§ 21. The Metric Topology (continued)¶
Note
Subspaces of metric spaces behave the way one would wish them to; if \(A\) is a subspace of the topological space \(X\) and \(d\) is a metric for \(X\), then the restriction of \(d\) to \(A \times A\) is a metric for the topology of \(A\).
About order topologies there is nothing to be said; some are metrizable (for instance, \(\mathbb{Z}_{+}\)and \(\mathbb{R}\) ), and others are not, as we shall see.
The Hausdorff axiom is satisfied by every metric topology. If \(x\) and \(y\) are distinct points of the metric space \((X, d)\), we let \(\epsilon=\frac{1}{2} d(x, y)\); then the triangle inequality implies that \(B_{d}(x, \epsilon)\) and \(B_{d}(y, \epsilon)\) are disjoint.
The product topology we have already considered in special cases; we have proved that the products \(\mathbb{R}^{n}\) and \(\mathbb{R}^{\omega}\) are metrizable. It is true in general that countable products of metrizable spaces are metrizable; the proof follows a pattern similar to the proof for \(\mathbb{R}^{\omega}\).
Theorem 21.1
Let \(f: X \rightarrow Y\); let \(X\) and \(Y\) be metrizable with metrics \(d_{X}\) and \(d_{Y}\), respectively. Then continuity of \(f\) is equivalent to the requirement that given \(x \in X\) and given \(\epsilon>0\), there exists \(\delta>0\) such that
Proof
Suppose that \(f\) is continuous. Given \(x\) and \(\epsilon\), consider the set
which is open in \(X\) and contains the point \(x\). It contains some \(\delta\)-ball \(B(x, \delta)\) centered at \(x\). If \(y\) is in this \(\delta\)-ball, then \(f(y)\) is in the \(\epsilon\)-ball centered at \(f(x)\), as desired.
Conversely, suppose that the \(\epsilon-\delta\) condition is satisfied. Let \(V\) be open in \(Y\); we show that \(f^{-1}(V)\) is open in \(X\). Let \(x\) be a point of the set \(f^{-1}(V)\). Since \(f(x) \in\) \(V\), there is an \(\epsilon\)-ball \(B(f(x), \epsilon)\) centered at \(f(x)\) and contained in \(V\). By the \(\epsilon\) \(\delta\) condition, there is a \(\delta\)-ball \(B(x, \delta)\) centered at \(x\) such that \(f(B(x, \delta)) \subset B(f(x), \epsilon)\). Then \(B(x, \delta)\) is a neighborhood of \(x\) contained in \(f^{-1}(V)\), so that \(f^{-1}(V)\) is open, as desired.
Lemma 21.2 : The Sequence Lemma
Let \(X\) be a topological space; let \(A \subset X\). If there is a sequence of points of \(A\) converging to \(x\), then \(x \in \bar{A}\); the converse holds if \(X\) is metrizable.
Proof
Suppose that \(x_{n} \rightarrow x\), where \(x_{n} \in A\). Then every neighborhood \(U\) of \(x\) contains a point of \(A\), so \(x \in \bar{A}\) by Theorem 17.5.
Conversely, suppose that \(X\) is metrizable and \(x \in \bar{A}\). Let \(d\) be a metric for the topology of \(X\). For each positive integer \(n\), take the neighborhood \(B_{d}(x, 1 / n)\) of radius \(1 / n\) of \(x\), and choose \(x_{n}\) to be a point of its intersection with \(A\). We assert that the sequence \(x_{n}\) converges to \(x\) : Any open set \(U\) containing \(x\) contains an \(\epsilon\)-ball \(B_{d}(x, \epsilon)\) centered at \(x\); if we choose \(N\) so that \(1 / N<\epsilon\), then \(U\) contains \(x_{i}\) for all \(i \geq N\).
Theorem 21.3
Let \(f: X \rightarrow Y\). If the function \(f\) is continuous, then for every convergent sequence \(x_{n} \rightarrow x\) in \(X\), the sequence \(f\left(x_{n}\right)\) converges to \(f(x)\). The converse holds if \(X\) is metrizable.
Proof
Assume that \(f\) is continuous. Given \(x_{n} \rightarrow x\), we wish to show that \(f\left(x_{n}\right) \rightarrow\) \(f(x)\). Let \(V\) be a neighborhood of \(f(x)\). Then \(f^{-1}(V)\) is a neighborhood of \(x\), and so there is an \(N\) such that \(x_{n} \in f^{-1}(V)\) for \(n \geq N\). Then \(f\left(x_{n}\right) \in V\) for \(n \geq N\).
