J. Munkres. Topology (2013) Chapter 2: Topological Spaces and Continuous Functions. 21: The Metric Topology (continued)

Proposition

Let \((X,d)\) be a metric space and \(Y \subseteq X\). Then \(d \restriction Y^2\) is a metric on \(Y\) that induces the subspace topology.

Proposition

Every metric space is Hausdorff.

Proposition

Let \(((X_n, d_n))\) be a sequence of metric spaces. Define \(D\) on \(prod_n X_n\) by \(D((x_n)) = \sup_n (\overline{d_n}(x_n, y_n)/n)\), where \(\overline{d_n}\) is the standard bounded metric corresponding to \(d_n\). Then \(D\) is a metric on \(\prod_n X_n\) that induces the product topology.

Theorem 21.1

Let \((X, d_X)\) and \((Y, d_Y)\) be metric spaces. Let \(f : X \rightarrow Y\). Then \(f\) is continuous if and only if, for all \(x \in X\) and \(\epsilon > 0\), there exists \(\delta > 0\) such that, for all \(y \in X\),

\[ d_X(x,y) < \delta \Rightarrow d_Y(f(x),f(y)) < \epsilon \enspace . \]

Definition

Let \(X\) be a topological space. Let \(x \in X\). A basis at the point \(x\) is a set \(\mathcal{B}\) of neighbourhoods of \(x\) such that every neighbourhood of \(x\) includes a member of \(\mathcal{B}\).

Definition

A topological space satisfies the first countability axiom iff every point has a countable basis.

Proposition

Every metric space is first countable.

Lemma 21.2 (The sequence lemma)

Let \(X\) be a topological space. Let \(A \subseteq X\). Let \(x \in X\). If there is a sequence of points of \(A\) converging to \(x\), then \(x \in \overline{A}\). The converse holds if \(X\) is first countable.

Theorem 21.3

Let \(X\) and \(Y\) be topological spaces. Let \(f : X \rightarrow Y\). If \(f\) is continuous, then whenever \(x_n \rightarrow l\) as \(n \rightarrow \infty\) in \(X\), we have \(f(x_n) \rightarrow f(l)\) as \(n \rightarrow \infty\) in \(Y\). The converse holds if \(X\) is first countable.

Lemma 21.4

Addition, subtraction and multiplication are continuous functions \(\mathbb{R}^2 \rightarrow \mathbb{R}\). Division is a continuous function \(\mathbb{R} \times (\mathbb{R} - \{0\}) \rightarrow \mathbb{R}\).

Definition

Let \(X\) be a set and \(Y\) a metric space. Let \((f_n)\) be a sequence of functions from \(X\) to \(Y\) and \(f : X \rightarrow Y\). Then \((f_n)\) converges uniformly to \(f\) iff, for all \(\epsilon > 0\), there exists \(N\) such that, for all \(n > N\) and \(x \in X\),

\[ d(f_n(x),f(x)) < \epsilon \enspace . \]

Proposition

Let \(X\) be a set and \(Y\) a metric space. Let \((f_n)\) be a sequence of functions from \(X\) to \(Y\) and \(f : X \rightarrow Y\). Then \((f_n)\) converges uniformly to \(f\) iff \((f_n)\) converges to \(f\) in \(Y^X\) under the uniform topology.

Theorem 21.6 (Uniform Limit Theorem)

Let \(X\) be a topological space and \(Y\) a metric space. Let \((f_n)\) be a sequence of continuous functions from \(X\) to \(Y\) and \(f : X \rightarrow Y\). If \((f_n)\) converges uniformly to \(f\), then \(f\) is continuous.

Example 1

\(\mathbb{R}^\omega\) in the box topology is not first countable, hence not metrizable.

Example 2

For \(J\) an uncountable set, we have \(\mathbb{R}^J\) in the product topology is not first countable, hence not metrizable.

From the Exercises

Definition

Let \(X\) and \(Y\) be metric spaces. An isometric embedding of \(X\) in \(Y\) is a function \(f : X \rightarrow Y\) such that, for all \(x,y \in X\), we have \(d(f(x),f(y)) = d(x,y)\).

Proposition

Every isometric embedding is an embedding.

Examples

\(\mathbb{R}_l\) and the ordered square are first countable.

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