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In the Stone-Čech compactification of a discrete space, the closure of each subset of $X$ is clopen, because its indicator extends to a continuous zero-one-valued function. For disjoint closed sets $A,B\subseteq \beta X$, choose $u=0$ on $A$ and $u=1$ on $B$ by Urysohn's lemma. Then $U=\overline{\{x\in X:u(x)<1/2\}}$ and $V=\beta X\setminus U$ are disjoint clopen sets containing $A$ and $B$, respectively. Density proves these containments.

Continuous functions $X\to [0,1]$ on a normal Hausdorff space separate points and closed sets by Urysohn's lemma. The evaluation map into the product of their codomains is injective and continuous. For each point in an open set, a coordinate equal to one there and zero outside the open set detects a smaller coordinate neighborhood, proving the inverse topology.

If $\mathcal{B}$ is a countable topological basis of $X$, then $\{B\cap Y:B\in \mathcal{B}\}$ is a countable topological basis of the subspace $Y$. Indeed every subspace-open neighborhood is $Y\cap U$ for an open $U$ of $X$, and a suitable basic set lies between its point and $U$. This proves the first assertion.

Urysohn's metrisation theorem: every second-countable space that is also a T3 space is a metrizable space. Here regularity of a topological space allows a neighborhood of a point to be shrunk to an open neighborhood whose closure is still inside the original one.

We first prove the separation facts needed in a complete proof. Second countability implies the Lindelöf property: for each basic set contained in some member of an open cover choose one such member; these countably many members cover. In particular closed subspaces are Lindelöf. For disjoint closed sets $F,G$, regularity of a topological space and this property give open covers $F\subseteq {\bigcup }_n{U}_n$, $G\subseteq {\bigcup }_n{V}_n$ with ${\overline{U}}_n\cap G=\varnothing$ and ${\overline{V}}_n\cap F=\varnothing$. DefineThese sets are open and contain $F,G$ respectively. They are disjoint: if a point belonged to a term indexed by $n$ in the first union and by $m$ in the second, the term with the larger index would have removed the closure of the other covering set. Thus the space is a normal space.

For completeness, Urysohn's lemma supplies continuous separating functions in a normal topological space. Its construction is as follows. Given disjoint closed $F,G$, choose open sets $O_r$ for dyadic $r\in [0,1]$, with $F\subseteq O_0$, $O_1=X\setminus G$, and ${\overline{O}}_r\subseteq O_s$ whenever $r<s$. Start by shrinking a neighborhood of $F$ inside $X\setminus G$, and insert the successive dyadic sets by the closed-neighborhood shrinking in a normal space. DefineThen $f=0$ on $F$, $f=1$ on $G$, and $0\le f\le 1$. The identities $\{f<a\}={\bigcup }_{r<a}{O}_r$ and $\{f>a\}={\bigcup }_{a<r<1}(X\setminus {\overline{O}}_r)$ prove continuity for $0<a<1$, with the endpoint cases immediate.

Let $\{B_n\}$ be a countable topological basis. For each pair $(n,m)$ with ${\overline{B}}_n\subseteq B_m$, choose a continuous function $f_{nm}:X\to [0,1]$ equal to one on ${\overline{B}}_n$ and zero outside $B_m$. These form a countable family. Given $x\in O$ open, choose $B_m$ with $x\in B_m\subseteq O$, shrink a neighborhood of $x$ inside $B_m$, and then choose $B_n$ inside that smaller neighborhood. ThusConsequently these functions generate the original topology. They also separate distinct points: use the $T_1$ open set $X\setminus \{y\}$ when $x\ne y$. Enumerate them as $f_j$, padding by zero functions if necessary. The map $x\mapsto (f_j(x){)}_j$ is a topological embedding into the countable cube. Pulling back its metric, or directly usinggives a compatible metric. The fact that the functions distinguish distinct points proves positivity for distinct points, the triangle inequality is termwise, and the uniformly bounded tails show that this metric induces exactly the topology generated by the coordinates. Empty and singleton spaces are trivially metrizable spaces. This proves the theorem.

For the application, a separable metric space has a countable topological basis consisting of balls with rational positive radii about a countable dense set, and it is a T3 space. The given product theorems make a countable product of these spaces second countable and a T3 space. Urysohn's theorem makes it a metrizable space. Finally every second countable space is separable: choose a point from each nonempty basic open set; the resulting countable set meets every nonempty open set. Hence the product is separable and a metrizable space.

