Eduardo Leon Garrido (1856-1949) was a Spanish painter.
We have sections
\begin{equation}{\label{a}}\tag{A}\mbox{}\end{equation}
Existence of Continuous Functions.
A space is normal when, for each disjoint pair of closed sets \(A\) and \(B\), there are disjoint open sets \(U\) and \(V\) satisfying \(A\subseteq U\) and \(B\subseteq V\). A \(T_{4}\)-space is a normal space which is \(T_{1}\) (i.e., the singleton \(\{x\}\) is closed for each \(x\)). If it is agreed that a set \(U\) is a neighborhood of a set \(A\) when \(A\) is a subset of the interior \(\mbox{int}(U)\) of \(U\), then the definition of normality can be stated: a space is normal if and only if disjoint closed sets have disjoint neighborhoods. A family of neighborhoods of a set is a base for the neighborhood system of the set when every neighborhood of the set contains a member of the family. If \(W\) is a neighborhood of a closed subset \(A\) of a normal space \(X\), then there are disjoint open sets \(U\) and \(V\) satisfying \(A\subseteq U\) and \(X\setminus \mbox{int}(W)\subseteq V\), and hence the arbitrary neighborhood \(W\) of \(A\) contains the closed neighborhood \(\mbox{cl}(U)\). Consequently, the family of closed neighborhoods of a closed set \(A\) is a base for the neighborhood system of \(A\) if and only if the space is normal. The converse is also true, for if \(A\) and \(B\) are disjoint closed sets and \(W\) is a closed neighborhood of \(A\) which is contained in \(X\setminus B\), then \(\mbox{int}(W)\) and \(X\setminus W\) are disjoint open neighborhoods of \(A\) and \(B\) respectively. Every discrete space and every indiscrete space is normal and consequently a normal space need not be Hausdorff and may fail to satisfy the first or second axiom of countability. However a \(T_{4}\)-space is surely a Hausdoff space. A closed subset of a normal space is, with the relative topology, normal. However, subspace, products, and quotient spaces of normal spaces may not be normal.
A topological space is regular when, for each point \(x\) and each neighborhood \(U\) of \(x\), there is a closed neighborhood \(V\) of \(x\) satisfying \(V\subseteq U\); that is, the family of closed neighborhoods of each point is a base for the neighborhood system of the point. An equivalent statement: for each point \(x\) and each closed set \(A\), if \(x\not\in A\), then there are disjoint open sets \(U\) and \(V\) satisfying \(x\in U\) and \(A\subseteq V\). A regular space which is also \(T_{1}\) is called a \(T_{3}\)-space. Recall that a Lindelöf space is a topological space that each open cover has a countable subcover.
Proposition. Each regular Lindelöf space is normal.
Proof. Suppose \(A\) and \(B\) are closed disjoint subsets of \(X\). Because \(X\) is regular, for each point of \(A\), there is a neighborhood whose closure fails to intersect \(B\) and consequently the family \({\cal U}\) of all open sets whose closures do not intersect \(B\) is a cover of \(A\). Similarly, the family \({\cal V}\) of all open sets whose closures do not inetrsect \(A\) is a cover of \(B\), and \({\cal U}\cup {\cal V}\cup\{X\setminus (A\cup B)\}\) is a cover of \(X\). There is then a sequence \(\{U_{n}\}\) of members of \({\cal U}\) which covers \(A\), and a sequence \(\{V_{n}\}\) of members of \({\cal V}\) which covers \(B\). Let \\[U’_{n}=U_{n}\setminus\bigcup\{\mbox{cl}(V_{p}):p\leq n\}\]
and let
\[V’_{n}=V_{n}\setminus\bigcup\{\mbox{cl}(U_{p}):p\leq n\}.\]
Since \(U’_{n}\cap V_{m}\) is empty for \(m\leq n\), it follows that \(U’_{n}\cap V’_{m}\) is empty for \(m\leq n\). Applying the same argument with the roles of \(U\) and \(V\) interchanged, \(U’_{n}\cap V’_{m}\) is empty for all \(n\) and \(m\) and consequently \(\bigcup\{U’_{n}\}\) is disjoint from \(\bigcup\{V’_{n}\}\). Finally, \(\mbox{cl}(V_{p})\cap A\) and \(\mbox{cl}(U_{p})\cap B\) are empty for all \(p\) and hence the open disjoint sets \(\bigcup\{U’_{n}\}\)
and \(\bigcup\{V’_{n}\}\) contains \(A\) and \(B\) respectively. \(\blacksquare\)
In particular, a regular topological space satisfying the second axiom of countability is always normal. We now begin the construction of continuous real-valued functions. If \(A\) and \(B\) are disjoint closed sets, we want to construct a continuous real-valued function which is zero on \(A\) and one on \(B\), with all values in the closed interval \([0,1]\).
Proposition. Let \(D\) be a dense subset of \(\mathbb{R}_{+}\). Suppose that the family of \(F_{t}\), \(t\in D\), of subsets of \(X\) such that
the following conditions are satisfied:
- if \(t<s\), then \(F_{t}\subseteq F_{s}\);
- \(\bigcup\{F_{t}:t\in D\}=X\).
For each \(x\) in \(X\), let
\[f(x)=\inf\{t:x\in F_{t}\}.\]
Then
\[\{x:f(x)<s\}=\bigcup\{F_{t}:t\in D\mbox{ for }t<s\}\]
and
\[\{x:f(x)\leq s\}=\bigcap\{F_{t}:t\in D\mbox{ for }t>s\}\]
for each real number \(s\). \(\sharp\)
Proposition. Let \(D\) be a dense subset of \(\mathbb{R}_{+}\). Suppose that the family of \(F_{t}\), \(t\in D\), of open subset of a topological space \(X\) such that the following conditions are satisfied:
- if \(t<s\), then the closure of \(F_{t}\) is a subset of \(F_{s}\);
- \(\bigcup\{F_{t}:t\in D\}=X\).
