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A subset $Y\subseteq X$ is C-embedded if every continuous real-valued function on its subspace topology extends continuously to $X$. Closed subsets of normal spaces have this property by the Tietze extension theorem. The condition also makes sense for nonclosed subsets.

A continuous function on a compact subset of a Tychonoff space extends to the whole space. Embed the space into a cube of closed unit intervals. The Tychonoff theorem makes the cube compact Hausdorff and thus normal. The compact subset is closed in that cube, so the Tietze extension theorem extends its bounded function to the cube. Restrict this extension to the embedded original space. Normality of the original space is unnecessary.

A T0 space distinguishes any two distinct points by an open set containing one and not the other. A Tychonoff space, also called a completely regular Hausdorff space, is a $T_1$ space in which a point outside a closed set can be separated from it by a continuous function: after rescaling, $f(x)=0$ and $f{|}_C=1$ with values in $[0,1]$.

The product topology has the subbasis of a topologyA map into the product is continuous exactly when each coordinate map is continuous. One direction follows by composition with the continuous product projections; the other follows by taking inverse images of the displayed subbasic sets.

Define the diagonal evaluation map $\Delta (x)=(f_{\lambda }(x){)}_{\lambda \in \Lambda }$. Its coordinates are continuous, so the map is continuous by continuity into a product space. Point separation makes it injective. For a basic open set $B=f_{\lambda }^{-1}(U)$,Every open subset of $X$ is a union of such $B$, so its image is open in $\Delta (X)$. The inverse on the image is continuous, and $\boxed{\Delta \text{ is a topological embedding}}$. If $X$ is $T_0$, any two distinct points are distinguished by an open set and then by one basic neighborhood within it. Equality of all coordinates would prevent that distinction. Thus the basis hypothesis alone implies point separation in a $T_0$ space. The unsubscripted $f$ in the printed separation hypothesis is understood as the chosen coordinate $f_{\lambda }$.

For the Sierpiński space $S=\{0,1\}$ with opens $\varnothing ,\{1\},S$, each open $U\subseteq X$ gives a continuous characteristic map ${\chi }_U:X\to S$ with ${\chi }_U^{-1}(\{1\})=U$. These inverse images form the whole topology, hence a basis. In a $T_0$ space they separate points, proving the embedding into $S^{\tau }$.

For a Tychonoff space, use all continuous maps $X\to [0,1]$. Given $x\in U$ with $U$ open, complete regularity supplies $f(x)=0$ and $f=1$ on $X\setminus U$. Then $f^{-1}([0,1/2))$ is an open neighborhood of $x$ contained in $U$. These inverse images form a basis. The diagonal theorem gives an embedding into a product of closed unit intervals.

A three-point space works even without any separation axiom. TakeFor each open $U$ of an arbitrary space, let $g_U=2$ on $U$ and $0$ elsewhere. It is continuous. For each point $a$, let $h_a=1$ at $a$ and $0$ elsewhere. It is also continuous, because its image lies in the subspace $\{0,1\}$ of $Y$, which has the indiscrete topology; equivalently $h_a^{-1}(\{2\})=\varnothing$. The $g_U$ supply every open set as an inverse image, and the $h_a$ distinguish every pair of points. Using the diagonal evaluation map for their combined family gives a universal three-point space for product embeddings:

Finally embed the given completely regular Hausdorff space into a cube $K=[0,1{]}^I$. The Tychonoff theorem makes $K$ compact; it is Hausdorff and hence normal. The image of the compact subset $C$ is compact, so closed in $K$. The Tietze extension theorem says that a continuous bounded real-valued function on a closed subset of a normal space extends continuously to the whole space. The given function on $C$ is bounded by compactness, so apply this theorem to its transported function on the image of $C$ in $K$. Restrict the extension to the image of $X$ and pull it back along the embedding. This yields $\boxed{F:X\to \mathbb{R}\text{ continuous},F{|}_C=f}$. This proves that compact subsets of Tychonoff spaces are C-embedded, using normality of the compact cube.

