Metric Spaces

Definition 1. A metric on a set \(X\) is a function \(d: X\times X \to \mathbb{R}\) such that

For all \(x,y\in X\), one has \(d(x,y)=0\) if and only if \(x=y\).
For all \(x,y\in X\), one has \(d(y,x)=d(x,y)\).
For all \(x,y,z\in X\), one has \(d(x,z)\leq d(x,y)+d(y,z)\).

If \(d\) is a metric on \(X\), we call \((X,d)\) a metric space. We may occasionaly say \(X\) is a metric space with metric \(d\).

Theorem 1. If \(X\) is a metric space with metric \(d\), then for all \(x,y\in X\) one has \(d(x,y)\geq 0\).

Proof. Let \(X\) be a metric space with metric \(d\) and let \(x,y\in X\). Then \[0 = d(x,x) \leq d(x,y) + d(y,x) = 2d(x,y)\] and it follows that \(0\leq d(x,y)\). ◻

Definition 2. Let \(X\) and \(Y\) be metric spaces with metrics \(d_X\) and \(d_Y\) respectively. A function \(f:X\to Y\) is an isometric embedding if for all \(x_1,x_2\in X\) \[d_Y(f(x_1), f(x_2)) = d_X(x_1, x_2)\] An isometry is a bijective isometric embedding.

Theorem 2. An isometric embedding is injective.

Proof. Let \(X\) and \(Y\) be metric spaces with metrics \(d_X\) and \(d_Y\) respectively. Let \(f:X\to Y\) be an isometric embedding. Suppose \(f(x_1)=f(x_2)\). Then \[0 = d_Y(f(x_1), f(x_2)) = d_X(x_1, x_2)\] and it follows that \(x_1 = x_2\). Hence \(f\) is injective. ◻

Definition 3. Let \(X\) be a metric space with metric \(d\). Given \(x\in X\) and \(r > 0\), the ball of radius \(r\) centered at \(x\) is the set \[B_r(x) = \{y\in X : d(x,y) < r\}\]

Definition 4. Let \((X,d)\) be a metric space. A subset \(U\subseteq X\) is an open set with respect to \(d\) if for each \(x\in U\), there is \(r> 0\) such that \(B_r(x)\subseteq U\).

Theorem 3. Let \((X,d)\) be a metric space. For each \(x\in X\) and \(r> 0\), the ball \(B_r(x)\) is an open set with respect to \(d\).

Proof. Let \(x\in X\) and \(r>0\). Let \(y\in B_r(x)\). Then \(d(x,y) < r\). Let \(R = r - d(x,y) > 0\). Let \(u\in B_R(y)\). Then \[d(x,u) \leq d(x,y) + d(y,u) < d(x,y) + R = r\] and so \(u\in B_r(x)\). It follows that \(B_R(y)\subseteq B_r(x)\). Hence \(B_r(x)\) is an open set with respect to \(d\). ◻

Theorem 4. Let \((X,d)\) be a metric space. Let \(\mathcal{B} = \{B_r(x) : x\in X,\ r>0\}\). Then \(\mathcal{B}\) is a basis for a topology on \(X\).

Proof. Let \(x\in X\). Then \(x\in B_1(x)\in \mathcal{B}\). Suppose there are \(B_1,B_2\in\mathcal{B}\) with \(x\in B_1\cap B_2\). Then there are \(a,b\in X\) and \(r,s>0\) with \(B_1=B_r(a)\) and \(B_2=B_s(b)\). Then \(d(x,a)<r\) and \(d(x,b)<s\). Let \(t=\min\{r-d(x,a), s-d(x,b)\}\). Since \(r>0\) and \(s>0\), we have \(t>0\). Let \(B = B_t(x)\). Then \(B\in\mathcal{B}\). Let \(y\in B\). Then \(d(x,y)<t\). Hence \[d(a,y) \leq d(a,x) + d(x,y) < r\] and \[d(b,y) \leq d(b,x) + d(x,y) < s.\] Then \(y\in B_1\cap B_2\) and consequently \(x\in B\subseteq B_1\cap B_2\). Then \(\mathcal{B}\) is a basis for a topology on \(X\). ◻

Theorem 5. Let \((X,d)\) be a metric space. Let \(\mathcal{B} = \{B_r(x) : x\in X,\ r>0\}\). Then \(U\in\tau_\mathcal{B}\) if and only if \(U\subseteq X\) is an open set with respect to \(d\).

