Proof of Elementary Properties of the Closed Ball in a Metric Space

Throughout, $d$ satisfies the axioms of Metric Space: $d(p,q)=0$ exactly when $p=q$, $d(p,q)=d(q,p)$, and $d(p,q)\le d(p,q')+d(q',q)$ for all $p,q,q'\in X$. The order $\le$ on $\mathbb{R}$ is in particular a total order.

Claim 1. We have $d(x,x)=0$ and $0\le r$, so $x\in\bar{B}_d(x,r)$. If $0<r$ and $y\in B_d(x,r)$, then $d(x,y)<r$, hence $d(x,y)\le r$ and $y\in\bar{B}_d(x,r)$.

Claim 2. Put $R=r+1$. By claim 6 of Elementary Order Arithmetic in an Ordered Field we have $0<1$, and $0\le r$, so claim 3 of that lemma gives $0<R$. Moreover $r\le R$, since $R-r=1$ and $0\le1$ by claim 1 of Elementary Arithmetic in an Ordered Field, so claim 3 of that lemma applies. If $y\in\bar{B}_d(x,r)$ then $d(x,y)\le r\le R$, hence $d(x,y)\le R$ by transitivity. Thus the point $x$ and the radius $R$ witness that $\bar{B}_d(x,r)$ is bounded in the sense of Bounded Subset of a Metric Space.

Claim 3. By Closed Subset of a Topological Space it suffices to prove that $U=X\setminus\bar{B}_d(x,r)$ belongs to $\mathcal{T}_d$, that is, is open in $(X,d)$. Let $y\in U$. Then $d(x,y)\le r$ fails, so $r<d(x,y)$ because $\le$ is a total order. Put $\rho=d(x,y)-r$; by claim 1 of Elementary Order Arithmetic in an Ordered Field, $0<\rho$, so the open ball $B_d(y,\rho)$ is defined.

Let $z\in B_d(y,\rho)$, so that $d(y,z)<\rho$. By claim 4 of Elementary Order Arithmetic in an Ordered Field we get $-\rho<-d(y,z)$, and adding $d(x,y)$ by claim 1 of that lemma,

$$d(x,y)-\rho<d(x,y)-d(y,z).$$

Here $d(x,y)-\rho=r$, by claims 3, 4 and 6 of Additive Cancellation and Elementary Additive Identities in a Field. The triangle inequality and the symmetry of $d$ give $d(x,y)\le d(x,z)+d(y,z)$, which by claim 3 of Elementary Arithmetic in an Ordered Field is equivalent to $d(x,y)-d(y,z)\le d(x,z)$. Combining the last two displays by claim 2 of Elementary Order Arithmetic in an Ordered Field gives $r<d(x,z)$, so $d(x,z)\le r$ fails and $z\in U$.

Thus $B_d(y,\rho)\subseteq U$ for every $y\in U$, so $U$ is open and $\bar{B}_d(x,r)$ is closed in $(X,\mathcal{T}_d)$.