Then, the function \(f\) such that \(f(x)=\inf\{t:x\in F_{t}\}\) is continuous. \(\sharp\)
\begin{equation}{\label{kelp4}}\tag{1}\mbox{}\end{equation}
Lemma \ref{kelp4}. (Urysohn). If \(A\) and \(B\) are disjoint closed subsets of a normal space \(X\), then there is a continuous function \(f\) on \(X\) to the interval \([0,1]\) such that \(f\) is zero on \(A\) and one on \(B\). \(\sharp\)
\begin{equation}{\label{b}}\tag{B}\mbox{}\end{equation}
Embedding in Cubes.
The Cartesian product of closed unit intervals with the product topology is called a cube. A cube is the set \(Q^{A}\) of all functions on a set \(A\) to the closed unit interval \(Q\) with the topology of pointwise, or coordinate-wise, convergence. Suppose that \(F\) is a family of functions such that each member \(f\) of \(F\) is defined on a topological space \(X\) to a space \(Y_{f}\) (the range may be different for different members of the family). There is a natural mapping of \(X\) into the product \(\prod\{Y_{f}:f\in F\}\) which is defined by mapping a point \(x\) of \(X\) into the member of the product whose \(f\)-th coordinate is \(f(x)\). Formally, the {\bf evaluation} map \(e\) is defined by \((e(x))_{f}=f(x)\). It turns out that \(e\) is continuous if the members of \(F\) are continuous and \(e\) is a homeomorphism if, in addition, \(F\) contains “enough functions”. A family \(F\) of functions on \(X\) distinguishes points when, for each pair of distinct points \(x\) and \(y\), there is \(f\) in \(F\) satifying \(f(x)\neq f(y)\). The family distinguishes points and closed sets when, for each closed subset \(A\) of \(X\) and each member \(x\) of \(X\setminus A\), there is \(f\) in \(F\) such that \(f(x)\) does not belong to the closure of \(f(A)\).
\begin{equation}{\label{top262}}\tag{2}\mbox{}\end{equation}
Lemma \ref{top262}. (Embedding Lemma). Let \(F\) be a family of continuous functions such that each member \(f\) of \(F\) is defined on a topological space \(X\) to a space \(Y_{f}\). Then, we have the following properties.
(i) The evaluation map \(e\) is a continuous function on \(X\) to the product space \(\prod\{Y_{f}:f\in F\}\).
(ii) The function \(e\) is an open map of \(X\) onto \(e(X)\) if \(F\) distinguishes points and closed sets.
(iii) The function \(e\) is one-to-one if and only if \(F\) distinguishes points.
Proof. The map \(e\) followed by projection \(P_{f}\) into the \(f\)-th coordinate space is continuous since \(P_{f}\circ e(x)=f(x)\). Consequently, by Proposition \ref{top217}, \(e\) is continuous. To prove (ii), it is sufficient to show that the image under \(e\) of an open neighborhood \(U\) of a point \(x\) contains the intersection of \(e(X)\) and a neighborhood of \(e(x)\) in the product. Choose a member \(f\) of \(F\) such that \(f(x)\) dose not belong to the closure of \(f(X\setminus U)\). The set of all \(y\) in the product such that \(y_{f}\not\in \mbox{cl}(f(X\setminus U))\) is open and evidently its intersection with \(e(X)\) is a subset of \(e(U)\). Hence \(e\) is an open map of \(X\) onto \(e(X)\). Statement (iii) is clear. \(\blacksquare\)
The preceding lemma reduces the problem of embedding a space topologically in a cube to the problem of finding a “rich” set of continuous real-valued functions on the space. There are topological spaces on which each continuous real-valued function is constant. For example, any indiscrete space has this property. There are less trivial examples; there are regular Hausdorff spaces on which every real continuous function is constant.
Definition. A topological space \(X\) is called completely regular when, for each member \(x\) of \(X\) and each neighborhood \(U\) of \(x\), there is a continuous function \(f\) on \(X\) to the closed unit interval such that \(f(x)=0\) and \(f\) is identically one on \(X\setminus U\). \(\sharp\)
It is clear that the family of all continuous functions on a completely regular space to the unit closed interval \([0,1]\) distinguishes points and closed sets, in the sense of the preceding lemma. (The converse statement is also true, but will not be needed here.) If a completely regular space is \(T_{1}\), then the family of continuous functions on the space to \([0,1]\) also distinguishes points. A completely regular \(T_{1}\)-space is called a Tychonoff space. If \(X\) is a Tychonoff space and \(F\) is the family of all continuous functions on \(X\) to \([0,1]\), the above embedding lemma shows that the evaluation map of \(X\) into the cube \(Q^{F}\) is a homoemorphism. Thus each Tychonoff space is homeomorphic to a subspace of a cube.
Each normal \(T_{1}\)-space is a Tychonoff space in view of Urysohn’s lemma \ref{kelp4}. Each completely regular is regular. For \(T_{1}\)-space there is a hierarchy of so-called separation axiom: Hausdorff, regular, completely regular, normal. Except for normality, these properties are hereditary, in the sense that each subspace of a space \(X\) enjoys the property if \(X\) does. The product spaces of each of these types is again of the same type, excepting, again, normality.
\begin{equation}{\label{top263}}\tag{3}\mbox{}\end{equation}
Proposition \ref{top263}. The product of Tychonoff spaces is a Tychonoff space.