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.

Start with the approximate surjectivity with geometric correction argument. For a fixed $y$, put $r_0=y$. Choose $x_j$ successively so thatInduction gives $\Vert r_j\Vert \le k^j\Vert y\Vert$ and $\Vert x_j\Vert \le R{k}^{j-1}\Vert y\Vert$. The series ${\sum }_j{x}_j$ is absolutely convergent in the Banach space $V$, because the sum of its norms is bounded by the convergent geometric series $R\Vert y\Vert /(1-k)$. Write its sum as $x$. The bounded linear operator $T$ commutes with this limit, while $T{\sum }_{j=1}^m{x}_j=y-r_m\to y$. ConsequentlyThe choices need not depend linearly on $y$; no bounded linear right inverse is being asserted.

For the bounded extension, use the restriction bounded linear operator $S:C_b(X;\mathbb{R})\to C_b(Y;\mathbb{R})$ on the Banach spaces of bounded continuous functions, each with its supremum norm. We will construct an approximation to any $g$ with $\Vert g{\Vert }_{\infty }=M>0$. The two setsare disjoint closed sets in $X$, since $Y$ is closed. If both are nonempty, the distance to a set construction defines a continuous functionThe denominator is positive at every point: zero distance to a closed set means membership in that set. Thus $\Vert u{\Vert }_{\infty }\le M/3$, with $u=-M/3$ on $A$ and $u=M/3$ on $B$. On $A$ and $B$, and also on the intervening range $|g|<M/3$, this impliesIf only $A$ is empty take $u=M/3$, if only $B$ is empty take $u=-M/3$, and if both are empty take $u=0$; the same bounds hold. For $M=0$ take $u=0$. This proves the approximation hypothesis with $R=1/3$ and $k=2/3$. Applying the argument above to $S$ yields an element $h$ of the bounded continuous functions with $h{|}_Y=g$ and $\Vert h{\Vert }_{\infty }\le \Vert g{\Vert }_{\infty }$. Restriction gives the reverse inequality, so the extension preserves the exact supremum norm:This proves the bounded metric-space case of the Tietze extension theorem rather than presupposing it.

For an arbitrary real continuous function $f$ on $Y$, let $g=\tanh f$. Its values lie strictly between $-1$ and $1$, although its supremum norm can be $1$. The bounded result supplies an extension $h$ with $|h|\le 1$. Let $D=\{x:|h(x)|=1\}$, a closed set disjoint from $Y$. If $D$ is nonempty and $Y$ is nonempty, putBoth factors are continuous functions; $\eta =1$ on $Y$ and $\eta =0$ on $D$. Off $D$ we have $|h|<1$, and on $D$ we have $q=0$, so $|q(x)|<1$ everywhere. Therefore a real continuous extension isOn $Y$ it equals $f$. If $D$ is empty use $q=h$; if $Y$ is empty simply take $F=0$. This handles unbounded $f$ without requiring a uniform margin between $|g|$ and $1$.

The group $G(K)$ consists of the nonvanishing continuous functions, and $E(K)$ is a subgroup because multiplication is commutative: $e^g{e}^h=e^{g+h}$. The objective is a logarithm decomposition with finitely many integer obstructions, one for each bounded complementary component. No local connectivity of $K$ is assumed.

We use two elementary finite-neighborhood facts, which make the outline independent of the general Jordan theorem. Given a neighborhood $V$ of $K$, take a finite grid neighborhood $N\subset V$ with $K\subset \mathrm{int}N$. It can be chosen to be a finite union of polygonal surfaces with boundary. Its complement has finitely many bounded holes. Holes that lie in the same component of $\mathbb{C}\setminus K$ can be joined by thin polygonal corridors disjoint from $K$, which are removed from $N$. A hole lying in the unbounded component is similarly joined to the exterior. For the remaining finitely many components, corridors may be continued to include their chosen representatives $a_i$. Small polygonal adjustments preserve $K\subset \mathrm{int}N$ and $N\subset V$. Thus every remaining bounded hole lies in a distinct bounded component of $\mathbb{C}\setminus K$ and contains its representative. One may also arrange this for any prescribed finite list of bounded components by first keeping their representative points out of the grid neighborhood. All paths used are compact and disjoint from $K$, so sufficiently thin corridors remain disjoint from $K$.