Urysohn's lemma states that, in a normal space, disjoint closed subsets $A,B$ can be separated by a continuous function $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

First use the shrinking consequence of the separation property of a normal space: if a closed set $F$ lies in an open set $G$, there is an open $W$ with $F\subseteq W\subseteq \overline{W}\subseteq G$. To prove it, separate the disjoint closed sets $F$ and $X\setminus G$ by disjoint open neighborhoods $W,V$. Then $\overline{W}\subseteq X\setminus V\subseteq G$.

Let $D$ be the dyadic rationals in $[0,1]$. Set $U_1=X\setminus B$, and shrink around $A$ to choose $U_0$ with $A\subseteq U_0\subseteq \overline{U_0}\subseteq U_1$. At each subsequent dyadic stage, insert between each adjacent pair $s<t$ an open set $U_{(s+t)/2}$ such thatThe shrinking lemma applies because the first set is closed and contained in the last open set. Induction gives $\overline{U_s}\subseteq U_t$ for all dyadic $s<t$. Part (a) now constructs a continuous $u$. Each point of $A$ lies in $U_0$, so has value zero. Each point of $B$ lies in none of the $U_t$, so has value one. This proves the lemma.

For the embedding, let $I=C(X,[0,1])$, a set of continuous functions, and define the evaluation mapEvery coordinate is continuous, so $e$ is continuous in the product topology. The Hausdorff assumption makes singletons closed. For distinct points, Urysohn's lemma applied to their singletons gives a coordinate with different values, so $e$ is injective.

To check the inverse topology, take $x\in O$ with $O$ open. Apply the lemma to $\{x\}$ and $X\setminus O$, choosing a coordinate $u$ with $u(x)=1$ and $u=0$ outside $O$. ThenThis is the preimage of an open coordinate cylinder under $e$. Thus every original neighborhood is detected by coordinate neighborhoods, and $e^{-1}:e[X]\to X$ is continuous. The evaluation map is a topological embedding.

For a strict-range modification of a continuous extension, the closed sets $Y$ and $Z=\{x:|f(x)|\ge 1\}$ are disjoint. Choose by Urysohn's lemma a continuous $u:X\to [0,1]$ with $u{|}_Y=0$ and $u{|}_Z=1$, and defineThis function is continuous and agrees with $f$ on $Y$. On $Z$ it vanishes. Off $Z$ we have $|f|<1$ and $0\le 1-u\le 1$, so $|\tilde{f}|\le |f|<1$. Hence its range lies strictly inside $(-1,1)$ on all of $X$. The construction also covers empty $Y$ or empty $Z$ by a constant separating function where appropriate.

A C-embedded subset $Y\subseteq X$ is one when every continuous real-valued function on $Y$ extends continuously to $X$. It is a Cb-embedded subset when this is required for every bounded continuous real-valued function, with a bounded extension. These definitions concern the subspace topology on $Y$.

Here is the bounded Tietze extension construction from Urysohn's lemma. Let $g:Y\to \mathbb{R}$ be continuous with $|g|\le M$. The case $M=0$ has the zero extension. For $M>0$, the sets $A=\{g\le -M/3\}$ and $B=\{g\ge M/3\}$ are disjoint and closed in $X$, since $Y$ is closed. Urysohn's lemma gives a continuous $u_0:X\to [-M/3,M/3]$ taking the endpoint values on $A,B$. On $Y$,Indeed on $A$ or $B$ this follows from the prescribed endpoint; on their complement both $|g|$ and $|u_0|$ are at most $M/3$.

Repeat on the residual. With $M_n=M(2/3{)}^n$, choose continuous $u_n$ satisfying $\Vert u_n{\Vert }_{\infty }\le M_n/3$ and reducing the residual on $Y$ to at most $M_{n+1}$. The series $F={\sum }_{n\ge 0}{u}_n$ converges uniformly by the geometric majorant. Its sum is continuous, has $|F|\le \sum M_n/3=M$, and restricts to $g$, since the residual tends uniformly to zero. Thus every bounded continuous function extends, with its bound preserved, proving $C^b$-embeddedness without assuming the extension theorem being proved.

Urysohn's lemma states that if $X$ is normal and $A,B$ are disjoint closed subsets, there is a continuous function $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. The values on the rest of $X$ remain in that interval. Interchanging $A,B$ or composing with an affine function gives any two prescribed real endpoint values. This is functional separation of closed sets, stronger than merely choosing disjoint open neighbourhoods.

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.