Proof. For convenience, let us agree that a continuous function \(f\) on a topological space \(X\) to the closed unit interval is for a pair \((x,U)\) if and only if \(x\) is a point, \(U\) is a neighborhood of \(x\), \(f(x)=0\), and \(f\) is identically one on \(X\setminus U\). If \(f_{1},\cdots ,f_{n}\) are functions for \((x,U_{1}),\cdots ,(x,U_{n})\), and if \(g(x)=\sup_{1\leq i\leq n}f_{i}(x)\), then \(g\) is a function for \((x,\bigcap_{i=1}^{n}U_{i})\). Consequently the space is completely regular if for each \(x\) and each neighborhood \(U\) of \(x\) belonging to some subbase for the topology there is a function for \((x,U)\). If \(X\) is the product \(\prod_{a\in A}X_{a}\) of Tychnoff space and \(x\in X\), let \(U_{a}\) be a neighborhood of \(x_{a}\) in \(X_{a}\). If \(f\) is a function for \((x_{a},U_{a})\), then \(f\circ P_{a}\), where \(P_{a}\) is the projection into the \(a\)-th coordinate space, is a function for \((x,P_{a}^{-1}(U_{a})\). The family of sets of the form \(P_{a}^{-1}(U_{a})\) is a subbase for the product topology and hence the product is completely regular. Since the product of \(T_{1}\)-space is a \(T_{1}\)-space, the result follows. \(\blacksquare\)
Theorem. (Embedding Theorem). In order that a topological space be a Tychonoff space it is necessary and sufficient that it be homeomorphic to a subspace of a cube.
Proof. The closed unit interval is a Tychonoff space, and hence a cube, being a product of unit intervals, is a Tychonoff space by Proposition \ref{top263}. Each subspace of a cube is therefore a Tychonoff space. It has already been observed that if \(X\) is a Tychonoff space and \(F\) the set of all continuous functions on \(X\) to the closed unit interval \(Q\), then (by the embedding Lemma \ref{top262}) the evaluation map is a homeomorphism of \(X\) into the cube \(Q^{F}\). \(\blacksquare\)
\begin{equation}{\label{c}}\tag{C}\mbox{}\end{equation}
Metric Topologies.
Let \((X,d)\) be a pseudo-metric space. If \(r\) is a positive number, the set \(B(x;r)=\{y:d(x,y)<r\}\) is the open sphere of radius \(r\) about \(x\), or briefly the open \(r\)-sphere about \(x\), and \(\bar{B}(x;r)=\{y:d(x,y)\leq r\}\) is the closed $r$-sphere about \(x\). The intersection of two spheres may not be a sphere. However, if \(d(x,y)<r\) and \(d(x,z)<s\), then each point \(w\) such that \(d(w,x)<\min\{r-d(x,y),s-d(x,z)\}\) is a member of both the open \(r\)-sphere about \(y\) and the open \(s\)-sphere about \(z\) because of the triangle inequality. Consequenly, the intersection of two open spheres contains an open sphere about each of its points. This says that the family of all open spheres is the base for a topology for \(X\) by Corollay 11 in page topological spaces. This topology is the pseudo-metric topology for \(X\). If \((X,d)\) is a metric space, then we can similarly generate the so-called metric topology. Observe that each closed sphere is closed relative to the pseudo-metric topology, and a set is open relative to the pseudo-metric topology if and only if it contains an open sphere about each of its points. We have the following observations.
- The metric topology is Hausdorff.
- A sequence \(\{x_{n}\}\) in \(X\) satisfies \(x_{n}\stackrel{\tau_{d}}{\rightarrow}\) if and only if \(d(x_{n},x)\rightarrow 0\), where \(\tau_{d}\) is the metric topology.
- The metric topology \(\tau_{d}\) is first countable, since, for each \(x\in X\), the countable family of open balls \(\{B(x;1/n):n\in\mathbb{N}\}\) is a base for the neighborhood system at \(x\). This also says that \(x\in\mbox{cl}(A)\) if and only there is a sequence \(\{x_{n}\}\) in \(A\) with \(d(x_{n},x)\rightarrow 0\).
- Note that whether a metric space is complete or a sequence is Cauchy depends on the metric, not just the metric topology. It is possible for two metrics to induce the same metric topologies, even though one is complete and the other one is not. It is clear that the space of irrational numbers in \((0,1)\) with its usual topology is not complete, since there are irrational sequences with rational limits.
Proposition. We have the following properties.
(i) A metrizable space is separable if and only if it is second countable.
(ii) A subset of a separable metrizable space is separable. \(\sharp\)
Two metrics generating the same metric topology are said to be equivalent.
Proposition. Let \((X,d)\) be a complete metric space. Then, the metric \(\rho\) defined by
\[\rho (x,y)=\frac{d(x,y)}{1+d(x,y)}\]
is a bounded equivalent metric taking values in \([0,1)\). Moreover, the metrics \(d\) and \(\rho\) have the same Cauchy sequences, and \((X,d)\) is complete if and only if \((X,\rho )\) is complete. \(\sharp\)
Let \(A\) be a subset of a metric space \((X,d)\). We recall that the diameter of \(A\) is defined by \(\delta (A)=\sup_{x,y\in M}d(x,y)\). We say that a sequence \(\{A_{n}\}\) of subsets of \(X\) has vanashing diameter if \(\delta (A_{n})\rightarrow 0\) as \(n\rightarrow\infty\).
Theorem. (Cantor Intersection Theorem). In a complete metric space, if a decreasing sequence of nonempty closed subsets has vanashing diameter, then the intersection of the sequence is a singleton. \(\sharp\)
Proposition. Let \(\{A_{n}\}\) be a sequence of subsets in a metric space \((X,d)\) satisfying \(\bigcap_{n=1}^{\infty}A_{n}\neq\emptyset\). If the function \(f:(X,d_{X})\rightarrow (Y,d_{Y})\) is continuous and \(\{A_{n}\}\) has vanashing \(d_{X}\)-diameter, then \(\{f(A_{n})\}\) has vanashing \(d_{Y}\)-diameter. \(\sharp\)
The distance from a point \(x\) to a nonempty subset \(A\) of a pseudo-metric space is defined to be \(D(A,x)=\inf\{d(x,y):y\in A\}\).
Proposition. If \(A\) is a fixed subset of a pseudo-metric space, then the distance from a point \(x\) to \(A\) is a continuous function of \(x\) with respect to the pseudo-metric topology.