For such a finite polygonal $N$, winding indices identify the homology group $H_1(N;\mathbb{Z})$ with the free abelian group on its bounded holes. Here is a finite-chain explanation. Triangulate a large disc with $N$ as a subcomplex. Every integral one-cycle bounds a two-chain in the disc. Its filling coefficient is constant on each component of the complement of $N$, is zero on the exterior, and is the winding index there. If all bounded-hole indices vanish, the filling is supported in $N$, so the cycle already bounds in $N$. Conversely the sum of the oriented faces in one hole has its chain boundary in $N$ and has index one in that hole and zero in the others. Thus cycles ${\gamma }_j$ can be chosen with $\mathrm{wind}({\gamma }_j,a_i)={\delta }_{ij}$. These can be integral sums of boundary curves, which matters when components of $N$ are nested. This proves the needed statement using finite triangulation, not separation of an arbitrary simple closed curve. Together these are finite-hole neighborhoods of a planar compact set.

Now extend $f\in G(K)$ to a continuous $F$ near $K$, using the Tietze extension theorem on its real and imaginary parts. Since ${\min }_K|f|>0$, shrink the neighborhood until $F$ has no zeros, and choose $N$ within it as above. The winding of $F$ on the cycles ${\gamma }_j$ defines integers $k_j$. Setwhere only the finitely many represented holes are used. The winding of $F/r$ on every cycle of $N$ is zero. On each component choose a local complex logarithm at a base point and continue it along paths. Zero winding number around every loop makes the continued value independent of the path; local logarithm charts make it continuous. The continuous logarithm lifting criterion therefore gives $F/r=e^g$ on $N$, and restriction gives $f=e^{g{|}_K}r{|}_K$. Thus $G(K)=E(K)G(\mathbf{a})$, even when $K$ has infinitely many holes: each individual function uses finitely many factors.

To prove uniqueness, suppose a finite product $r=\prod (z-a_i{)}^{k_i}$ equals $e^g$ on $K$. Extend $g$ continuously to a neighborhood. There $r{e}^{-g}$ is close to $1$ after shrinking, so it has its small-branch logarithm. Hence $r$ itself has a continuous function giving a complex logarithm on that neighborhood. Choose $N$ there detecting all the finitely many $a_i$ as separate holes. Its dual cycles give $\mathrm{wind}(r\circ {\gamma }_j,0)=k_j$, while a function with a logarithm has zero winding number. Therefore every $k_j=0$. This proves both freeness of the listed rational functions and $E(K)\cap G(\mathbf{a})=\{1\}$. All subgroups commute, soThis is planar logarithm factorization. If there are no bounded components the second factor is trivial.

For the Jordan curve deduction, let $J$ be homeomorphic to the circle. Parametrize it by $\gamma :[0,1]\to J$ with just the two endpoints identified. For $f\in G(J)$, lift a logarithm of $f\circ \gamma$ along the interval. Its endpoint increment is $2\pi in$. The integer $n$ is additive, every integer is attained by a function with $f(\gamma (t))=e^{2\pi int}$, and $n=0$ precisely when the lift descends to a logarithm on $J$. Hence $G(J)/E(J)\cong \mathbb{Z}$. The factorization says this is free abelian on the bounded complementary components; comparison of ranks, or tensoring with $\mathbb{Q}$, shows that exactly one bounded component exists. There is also the unique unbounded component, so the complement has exactly two components.

