Proof of The Set of Symmetric Real Matrices is a Metric Space

Let $X,Y,Z\in\mathcal{S}(n)$. Two real $n\times n$ matrices are equal exactly when all their entries agree, and by Difference of Real Matrices and Sum of Real Matrices the entries of a difference and of a sum are the corresponding differences and sums of entries. Write $0_{n}$ for the real $n\times n$ matrix all of whose entries are $0$. We verify the four conditions of Metric Space for $d_{\mathcal{S}(n)}$.

Condition 1 (nonnegativity). $0\le\lVert X-Y\rVert$ by claim 1 of Properties of the Norm of a Symmetric Real Matrix, that is $0\le d_{\mathcal{S}(n)}(X,Y)$.

Condition 2 (vanishing). By claim 4 of Properties of the Norm of a Symmetric Real Matrix, $\lVert X-Y\rVert=0$ holds if and only if $X-Y=0_{n}$. Comparing entries, $X-Y=0_{n}$ holds if and only if $X_{ij}-Y_{ij}=0$ for all indices $i,j$, and by claim 3 of Additive Cancellation and Elementary Additive Identities in a Field this holds if and only if $X_{ij}=Y_{ij}$ for all $i,j$, that is if and only if $X=Y$. Hence $d_{\mathcal{S}(n)}(X,Y)=0$ if and only if $X=Y$.

Condition 3 (symmetry). Let $\mu X$ denote the scalar multiple of a matrix, whose entries are $\mu X_{ij}$. For all $i,j$,

$$\bigl((-1)(X-Y)\bigr)_{ij}=(-1)\,(X_{ij}-Y_{ij})=-(X_{ij}-Y_{ij})=Y_{ij}-X_{ij}=(Y-X)_{ij},$$

using claim 2 of Zero Products and Elementary Identities in a Field together with the identity $1\cdot t=t$ for the second equality and claim 6 of Additive Cancellation and Elementary Additive Identities in a Field for the third. Hence $Y-X=(-1)(X-Y)$, and claim 5 of Properties of the Norm of a Symmetric Real Matrix gives

$$\lVert Y-X\rVert=|-1|\,\lVert X-Y\rVert .$$

Here $|-1|=|1|=1$, by claim 2 of Properties of the Absolute Value in an Ordered Field and by Absolute Value in an Ordered Field applied to $1$, which satisfies $0\le1$ by claim 1 of Elementary Arithmetic in an Ordered Field. Therefore $d_{\mathcal{S}(n)}(Y,X)=d_{\mathcal{S}(n)}(X,Y)$.

Condition 4 (triangle inequality). For all $i,j$, the field axioms give

$$X_{ij}-Z_{ij}=(X_{ij}-Y_{ij})+(Y_{ij}-Z_{ij}),$$

since the right-hand side equals $X_{ij}+\bigl((-Y_{ij})+Y_{ij}\bigr)+(-Z_{ij})$ by associativity and commutativity of addition, and $(-Y_{ij})+Y_{ij}=0$ is absorbed by the additive identity. Comparing entries, $X-Z=(X-Y)+(Y-Z)$. The matrices $X-Y$ and $Y-Z$ are symmetric by claim 1 of The Positive Semidefinite Ordering is a Partial Order Compatible with the Linear Structure, so claim 5 of Properties of the Norm of a Symmetric Real Matrix applies and gives

$$\lVert X-Z\rVert=\bigl\lVert (X-Y)+(Y-Z)\bigr\rVert\le\lVert X-Y\rVert+\lVert Y-Z\rVert ,$$

that is $d_{\mathcal{S}(n)}(X,Z)\le d_{\mathcal{S}(n)}(X,Y)+d_{\mathcal{S}(n)}(Y,Z)$.

All four conditions hold, so $d_{\mathcal{S}(n)}$ is a metric on $\mathcal{S}(n)$ and $\bigl(\mathcal{S}(n),d_{\mathcal{S}(n)}\bigr)$ is a metric space.