Proof. Since \(d(x,z)\leq d(x,y)+d(y,z)\), it follows, taking lower bounds for \(z\) in \(A\), that \(D(A,x)\leq d(x,y)+D(A,y)\). The same inequality holds with \(x\) and \(y\) interchanged and hence \(|D(A,x)-D(A,y)|\leq d(x,y)\). Consequently, if \(y\) is in the open \(r\)-sphere about \(x\), then \(|D(A,x)-D(A,y)|<r\) and continuity follows. \(\blacksquare\)
\begin{equation}{\label{kelp5}}\tag{4}\mbox{}\end{equation}
Proposition \ref{kelp5}. The closure of a set \(A\) in a pseudo-metric space is the set of all points which are zero distance from \(A\).
Proof. Since \(D(A,x)\) is continuous in \(x\), the set \(\{x:D(A,x)=0\}\) is closed and contains \(A\) and hence contains \(\mbox{cl}(A)\). If \(y\not\in \mbox{cl}(A)\), then there is a neighborhood of \(y\), which may be taken to be an open \(r\)-sphere, which does not intersect \(A\). Consequently \(D(A,y)\geq r\) and hence \(\{x:D(A,x)=0\}\) and hence \(\{x:D(A,x)=0\}\subseteq \mbox{cl}(A)\). Therefore \(\mbox{cl}(A)=\{x:D(A,x)=0\}\). \(\blacksquare\)
Proposition. Each pseudo-metric space is normal.
Proof. Let \(A\) and \(B\) be disjoint closed subsets of a pseudo-metric space \(X\), and let \(D(A,x)\) and \(D(B,x)\) be the distonce from \(x\) to \(A\) and \(B\) respectively. Let
\[U=\{x:D(A,x)-D(B,x)<0\}\mbox{ and }V=\{x:D(A,x)-D(B,x)>0\}.\]
The function \(D(A,x)-D(B,x)\) is continuous in \(x\) and therefore \(U\) and \(V\) are open. Clearly \(U\) is disjoint from \(V\), and Proposition \ref{kelp5} it follows \(A\subseteq U\) ($x\in A$ implies \(0=D(A,x)-D(B,x)<0\), i.e. \(x\in U\)) and \(B\subseteq V\). \(\blacksquare\)
\begin{equation}{\label{kelt411}}\tag{5}\mbox{}\end{equation}
Proposition \ref{kelt411}. Every pseudo-metric space satisfies the first axiom of countability. The second axiom of countability is satisfied if and only if the space is separable.
Proof. A set is open relative to the pseudo-metric topology if and only if it contains an open sphere about each of its points. Therefore the family of open spheres about a point \(x\) is a base for the neighborhood system of \(x\). Since each open sphere about \(x\) contains a sphere with rational radius, there is a countable base for the neighborhood system and the space satisfies the first axiom of countability. Each space which satisfies the second axiom of countability is separable, so it remains to prove that a separable pseudo-metric space has a countable base for its topology. Let \(Y\) be a countable dense subset and let \({\cal U}\) be the family of all open spheres with rational radii about members of \(Y\). Surely \({\cal U}\) is countable. If \(U\) is a neighborhood of a point \(x\), there is, for some positive \(r\), an open \(r\)-sphere about \(x\) which is contained in \(U\). Let \(s\) be a positive rational number less than \(r\), let \(y\) be a point of \(Y\) such that \(d(x,y)<s/3\) (since \(Y\) is a dnse subset), and let \(V\) be the open \(2s/3\)-sphere about \(y\). Then \(x\in V\subseteq U\) and hence \({\cal U}\) is a base for the topology. \(\blacksquare\)
Proposition. A net \(\{x_{\alpha}:\alpha\in D\}\) in a pseudo-metric space \((X,d)\) converges to a point \(x\) if and only if \(\{d(x_{\alpha},x):\alpha\in D\}\) converges to zero.
Proof. A net \(\{x_{\alpha}:\alpha\in D\}\) converges to \(x\) if and only if the net is eventually in each open \(r\)-sphere about \(x\), but this is true if and only if \(\{d(x_{\alpha},x):\alpha\in D\}\) is eventually in each \(r\)-sphere about \(0\) in \(\mathbb{R}\) with the usual metric. \(\blacksquare\)
The diameter of a subset \(A\) of a pseudo-metric space \((X,d)\) is \(\sup\{d(x,y):x,y\in A\}\). If this supremum does not exist, the diameter is said to be infinite. It is interesting to notice that the property of having a finite diameter is not a topological invariance.
Proposition. Let \((X,d)\) be a pseudo-metric space and let \(\hat{d}(x,y)=\min\{1,d(x,y)\}\). Then \((X,\hat{d})\) is a pseudo-metric space whose topology is identical with that of \((X,d)\). Consequently, each pseudo-metric space is homeomorphic to a pseudo-metric space of diameter at most one. \(\sharp\)
\begin{equation}{\label{kelp7}}\tag{6}\mbox{}\end{equation}
Proposition \ref{kelp7}. Let \(\{(X_{n},d_{n}):n\in\mathbb{N}\}\) be a sequence of pseudo-metric spaces such that each of diameter is at most one. Let \({\bf X}=\prod_{n\in\mathbb{N}}X_{n}\) be the Cartesian product. For any \({\bf x},{\bf y}\in {\bf X}\), we define
\[d({\bf x},{\bf y})=\sum_{n}\frac{1}{2^{n}}\cdot d_{n}(x_{n},y_{n}).\]
Then \(d\) is a pseudo-metric for the Cartesian product \({\bf X}\), and the pseudo-metric topology for \({\bf X}\) is the product topology. \(\sharp\)
Let \((X,d)\) be a pseudo-metric space and let \(A\) and \(B\) be two subsets of \(X\). The distance between \(A\) and \(B\) is defined by
\[D(A,B)=\mbox{dist}(A,B)=\inf\{d(x,y):x\in A\mbox{ and }y\in B\}.\]
We see that \(D\) is not a pseudo-metric in general. From Proposition \ref{kelp5}, we see that \(\mbox{cl}(\{x\})\) is the set of all points \(y\) with \(d(x,y)=0\). Let \({\cal D}\) be the family of all sets of the form \(\mbox{cl}(\{x\})\). Then \({\cal D}\) is the quotient set \(X/R\), where \(R\) is the relation \(\{(x,y):d(x,y)=0\}\).