We still need the common-boundary assertion. A proper closed arc has trivial exponential quotient because every nonvanishing function on an interval admits a logarithm. Factorization consequently makes the complement of such an arc connected, and since it is open it is path-connected. Fix $p\in J$ and any neighborhood $V$ of $p$, and choose a small open arc $A\subset J\cap V$ containing $p$. The remainder $J\setminus A$ is a proper closed arc. Join a point in each of the two complementary components by a path avoiding $J\setminus A$. This path must meet $J$, and every such meeting lies in $A$. The segment before its first meeting lies in the first region, and the segment after its last meeting lies in the second; continuity makes both regions meet $V$. Thus every point of $J$ lies on both boundaries. Conversely either boundary is contained in the closed set $J$. There are exactly two regions, one bounded and one unbounded, with common boundary $J$, the Jordan curve theorem. This is Jordan separation from the exponential quotient; it does not assume the theorem in the factorization proof.

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.

The real Stone-Weierstrass theorem states that if $X$ is a compact Hausdorff space and $A\subseteq C(X,\mathbb{R})$ is a subalgebra containing the constants and separating points of $X$, then $A$ is dense in $C(X,\mathbb{R})$ in the uniform norm.

The Tietze extension theorem states that if $X$ is a normal topological space, $F\subseteq X$ is closed, and $f:F\to \mathbb{R}$ is continuous, then $f$ extends to a continuous function on $X$. In its bounded form, an $f$ taking values in $[-M,M]$ has an extension with the same range bound. The underlying separation step is the Urysohn lemma.

The Urysohn lemma states that if $A$ and $B$ are disjoint closed subsets of a normal space $X$, there is a continuous $u:X\to [0,1]$ with $u=0$ on $A$ and $u=1$ on $B$.

The Tietze extension theorem states that if $A$ is a closed subset of a normal space $X$, then every continuous $f:A\to \mathbb{R}$ has a continuous extension $F:X\to \mathbb{R}$. In the bounded form, if $f(A)\subseteq [a,b]$, the extension can be chosen with $F(X)\subseteq [a,b]$.

If $X$ is compact and $f:X\to \mathbb{R}$ is continuous, then $f(X)$ is compact by the continuous image of a compact space theorem. Every compact subset of $\mathbb{R}$ is bounded, so $f$ is bounded.

Conversely, suppose the metric space $X$ is not compact. For metric spaces, compactness is equivalent to sequential compactness, so there is a sequence of distinct points $(x_n)$ having no convergent subsequence. The setis closed: an accumulation point would supply a convergent subsequence. It is also discrete for the same reason. Consequently the functionis continuous. By part (a), $X$ is normal, and the Tietze extension theorem extends $g$ to a continuous $G:X\to \mathbb{R}$. Since $G(x_n)=n$, this extension is unbounded. Therefore, if every continuous real-valued function on $X$ is bounded, $X$ must be compact. This is the bounded-continuous-function characterization of compact metric spaces.

The Tietze extension theorem states: if $X$ is a normal topological space, $A\subseteq X$ is closed, and $f:A\to [-M,M]$ is continuous, then there is a continuous $F:X\to [-M,M]$ with $F{|}_A=f$.

We first prove an approximation lemma. Given continuous $h:A\to [-r,r]$, the closed subsetsare disjoint and closed in $X$. By the Urysohn lemma, there is a continuous $g:X\to [-r/3,r/3]$ equal to $-r/3$ on $A_{-}$ and $r/3$ on $A_{+}$. Then

Starting with $h_0=f$ and $r_0=M$, apply the lemma recursively to obtain continuous $g_j:X\to \mathbb{R}$ such thatwhere $r_j=M(2/3{)}^j$. The seriesconverges uniformly by the Weierstrass test, so its sum is continuous. On $A$, its remainder after $j$ terms is $h_{j+1}$, which tends uniformly to zero; hence $F{|}_A=f$. The bounds also keep $F$ in $[-M,M]$ after the standard endpoint-preserving version of the construction, proving the theorem.

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.