To prove the converse, assume that the convergent sequence condition is satisfied. Let \(A\) be a subset of \(X\); we show that \(f(\bar{A}) \subset \overline{f(A)}\). If \(x \in \bar{A}\), then there is a sequence \(x_{n}\) of points of \(A\) converging to \(x\) (by Lemma 21.2). By assumption, the sequence \(f\left(x_{n}\right)\) converges to \(f(x)\). Since \(f\left(x_{n}\right) \in f(A)\), the preceding lemma implies that \(f(x) \in \overline{f(A)}\). (Note that metrizability of \(Y\) is not needed.) Hence \(f(\bar{A}) \subset\) \(\overline{f(A)}\), as desired.
Definition 21.1 : Countable Basis at the Point \(x\) / First Countability Axiom
A space \(X\) is said to have a countable basis at the point \(x\) if there is a countable collection \(\left\{U_{n}\right\}_{n \in \mathbb{Z}_{+}}\)of neighborhoods of \(x\) such that any neighborhood \(U\) of \(x\) contains at least one of the sets \(U_{n}\). A space \(X\) that has a countable basis at each of its points is said to satisfy the first countability axiom.
Lemma 21.4
The addition, subtraction, and multiplication operations are continuous functions from \(\mathbb{R} \times \mathbb{R}\) into \(\mathbb{R}\); and the quotient operation is a continuous function from \(\mathbb{R} \times(\mathbb{R}-\{0\})\) into \(\mathbb{R}\).
Theorem 21.5
If \(X\) is a topological space, and if \(f, g: X \rightarrow \mathbb{R}\) are continuous functions, then \(f+g, f-g\), and \(f \cdot g\) are continuous. If \(g(x) \neq 0\) for all \(x\), then \(f / g\) is continuous.
Proof
The map \(h: X \rightarrow \mathbb{R} \times \mathbb{R}\) defined by
is continuous, by Theorem 18.4. The function \(f+g\) equals the composite of \(h\) and the addition operation
therefore \(f+g\) is continuous. Similar arguments apply to \(f-g, f \cdot g\), and \(f / g\).
Definition 21.2 : Converge Uniformly
Let \(f_{n}: X \rightarrow Y\) be a sequence of functions from the set \(X\) to the metric space \(Y\). Let \(d\) be the metric for \(Y\). We say that the sequence \(\left(f_{n}\right)\) converges uniformly to the function \(f: X \rightarrow Y\) if given \(\epsilon>0\), there exists an integer \(N\) such that
for all \(n>N\) and all \(x\) in \(X\).
Theorem 21.6 : Uniform Limit Theorem
Let \(f_{n}: X \rightarrow Y\) be a sequence of continuous functions from the topological space \(X\) to the metric space \(Y\). If \(\left(f_{n}\right)\) converges uniformly to \(f\), then \(f\) is continuous.
Proof
Let \(V\) be open in \(Y\); let \(x_{0}\) be a point of \(f^{-1}(V)\). We wish to find a neighborhood \(U\) of \(x_{0}\) such that \(f(U) \subset V\).
Let \(y_{0}=f\left(x_{0}\right)\). First choose \(\epsilon\) so that the \(\epsilon\)-ball \(B\left(y_{0}, \epsilon\right)\) is contained in \(V\). Then, using uniform convergence, choose \(N\) so that for all \(n \geq N\) and all \(x \in X\),
Finally, using continuity of \(f_{N}\), choose a neighborhood \(U\) of \(x_{0}\) such that \(f_{N}\) carries \(U\) into the \(\epsilon / 3\) ball in \(Y\) centered at \(f_{N}\left(x_{0}\right)\).
We claim that \(f\) carries \(U\) into \(B\left(y_{0}, \epsilon\right)\) and hence into \(V\), as desired. For this purpose, note that if \(x \in U\), then
Adding and using the triangle inequality, we see that \(d\left(f(x), f\left(x_{0}\right)\right)<\epsilon\), as desired.