\begin{equation}{\label{kelp21}}\tag{7}\mbox{}\end{equation}
Proposition \ref{kelp21}. Let \((X,d)\) be a pseudo-metric space and let \({\cal D}\) be the family of all sets \(\mbox{cl}(\{x\})\) for \(x\) in \(X\). For any \(A,B\in {\cal D}\), we define \(D(A,B)=\mbox{\em dist}(A,B)\). Then \(({\cal D},D)\) is a metric space whose metric topology is the quotient topology for \({\cal D}\), and the projection of \(X\) onto \({\cal D}\) is an isometry. \(\sharp\)
\begin{equation}{\label{d}}\tag{D}\mbox{}\end{equation}
Metrization.
Given a topological space \((X,\tau )\), it is natural to ask whether there is a metric for \(X\) such that \(\tau\) is a metric topology. Such a metric metrizes the topological space and the space is said to be metrizable. A metric generating a topology is called compatible or consistent for the topology. Similarly, a topological space is pseudo-metrizable when there is a pseudo-metric such that the topology is the pseudo-metric topology. A pseudo-metric is a metric if and only if the topology is \(T_{1}\); that is, if and only if the singleton \(\{x\}\) is closed or each point \(x\). This says that a space is metrizable if and only if it is \(T_{1}\) and pseudo-metrizable. Given a nonempty set \(A\) in a metric space \((X,d)\), the distance function \(\mbox{dis}(\cdot ,A)\) is defined by
\[\mbox{dis}(x,A)=\inf_{y\in A}d(x,y).\]
Then, the distance function is Lipschitz continuous.
Proposition. In a metrizable space, every closed set is a \({\cal G}_{\delta}\)-set and every open set is an \({\cal F}_{\sigma}\)-set.
Proof. Let \(F\) be a closed subset of the metric space \((X,d)\), and let
\[G_{n}=\{x:\mbox{dis}(x,F)<\frac{1}{n}\}.\]
It is clearly \(F=\bigcap_{n=1}^{\infty}G_{n}\). Since the distance function is continuous, it says that \(G_{n}\) is open. Therefore, \(F\) is a \({\cal G}_{\delta}\)-set. Since every open set is the complement of a closed set, de Morgan’s law imply that every open set is an \({\cal F}_{\sigma}\)-set. \(\blacksquare\)
Lemma. Let \((X,d)\) be a metric space and let \(A\) and \(B\) be disjoint nonempty closed sets. We define the function \(f:X\rightarrow [0,1]\) by
\[f(x)=\frac{\mbox{dis}(x,A)}{\mbox{ dis}(x,A)+\mbox{dis}(x,B)}.\]
Then \(f\) is continuous on \(X\) and satisfies \(f{-1}(0)=A\) and \(f^{-1}(1)=B\). Moreover, if \(\inf_{x\in A,y\in B}d(x,y)>0\), then the function \(f\) is Lipschitz continuous, i.e., uniformly continuous. \(\sharp\)
Proposition. Every metrizable space is perfectly normal. \(\sharp\)
A topological space \(X\) is completely metrizable when there exists a compatible metric \(d\) on \(X\) such that \((X,d)\) is complete. Let \((X,d)\) be a metric space. A subset \(Y\) of \(X\) is said to be completely metrizable when there exists a compatible metric \(\rho\) such that \((Y,\rho )\) is complete. We recall that a \({\cal G}_{\delta}\)-set is a countable intersection of open sets.
Proposition. We have the following properties.
(i) A completely metrizable subset of a metric space is a \({\cal G}_{\delta}\)-set.
(ii) Every \({\cal G}_{\delta}\)-set in a complete metric space is completely metrizable.
(iii) Every open subset of a complete metric space is completely metrizable. \(\sharp\)
Given a metric space \((X,d)\), we denote by \(U_{d}(X)\) the family of all bounded and uniformly continuous real-valued functions defined on \(X\), and by \(C_{b}(X)\) the family of bounded continuous real-valued functions defined on \(X\).
Proposition. Let \(X\) be a metrizable space and \(d\) be a compatible metric on \(X\). The set \(U_{d}(X)\) is a closed subspace of \(C_{b}(X)\). Thus if \(U_{d}(X)\) is equipped with the uniform metric, then it is a complete metric space. \(\sharp\)
Let \(\{X_{1},X_{2}\cdots\}\) be a countable collection of topological space. Suppose that each \(X_{n}\) is metrizable with the compatible metric \(d_{n}\). Then, we can define a metric on the product space \(X=\prod_{n=1}^{\infty}X_{n}\) by
\begin{equation}{\label{topeq5}}\tag{8}
d(x,y)=\sum_{n=1}^{\infty}\frac{1}{2^{n}}\cdot\frac{d_{n}(x_{n},y_{n})}{1+d_{n}(x_{n},y_{n})},
\end{equation}
where \(x=(x_{n}),y=(y_{n})\in X\). The routine arguments can show that \(d\) is indeed a metric on \(X\). Let \(\{x_{\alpha}\}\) be a net in \(X\) satisfying \(d(x_{\alpha},x)\rightarrow 0\), where \(x_{\alpha}=(x_{n}^{\alpha})\) and \(x=(x_{n})\). Then, we see that \(d(x_{\alpha},x)\rightarrow 0\) if and only if \(d_{n}(x_{n}^{\alpha},x_{n})\stackrel{\alpha}{\rightarrow}0\) for each \(n\). This shows that the product topology for \(X\) coincides with the metric topology generated by \(d\).
Proposition. Let \(\{X_{1},X_{2}\cdots\}\) be a countable collection of topological space. Then, we have the following properties.
(i) The product topology for the product space \(\prod_{n=1}^{\infty}X_{n}\) is metrizable if and only if each topological space \(X_{n}\) is metrizable.