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

Let $\mathcal{F}=C(X,[0,1])$. Form the evaluation map into a product spaceThe cube is compact by Tychonoff's theorem and Hausdorff, so its closed subspace $\beta X$ is a compact Hausdorff space. A completely regular Hausdorff space has enough continuous functions to separate points and closed sets. Consequently $e$ is injective and continuous, and is a topological embedding: given $x\in O$ open, choose $f$ with $f(x)=0$ and $f=1$ off $O$; the coordinate neighborhood $f<1/2$ pulls back to a neighborhood of $x$ inside $O$. Identify $X$ with this dense embedded subspace. This constructs the Stone-Čech compactification.

Each $f\in \mathcal{F}$ extends as its coordinate function. Rescaling therefore extends every bounded real continuous function; real and imaginary parts extend bounded complex continuous functions. The extension is unique because $X$ is dense and the target is Hausdorff. Its supremum norm is unchanged by density, so restriction is an isometric isomorphism of normed spaces of Banach spacesThese are the space of continuous functions on a compact space and the space of bounded continuous functions, respectively.

More generally, embed a compact Hausdorff space $K$ in its own evaluation cube $[0,1{]}^{C(K,[0,1])}$. This is an embedding by complete regularity, and its image is closed by compactness. For a continuous map $u:X\to K$, extend all the bounded coordinates $f\circ u$ to $\beta X$. The resulting continuous map into the cube takes its values in the closed copy of $K$: it does so on the dense subset $X$, and the inverse image of that closed copy is closed. ThusUniqueness again follows from density. This is the universal property of the Stone-Čech compactification. It implies uniqueness of the compactification up to a homeomorphism fixing $X$, by extending the identity in both directions. Every Hausdorff compactification of $X$ is the image of $\beta X$ under a continuous surjection fixing $X$, since that image is compact and contains the dense copy of $X$. A continuous map $X\to Y$ between completely regular Hausdorff spaces likewise extends uniquely to $\beta X\to \beta Y$ after composing with the embedding of $Y$; these extensions preserve identities and composition. If $X$ is already compact, its dense copy is closed, so $\beta X=X$.

For discrete $X$, every indicator $1_D$, $D\subseteq X$, is continuous. Its extension to $\beta X$ takes only values $0,1$, since this closed condition holds on the dense subset $X$. Its one-set is precisely $\overline{D}$ and is clopen: it is closed, and every neighborhood of one of its points meets $D$ by density and continuity. This also gives the discrete Stone-Čech compactifications are extremally disconnected property: for open $O$, density implies $\overline{O}=\overline{O\cap X}$, which is clopen.

For the requested separation, a compact Hausdorff space is normal, so Urysohn's lemma supplies $u:\beta X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$. Put $D=\{x\in X:u(x)<1/2\}$ andThe preceding indicator argument makes both sets clopen. At a point of $A$, the neighborhood $u<1/2$ has its dense $X$-part inside $D$, so the point lies in $U$. At a point of $B$, the neighborhood $u>1/2$ misses $D$, so the point lies outside $U$. We have proved the stronger clopen separation in discrete Stone-Čech compactifications conclusion

Finally take $\boxed{K=\beta \mathbb{N}}$, the Stone-Čech compactification of the natural numbers. It is a compact Hausdorff space and is separable, because its embedded copy of $\mathbb{N}$ is countable and dense. For every $D\subseteq \mathbb{N}$, the indicator extends to an element $h_D\in C(K)$. For distinct $D,D^{\prime }$, evaluating at an integer in their symmetric difference gives $\Vert h_D-h_{D^{\prime }}{\Vert }_{\infty }=1$. There are uncountably many such indicators. A separable metric space cannot contain an uncountable family at pairwise distance one: disjoint balls of radius $1/3$ would require distinct members of a countable dense set. Hence $C(K)$ is not separable, despite separability of $K$.

Urysohn's lemma states that disjoint closed subsets $A,B$ of a normal space admit a continuous $u:X\to [0,1]$ with $u{|}_A=0$ and $u{|}_B=1$. Using the usual convention, normal spaces are also $T_1$.

The Tietze extension theorem states that a continuous real-valued function on a closed subset of a normal space extends continuously to the whole space. If its range lies in a closed bounded interval, the extension may be chosen in that same interval.