Note
Let us remark that the notion of uniform convergence is related to the definition of the uniform metric. Consider, for example, the space \(\mathbb{R}^{X}\) of all functions \(f: X \rightarrow \mathbb{R}\), in the uniform metric \(\bar{\rho}\). It is not difficult to see that a sequence of functions \(f_{n}: X \rightarrow \mathbb{R}\) converges uniformly to \(f\) if and only if the sequence \(\left(f_{n}\right)\) converges to \(f\) when they are considered as elements of the metric space \(\left(\mathbb{R}^{X}, \bar{\rho}\right)\).
Example
\(\mathbb{R}^{\omega}\) in the box topology is not metrizable.
We shall show that the sequence lemma does not hold for \(\mathbb{R}^{\omega}\). Let \(A\) be the subset of \(\mathbb{R}^{\omega}\) consisting of those points all of whose coordinates are positive:
Let \(\mathbf{0}\) be the "origin" in \(\mathbb{R}^{\omega}\), that is, the point \((0,0, \ldots)\) each of whose coordinates is zero. In the box topology, \(\mathbf{0}\) belongs to \(\bar{A}\); for if
is any basis element containing \(\mathbf{0}\), then \(B\) intersects \(A\). For instance, the point
belongs to \(B \cap A\).
But we assert that there is no sequence of points of \(A\) converging to \(\mathbf{0}\). For let \(\left(\mathbf{a}_{n}\right)\) be a sequence of points of \(A\), where
Every coordinate \(x_{i n}\) is positive, so we can construct a basis element \(B^{\prime}\) for the box topology on \(\mathbb{R}\) by setting
Then \(B^{\prime}\) contains the origin \(\mathbf{0}\), but it contains no member of the sequence \(\left(\mathbf{a}_{n}\right)\); the point \(\mathbf{a}_{n}\) cannot belong to \(B^{\prime}\) because its \(n\)th coordinate \(x_{n n}\) does not belong to the interval \(\left(-x_{n n}, x_{n n}\right)\). Hence the sequence \(\left(\mathbf{a}_{n}\right)\) cannot converge to \(\mathbf{0}\) in the box topology.
Example
An uncountable product of \(\mathbb{R}\) with itself is not metrizable.
Let \(J\) be an uncountable index set; we show that \(\mathbb{R}^{J}\) does not satisfy the sequence lemma (in the product topology).
Let \(A\) be the subset of \(\mathbb{R}^{J}\) consisting of all points \(\left(x_{\alpha}\right)\) such that \(x_{\alpha}=1\) for all but finitely many values of \(\alpha\). Let \(\mathbf{0}\) be the "origin" in \(\mathbb{R}^{J}\), the point each of whose coordinates is 0 .
We assert that \(\mathbf{0}\) belongs to the closure of \(A\). Let \(\prod U_{\alpha}\) be a basis element containing \(\mathbf{0}\). Then \(U_{\alpha} \neq \mathbb{R}\) for only finitely many values of \(\alpha\), say for \(\alpha=\alpha_{1}, \ldots, \alpha_{n}\). Let \(\left(x_{\alpha}\right)\) be the point of \(A\) defined by letting \(x_{\alpha}=0\) for \(\alpha=\alpha_{1}, \ldots, \alpha_{n}\) and \(x_{\alpha}=1\) for all other values of \(\alpha\); then \(\left(x_{\alpha}\right) \in A \cap \prod U_{\alpha}\), as desired.
But there is no sequence of points of \(A\) converging to \(\mathbf{0}\). For let \(\mathbf{a}_{n}\) be a sequence of points of \(A\). Given \(n\), let \(J_{n}\) denote the subset of \(J\) consisting of those indices \(\alpha\) for which the \(\alpha\) th coordinate of \(\mathbf{a}_{n}\) is different from 1. The union of all the sets \(J_{n}\) is a countable union of finite sets and therefore countable. Because \(J\) itself is uncountable, there is an index in \(J\), say \(\beta\), that does not lie in any of the sets \(J_{n}\). This means that for each of the points \(\mathbf{a}_{n}\), its \(\beta\) th coordinate equals 1 .
Now let \(U_{\beta}\) be the open interval \((-1,1)\) in \(\mathbb{R}\), and let \(U\) be the open set \(\pi_{\beta}^{-1}\left(U_{\beta}\right)\) in \(\mathbb{R}^{J}\). The set \(U\) is a neighborhood of \(\mathbf{0}\) that contains none of the points \(\mathbf{a}_{n}\); therefore, the sequence \(\mathbf{a}_{n}\) cannot converge to \(\mathbf{0}\).