Proof. Suppose \(U\in\tau_\mathcal{B}\). If \(U\) is empty, it vacuously follows that \(U\) is an open set. Suppose \(U\) is nonempty and let \(x\in U\). Then there is \(B\in\mathcal{B}\) such that \(x\in B\subseteq U\). Moreover, there is \(a\in X\) and \(r>0\) such that \(B=B_r(a)\). If \(a=x\), then \(U\) is an open set. Suppose \(a\neq x\). Then \(d(a,x) > 0\). Let \(s=r-d(a,x)\). Since \(x\in B\), we have \(d(a,x)<r\) and so \(s>0\). If \(d(x,y)<s\), then \[d(a,y) \leq d(a,x) + d(x,y) < r.\] It follows that \(x\in B_s(x) \subseteq B_r(a) = B \subseteq U\). Hence \(U\) is an open set with respect to \(d\).

Suppose \(V\subseteq X\) is an open set with respect to \(d\). Let \(x\in V\). Then there is \(r>0\) such that \(B_r(x)\subseteq V\). Let \(B = B_r(x)\). Then \(B\in\mathcal{B}\) and \(x\in B\subseteq V\). Hence \(V\in\tau_\mathcal{B}\). ◻

It follows that there is no ambiguity by referring to an open set with respect to \(d\) as, simply, an open set.

Theorem 6. Let \((X,d)\) be a metric space. Let \(\mathcal{B} = \{B_r(x) : x\in X,\ r>0\}\). Then \((X, \tau_\mathcal{B})\) is Hausdorff.

Proof. Let \(x,y\in X\) with \(x\neq y\). Then \(d(x,y)>0\). Let \(r=d(x,y)/2\). Then \(r>0\). Let \(U=B_r(x)\) and \(V=B_r(y)\). Then \(U,V\in\mathcal{B}\) so that \(U,V\in\tau_\mathcal{B}\). If \(a\in U\), then \(d(a,x)<r\) and \[d(a,y) \geq d(x,y) - d(a,x) > d(x,y) - r = r\] so that \(a\notin V\). If \(b\in V\), then \(d(b,y) < r\) and \[d(b,x) \geq d(x,y) - d(b,y) > d(x,y) - r = r\] so that \(b\notin U\). Then \(x\in U\), \(y\in V\), and \(U\cap V=\emptyset\). It follows that \((X,\tau_\mathcal{B})\) is Hausdorff. ◻

Definition 5. Let \((X,\tau)\) be a topological space. If there is a metric \(d\) such that \((X,d)\) is a metric space, then \((X,\tau)\) is metrizable.

Example 1. If \(X\) is an infinite set with the finite complement topology, then \(X\) is not metrizable.

Proof. Let \(X\) be an infinite set with the finite complement topology. Then \(X\) is not Hausdorff. Hence \(X\) is not metrizable. ◻

Definition 6. Let \(X\) be a metric space. Let \(A\subseteq X\). The closure of a \(A\), denoted \(\overline{A}\), is the intersection of all closed sets containing \(A\). The interior of \(A\), denoted \(\text{int}A\), is the union of all open sets contained in \(A\). The boundary of \(A\), denoted \(\text{bd}A\), is the set \(\overline{A}\setminus\text{int}A\).

Definition 7. Let \(X\) and \(Y\) be metric spaces. A function \(f:X\to Y\) is continuous at \(x\in X\) if for all \(\varepsilon > 0\) there is \(\delta > 0\) so that \(s\in B_\delta(x) \implies f(s) \in B_\varepsilon(f(x))\). We say \(f\) is continuous if it is continuous at each \(x\in X\).

Definition 8. Let \(X\) be a metric space. A sequence \((x_n)_{n=1}^\infty \subseteq X\) converges to \(x\in X\) if for all \(\varepsilon > 0\) there is \(N\in\mathbb{N}\) such that \(n\geq N \implies x_n \in B_\varepsilon(x)\). We denote \((x_n)\) converges to \(x\) by \(\lim x_n = x\) or \(x_n \to x\).

Definition 9. Let \(X\) and \(Y\) be metric spaces. A function \(f:X\to Y\) is sequentially continuous if whenever \((x_n)_{n=1}^\infty \subseteq X\) is a sequence converging to \(x\in X\), it follows that \(\lim f(x_n) = f(x)\).

Theorem 7. Let \(X\) and \(Y\) be metric spaces. Let \(f:X\to Y\) be a function. The following are equivalent.

\(f\) is continuous.
\(f\) is sequentially continuous.
For any open set \(V\subseteq Y\), the set \(f^{-1}(V)\) is open in \(X\).
For any set \(A\subseteq X\), one has \(f(\overline{A}) \subseteq \overline{f(A)}\).

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