$ lies in every clopen neighborhood of $x$. Hence $Q_x$ is connected. Conversely every connected subset containing $x$ lies in every clopen neighborhood of $x$, so the connected component is contained in $Q_x$. Equality follows.

If $K$ is totally disconnected, then $Q_x=\{x\}$. Given an open neighborhood $O$ of $x$, for each $y\in K\setminus O$ choose a clopen neighborhood of $x$ missing $y$. Compactness again gives a finite intersection $N$ with $x\in N\subseteq O$. These $N$ form a clopen base. Therefore

The Stone-Čech characterization. The stated clopen separation property is strong zero-dimensionality in this normal Hausdorff setting. A normal T1 space is a Tychonoff space, so its Stone-Čech compactification exists. We use its extension property for maps into compact Hausdorff spaces and the density of $X$.

First assume $\beta X$ is a zero-dimensional space, and let $C,D$ be disjoint closed subsets of $X$. Urysohn's lemma gives $f:X\to [0,1]$ equal to zero on $C$ and one on $D$. Its continuous extension to $\beta X$ puts ${\overline{C}}^{\beta X}$ and ${\overline{D}}^{\beta X}$ in disjoint zero and one level sets. Using a clopen base, cover the first compact set by finitely many clopen sets missing the second; their union $H$ is clopen and separates them. Thus $U=H\cap X$ and $V=X\setminus U$ are disjoint clopen sets containing $C,D$.

Conversely, assume the stated clopen separation property on $X$. Let $A,B$ be disjoint closed subsets of $\beta X$. Normality supplies a continuous function $f:\beta X\to [0,1]$ with $f{|}_A=0$, $f{|}_B=1$. The setsare disjoint and closed in $X$. Select clopen $U\subseteq X$ containing $C$ and missing $D$; the given disjoint pair of clopen neighborhoods supplies such a $U$. Its characteristic function, one on $U$ and zero outside, extends from $X$ to a continuous map $\beta X\to \{0,1\}$. Let $H$ be its one level set, a clopen set.

Every $a\in A$ is in ${\overline{C}}^{\beta X}$: any neighborhood of $a$, intersected with $f^{-1}([0,1/3))$, is an open neighborhood meeting the dense set $X$. Likewise $B\subseteq {\overline{D}}^{\beta X}$. Since $C\subseteq H$ and $D\subseteq \beta X\setminus H$, it follows that $A\subseteq H$ and $B\cap H=\varnothing$.

Apply this to $A=\{z\}$ and $B=\beta X\setminus O$, for any point $z$ and open neighborhood $O$. It yields a clopen $H$ with $z\in H\subseteq O$, proving that $\beta X$ has a clopen base. Hence the strong zero-dimensionality and Stone-Čech compactification criterion is

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 $\boxed{\mathbb{Q}\text{ is not }{G}_{\delta }\text{ in }\mathbb{R}}$.

The extension. Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal topological space can be separated by a continuous function $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$.

For the bounded Tietze extension theorem, first suppose $g:C\to [-M,M]$ is continuous. The subsets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ are closed in $X$, because $C$ is closed. Urysohn's lemma supplies a continuous function $h:X\to [-M/3,M/3]$ with $h=-M/3$ on $A$ and $h=M/3$ on $B$. Directly checking the two outer ranges and the middle range of $g$ givesEmpty level sets cause no difficulty: take the appropriate constant, or zero if both are empty. This is the Urysohn two-thirds approximation.

Apply this successively to the residual on $C$, beginning with $f$ and $M_0=1$. We get $h_n$ with $\Vert h_n{\Vert }_{\infty }\le \frac{1}{3}(\frac{2}{3}{)}^n$ and residual after the first $n$ terms at most $(2/3{)}^n$. The Weierstrass M-test gives uniform convergence of $F={\sum }_{n\ge 0}{h}_n$. By the uniform limit theorem, $F$ is continuous; the residual bound proves $F{|}_C=f$, and the sum of the norm bounds is one. Thus

Closed sets as exact zero sets. If $X$ is a perfectly normal space, write a closed $C$ as ${\bigcap }_{n\ge 0}{U}_n$ with each $U_n$ open. Apply Urysohn's lemma to $C$ and $X\setminus U_n$ to obtain $f_n:X\to [0,1]$ vanishing on $C$ and equal to one on $X\setminus U_n$. The uniformly convergent seriesIndeed all summands vanish on $C$; outside $C$, at least one summand is strictly positive. The uniform limit theorem proves continuity. This realizes the closed G-delta set as a zero set construction.

Conversely, suppose each closed set is the zero set of a continuous function. For disjoint closed $C,D$, select $f,g:X\to [0,1]$ with zero sets $C,D$. Their sum never vanishes, andThe quotient is continuous. This also proves normality by inverse images of disjoint endpoint neighborhoods. Each zero set is a G-delta set, since $f^{-1}(0)={\bigcap }_{n\ge 1}{f}^{-1}([0,1/n))$; thus this property characterizes perfectly normal spaces among T1 spaces.

Subspaces. Let $Y\subseteq X$ and let $A,B$ be disjoint closed subsets of $Y$. Put $C={\overline{A}}^X$ and $D={\overline{B}}^X$. Since $C\cap Y=A$ and $D\cap Y=B$, the intersection $C\cap D$ misses $Y$. Choose continuous functions $f,g$ on $X$ whose zero sets are $C,D$. On the open neighborhood $W=X\setminus (C\cap D)$ of $Y$, the same quotient $f/(f+g)$ is continuous and is zero on $A$, one on $B$. Its inverse images of $[0,1/3)$ and $(2/3,1]$, restricted to $Y$, are disjoint open neighborhoods. Therefore every subspace is a normal topological space: perfect normality implies hereditary normality.

A strict counterexample to the converse. Let $D$ be an uncountable discrete space, and let $X=D\cup \{\infty \}$ be its one-point compactification. Its neighborhoods of $\infty$ have finite complement in $D$, because compact subsets of a discrete space are finite. The space is compact Hausdorff, hence a T1 space and normal. A subspace missing $\infty$ is discrete. A subspace containing $\infty$ is closed in $X$, since its complement is a subset of the open discrete part; it is therefore compact Hausdorff and normal. Thus all subspaces are normal.

But $\{\infty \}$ is not a G-delta set. A countable intersection of its open neighborhoods omits at most countably many elements of $D$, and cannot omit all of uncountable $D$. Hence $X$ is not a perfectly normal space. This is the uncountable discrete one-point compactification counterexample.