(ii) The product topology for the product space \(\prod_{n=1}^{\infty}X_{n}\) is completely metrizable if and only if each topological space \(X_{n}\) is completely metrizable.
Proof. Suppose that each topological space \(X_{n}\) is metrizable with the compatible metric \(d_{n}\) for \(X_{n}\). The metric \(d\) defined in (\ref{topeq5}) generates the metric topology that coincides with the product topology for the product space \(X\). This shows that the product topology is metrizable. Conversely, let \(d\) be the compatible metric for the product space \(X\). We consider the topological space \(X_{k}\). Also, for each \(n\in\mathbb{N}\), we fix some \(u_{n}\in X_{n}\). Now, for \(x_{k}\in X_{k}\), we define \(\hat{x}=(x_{1},x_{2},\cdots )\in X\), where \(x_{n}=u_{n}\) for \(n\neq k\). We define a metric \(d_{k}\) on \(X_{k}\) by \(d_{k}(x_{k},y_{k})=d(\hat{x},\hat{y})\). Since \(d\) is metric on \(X\), we can show that \(d_{k}\) is also a metric on \(X_{k}\). Since the \(d\)-convergence in the product space \(X\) is equivalent to pointwise convergence. The routine arguments can show that the metric \(d_{k}\) generates the topology for \(X_{k}\). The similar arguments can also be employed in the case of completely metrizable spaces. \(\blacksquare\)
A separable topological space that is completely metrizable is called a Polish space and such a topology is called a Polish topology.
Proposition. We have the following properties.
(i) The product of a countable collection of metrizable spaces is separable if and only if each factor is separable.
(ii) The product of a countable collection of Polish spaces is a Polish space. \(\sharp\)
The Hilbert cube \({\cal H}\) is the set of all real sequences with values in \([0,1]\), i.e., \({\cal H}=[0,1]^{\mathbb{N}}\). The Hilbert cube can be endowed with the product topology. By the Tychonoff’s theorem, the Hilbert cube is compact in the product topology. We can see that the following metric
\[d_{\cal H}(x,y)=\sum_{n=1}^{\infty}\frac{1}{2^{n}}\cdot\left |x_{n}-y_{n}\right |,\]
where \(x=(x_{n}),y=(y_{n})\in {\cal H}\), induces the product topology for \({\cal H}\). In other words, the Hilbert cube is metrizable. We also see that the Hilbert cube \(({\cal H},d_{\cal H})\) is a compact metric space.
\begin{equation}{\label{topt6}}\tag{9}\mbox{}\end{equation}
Theorem \ref{topt6}. (Urysohn Metrization Theorem). Let \(X\) be a Hausdorff space. The following statements are equivalent.
(a) \(X\) can be isometrically embedded in the Hilbert cube.
(b) \(X\) is a separable metrizable space.
(c) \(X\) is regular and second countable. \(\sharp\)
Corollary. We have the following properties..
(i) Every separable metrizable space \(X\) admits a compatible metric \(\rho\) such that \((X,\rho )\) is totally bounded. Consequently, every separable metrizable space has a metrizable compactification (the completion of its totally bounded metric space).
(ii) Every Polish space is a \({\cal G}_{\delta}\)-set in some metrizable compactification.
(iii) The continuous image of a compact metric space in a Hausdorff space is metrizable.
Proof. To prove part (i), let \(X\) be a separable metrizable space. By Theorem \ref{topt6}, there is an embedding \(\pi :X\rightarrow {\cal H}\) satisfying \(\rho (x,y)=d_{\cal H}(\phi (x),\pi (y))\), where \(\rho\) is a compatible metric for \(X\). Since the Hilbert cube \(({\cal H},d_{\cal H})\) is a compact metric space, i.e., totally bounded, we conclude that \((X,\rho )\) is totally bounded by the isometry \(\rho (x,y)=d_{\cal H}(\phi (x),\pi (y))\). \(\blacksquare\)
Theorem. We have the following properties.
(i) The one-point compactification \(X_{\infty}\) of a non-compact locally compact Hausdorff space \(X\) is metrizable if and only if \(X\) is second countable.
(ii) The one-point compactification of a non-compact locally compact separable metrizable space is metrizable. \(\sharp\)
The Cantor set is the countable product \({\cal C}=\{0,1\}^{\mathbb{N}}\), where the two-point set \(\{0,1\}\) is endowed with the discrete topology. In this case, the Cantor set \(\Delta\) is endowed with the product topology. We remark that the set \(\{0,1\}\) can be replaced by \(\{-1,1\}\) or by \(\{0,2\}\), or by any two-point set. The choice of the two-point set often simplifies the proof. For any \(a=(a_{n}),b=(b_{n})\in {\cal C}\), we define the metric
\[d_{\cal C}(a,b)=\sum_{n=1}^{\infty}\frac{1}{3^{n}}\cdot |a_{n}-b_{n}|.\]
Then, this metric \(d_{\cal C}\) generates the product topology for the Cantor set \({\cal C}\), i.e., the Cantor set \({\cal C}\) is metrizable. Also, the Tychonoff’s theorem shows that the Cantor set \({\cal C}\) is compact in the product topology, i.e., \(({\cal C},d_{\cal C})\) is a compact metric space. The Cantor set can be thought of as a subset of \([0,1]\), which can be constructed from the closed interval \([0,1]\) by removing the “open middle third-intervals” inductively as follows. Starting with \(C_{0}=[0,1]\), we divide \([0,1]\) into three equal sub-intervals \([0,\frac{1}{3}]\), \((\frac{1}{3},\frac{2}{3})\) and \([\frac{2}{3},1]\). We remove the open middle interval, i.e., \((\frac{1}{3},\frac{2}{3})\), and let \(C_{1}=[0,\frac{1}{3}]\cup [\frac{2}{3},1]\). Inductively, if \(C_{n}\) consists of \(2^{n}\) closed intervals, we subdivide each one of them into three equal subintervals and delete from each one of them the open middle sub-interval. The union of the remaining \(2^{n+1}\) closed sub-intervals is \(C_{n+1}\). In this process, the Cantor set is the compact set defined by $latex C=\bigcap_{n=1}^{\infty}
C_{n}$. From the above construction, we see that
\begin{equation}{\label{topeq8}}\tag{10}
C=\left\{\sum_{n=1}^{\infty}\frac{a_{n}}{3^{n}}:a_{n}=0\mbox{ or }a_{n}=2\right\}.