Prove first the interval version for $f:S\to [-1,1]$. If a continuous residual $r$ on $S$ obeys $|r|\le M$, the sets $\{r\le -M/3\}$ and $\{r\ge M/3\}$ are disjoint closed subsets of $X$. Urysohn's lemma gives a continuous $h:X\to [-M/3,M/3]$ taking the endpoint values on those two sets. On each of the three ranges of $r$, it follows that $|r-h|\le 2M/3$ on $S$. Repeating with $M_j=(2/3{)}^j$ produces $h_j$ with $\Vert h_j{\Vert }_{\infty }\le M_j/3$ and a residual bounded by $M_{j+1}$. The series $\tilde{f}={\sum }_{j\ge 0}{h}_j$ converges uniformly, has absolute value at most $\sum M_j/3=1$, is continuous and restricts to $f$ as the residual tends to zero. Rescaling proves every closed bounded interval version.

For unbounded $f$, first extend $f/(1+|f|)$ to $g:X\to [-1,1]$. The closed set $B=g^{-1}(\{-1,1\})$ is disjoint from $S$. Urysohn's lemma gives $\psi :X\to [0,1]$ with $\psi {|}_S=1$ and $\psi {|}_B=0$. Then $h=\psi g$ takes values strictly between minus one and one everywhere and equals $f/(1+|f|)$ on $S$. Thus $h/(1-|h|)$ is a continuous real-valued extension of $f$, proving the full theorem.

Statement 1 is true: metric spaces are normal, and the bounded interval form applies. Statement 2 is true: compact Hausdorff spaces are normal, and the unrestricted real-valued form applies. Statement 3 is false. For a counterexample, let $X=\{a,b,c\}$ have open sets $\varnothing ,\{a\},\{a,b\},\{a,c\},X$. The subset $S=\{b,c\}$ is closed and discrete, so $f(b)=-1$, $f(c)=1$ is continuous on it. Every open neighborhood of $b$ or $c$ in $X$ contains $a$. A continuous extension would pull back disjoint neighborhoods of minus one and one to disjoint open sets containing $b$ and $c$, yet both would contain $a$. This is impossible.

Suppose $Y$ is closed in a normal space $X$ and $|f|<1$ on $Y$ for a continuous real-valued $f$ on $X$. The closed set $Z=\{|f|\ge 1\}$ misses $Y$. Urysohn's lemma supplies $u:X\to [0,1]$ equal to zero on $Y$ and one on $Z$. Then $(1-u)f$ agrees with $f$ on $Y$ and has range in $(-1,1)$. This permits the inverse transform $s\mapsto s/(1-|s|)$ in the proof of unbounded Tietze extension.

In a normal Hausdorff space, strong zero-dimensionality means that any disjoint closed sets can be separated by a clopen set, equivalently by disjoint clopen neighborhoods. For general Tychonoff spaces, the usual definition asks this for completely separated sets. In the normal setting Urysohn's lemma makes the formulations agree. The condition is equivalent to zero-dimensionality of the Stone-Čech compactification.

For normal Hausdorff $X$, if $\beta X$ has a clopen base, extend a Urysohn's lemma separator of two closed subsets of $X$. Their closures in $\beta X$ are separated; compactness gives a clopen separator, whose trace works in $X$. Conversely separate the traces of the low and high level sets of a continuous $f:\beta X\to [0,1]$ by clopen $U\subseteq X$. The characteristic function of $U$ extends into $\{0,1\}$ on $\beta X$. Density ensures its one level contains the zero level of $f$ and misses the one level. Applying this with $f$ separating a point from a closed set gives a clopen base for $\beta X$.

Every second-countable space that is also a T3 space is a metrizable space. First regularity of a topological space and the Lindelöf property yield a normal space: cover each of two disjoint closed sets by countably many open sets whose closures avoid the other, and subtract the first $n$ opposite closures from the $n$th neighborhood. The resulting unions are disjoint open neighborhoods. Urysohn's lemma then gives, for each pair of basic sets ${\overline{B}}_n\subseteq B_m$, a continuous function equal to one on the smaller closure and zero outside the larger set. There are countably many such pairs. The shrinking form of regularity of a topological space shows these functions generate the topology, and $T_1$ makes them separate points. Their coordinate map embeds the space into the countable cube, with compatible metric $d(x,y)={\sum }_j{2}^{-j}|f_j(x)-f_j(y)|$. The bounded tails prove compatibility with the product topology. Thus a countable base converts separation axioms into a numerical distance.

Let $C$ be closed in a normal topological space and $g:C\to [-M,M]$ continuous. Separate its closed level sets $g\le -M/3$ and $g\ge M/3$ by Urysohn's lemma to obtain $h:X\to [-M/3,M/3]$ taking the corresponding endpoint values. Checking the outer and middle ranges gives the displayed bounds. Iteration on residuals gives a uniformly convergent series with total supremum norm at most $M$. The uniform limit theorem gives the bounded Tietze extension theorem.