\end{equation}
A nonempty set \(A\) in a topological space \(X\) is a retract of \(X\) when there is a continuous function \(f:X\rightarrow A\) satisfying \(f(x)=x\) for each \(x\in A\). The function \(f\) os called a retraction of \(X\) onto \(A\). Note that is \(A\) is a retract of \(X\) and \(A\subseteq B\subseteq X\), then \(A\) is also a retract of \(B\) under the retraction \(f|_{B}\).
\begin{equation}{\label{topp7}}\tag{11}\mbox{}\end{equation}
Proposition \ref{topp7}. We have the following properties.
(i) The Cantor set defined in (\ref{topeq8}) is an uncountable, perfect and nowhere dense set of Lebesgue measure zero.
(ii) The Cantor set \({\cal C}=\{0,2\}^{\mathbb{N}}\) is homeomorphic \(C\).
(iii) The Cantor set \({\cal C}\) is homeomorphic to \({\cal C}^{\mathbb{N}}\).
(iv) Both the closed interval \([0,1]\) and the Hilbert cube \({\cal H}\) are continuous images of the Cantor set \({\cal C}=\{0,1\}^{\mathbb{N}}\).
(v) Any nonempty closed subset of the Cantor set \({\cal C}\) is a retract of \({\cal C}\).
Proof. To prove part (ii), we define
\[\phi :\{0,2\}^{\mathbb{N}}\rightarrow C\mbox{ by }\phi (a_{1},a_{2},\cdots )=\sum_{n=1}^{\infty}\frac{a_{n}}{3^{n}}.\]
We can verify that \(\phi\) is one-to-one, onto and continuous. \(\blacksquare\)
Theorem. Every compact metrizable space is a continuous image of the Cantor set \({\cal C}\).
Proof. Let \(X\) be a compact metrizable space. By Theorem \ref{topt6}, \(X\) is homeomorphic to a closed subset \(Y\) of the Hilbert cube \({\cal H}\). Let \(\phi :Y\rightarrow X\) be such a homeomorphism. By part (iv) of Proposition \ref{topp7}, there exists a continuous function \(\psi\) from the Cantor set \({\cal C}\) onto the Hilbert cube \({\cal H}\). This says that the inverse image \(\psi^{-1}(Y)\) is a closed subset of \({\cal C}\). By part (v) of Proposition \ref{topp7}, there exists a continuous retraction \(f:{\cal C}\rightarrow\psi^{-1}(Y)\) satisfying \(f(z)=z\) for each \(z\in\psi^{-1}(Y)\). Therefore, we have
\[{\cal C}\stackrel{f}{\rightarrow}\psi^{-1}(Y)\stackrel{\psi}{\rightarrow}Y\stackrel{\phi}{\rightarrow}X,\]
which says that \(\phi\circ\psi\circ f\) is a continuous function from \({\cal C}\) onto \(X\). \(\blacksquare\)
According to Proposition \ref{kelp7}, the product of countably many pseudo-metric spaces is pseudo-metrizable. According to the embedding Lemma \ref{top262}, if \(F\) is a family of continuous functions on a \(T_{1}\)-space \(X\), where a member \(f\) of \(F\) maps \(X\) into a space \(Y_{f}\), then the evaluation map of \(X\) into \(\prod\{Y_{f}:f\in F\}\) is a homeomorphism whenever \(F\) distinguishes points and closed sets (that is, if \(A\) is a closed subset of \(X\) and \(x\) is a member of \(X\setminus A\), then \(f(x)\not\in \mbox{cl}(f(A))\) for some member \(f\) of \(F\)). The problem of metrizing a \(T_{1}\)-space \(X\) then reduces to that of finding a countable family of continuous functions, each on \(X\) to some pseudo-metrizable space, such that \(F\) distinguishes points and closed sets. (A pseudo-metrizable \(T_{1}\)-space is necessarily metrizable). For convenience, let \(Q^{\mathbb{N}}\) denote the product of the closed unit interval with itself countably many times; that is, \(Q^{\mathbb{N}}\) is the set of all functions on the nonnegative integers to the closed unit interval \(Q\), with the product topology.
\begin{equation}{\label{kelt416}}\tag{12}\mbox{}\end{equation}
Theorem \ref{kelt416}. (Urysohn Metrization Theorem). A regular \(T_{1}\)-space whose topology has a countable base is homeomorphic to a subspace of the cube \(Q^{\mathbb{N}}\) and is hence metrizable. \(\sharp\)
Proposition. Suppose that \(X\) is a \(T_{1}\)-space. Then, the following statements are equivalent.
(a) \(X\) is regular and there is a countable base for its topology.
(b) \(X\) is homeomorphic to a subspace of the cube \(Q^{\mathbb{N}}\).
(c) \(X\) is metrizable and separable.
Proof. Theorem \ref{kelt416} shows that (a) implies (b). The cube \(Q^{\mathbb{N}}\) is metrizable by Proposition \ref{kelp7}, and satisfies the second axiom of countability (the product topology has a countable base if and only if the topology of each coordinate space has a countable base and all but a countable number of the coordinate spaces are indiscrete). Hence each subspace is metrizable and satisfies the second axiom of countability and is therefore separable. Hence (b) implies (c). (Caution: it is not true that a subspace of a separable space is necessarily separable). Finally, to prove that (c) implies (a),
if \(X\) is metrizable and separable, then it is surely regular and by Proposition \ref{kelt411} it satisfies the second axiom of countability. \(\blacksquare\)
A family \({\cal A}\) of subsets of a topological space is locally finite when each point of the space has a neighborhood which intersects only finitely many elements of \({\cal A}\). It follows immediately from the definition that a point is an accumulation point of the union \(\bigcup\{A:A\in {\cal A}\}\) if and only if it is an accumulation point of some member of \({\cal A}\), and hence the closure of the union is the union of the closures; that is,
\[\mbox{cl}(\bigcup\{A:A\in {\cal A}\})=\bigcup\{\mbox{cl}(A):A\in {\cal A}\}.\]
It is also evident that the family of all closures of members of \({\cal A}\) is locally finite. A family \({\cal A}\) is {\bf discrete} if each point of the space has a neighborhood which intersects at most one member of \({\cal A}\). A discrete family is locally finite, and if \({\cal A}\) is discrete, then the family of closures of members of \({\cal A}\) is also discrete. Finally, a family \({\cal A}\) is \(\sigma\)-locally finite (resp. \(\sigma\)-locally dsicrete) when it is the union of a countable number of locally finite (resp. discrete) subfamilies.
Proposition. A regular space whose topology has a \(\sigma\)-locally finite base is normal. \(\sharp\)
Proposition. A regular \(T_{1}\)-space whose topology has a \(\sigma\)-locally finite base is metrizable. \(\sharp\)
A cover \({\cal B}\) of a set \(X\) is a refinement of a cover \({\cal A}\) when each member of \({\cal B}\) is a subset of a member of \({\cal A}\). For example, in a metric space, the family of all open spheres of radius one half is a refinement of the family of all open spheres of radius one.
\begin{equation}{\label{kelp14}}\tag{13}\mbox{}\end{equation}
Proposition \ref{kelp14}. Each open cover of a pseudo-metrizable space has an open \(\sigma\)-discrete refinement. \(\sharp\)
Theorem. (Metrization Theorem). The following three conditions on a topological space are equivalent.
(a) The space is metrizable.
(b) The space is \(T_{1}\) and regular, and the topology has a \(\sigma\)-locally finite base.
(c) The space is \(T_{1}\) and regular, and the topology has a \(\sigma\)-discrete base. \(\sharp\)
Let \(X\) be a compact space and \(Y\) be a metrizable space. We denote by \(C(X,Y)\) the set of all continuous functions from \(X\) to \(Y\). If \(\rho\) is a compatible metric on \(Y\), then, for any \(f,g\in C(X,Y)\), the following formula
\[d_{\rho}(f,g)=\sup_{x\in X}\rho (f(x),g(x))\]
defines a metric on \(C(X,Y)\). Since \(X\) is compact, we have \(d_{\rho}(f,g)<\infty\) for any \(f,g\in C(X,Y)\). This says that \((C(X,Y),d_{\rho})\) is a metric space. The metric \(d_{\rho}\) characterizes the topology of uniform convergence on \(X\) of functions in \(C(X,Y)\).
\begin{equation}{\label{topp10}}\tag{14}\mbox{}\end{equation}
Proposition \ref{topp10}. Let \(X\) be a compact space and \(Y\) be a metrizable space such that \(\rho\) is a compatible metric on \(Y\). The metric space \((C(X,Y),d_{\rho})\) is complete if and only if \((Y,\rho )\) is complete. \(\sharp\)
\begin{equation}{\label{topp9}}\tag{15}\mbox{}\end{equation}
Proposition \ref{topp9}. Let \(X\) be a compact space and \(Y\) be a metrizable space. If \(\rho_{1}\) and \(\rho_{2}\) are two compatible metrics on \(Y\), then \(d_{\rho_{1}}\) and \(d_{\rho_{2}}\) are equivalent metrics on \(C(X,Y)\); that is, the metrics \(d_{\rho_{1}}\) and \(d_{\rho_{2}}\) generate the same metric topology for \(C(X,Y)\). \(\sharp\)
Proposition \ref{topp9} says that the metric topology on \(C(X,Y)\) induced by \(d_{\rho}\) depends only on the topology of \(Y\), not on the particular metric \(\rho\) on \(Y\). Therefore, we can regard \(C(X,Y)\) as a topological space without specifying a metric for \(Y\). In this case, we can refer simply to the topology of uniform convergence for \(C(X,Y)\). If \((Y,\parallel\cdot\parallel )\) is a normed space, then, under the usual pointwise algebraic operations, \(C(X,Y)\) is a vector space that becomes a normed space under the norm
\[\parallel f\parallel =\sup_{x\in X}\parallel f(x)\parallel.\]
If \(Y\) is a Banach space, then \(C(X,Y)\) is a Banach space by Proposition \ref{topp10}.
Proposition. Let \(X\) be a compact and metrizable space and \(Y\) be a separable and metrizable space. Then, the metric space \(C(X,Y)\) is separable. \(\sharp\)
Proposition. Let \((X,d)\) be a metric space. Then, we have the following properties.
(i) Let \(f:(X,d)\rightarrow\mathbb{R}\) be bounded below. Then \(f\) is lower semicontinuous if and only if it is the pointwise limit of an increasing sequence of Lipschitz continuous functions.
(ii) Let \(g:(X,d)\rightarrow\mathbb{R}\) be bounded above. Then \(f\) is upper semicontinuous if and only if it is the pointwise limit of a decreasing sequence of Lipschitz continuous functions.
(iii) Let \(F\) be a closed subset of \(X\). Then, there is a sequence \(\{f_{n}\}\) of Lipschitz continuous functions taking values on \([0,1]\) satisfying \(f_{n}(x)\downarrow\chi_{F}(x)\) for all \(x\in X\).
(iv) Let \(f:(X,d)\rightarrow\mathbb{R}\) be a bounded and continuous function. Then, there exists sequences of bounded Lipschitz continuous functions \(\{g_{n}\}\) and \(\{h_{n}\}\) satisfying \(g_{n}(x)\uparrow f(x)\) and \(h_{n}(x)\downarrow f(x)\) for all \(x\in X\). \(\sharp\)
