Proof of Borel Sets of a Metric Space: Open and Closed Sets, Points, Open Sets of Subspaces, and Subspaces

Throughout, $\mathcal{B}(X)=\sigma(\mathcal{T}_{d})$ and $\mathcal{B}(Y)=\sigma(\mathcal{T}_{d_{Y}})$ by Borel Sigma-Algebra of a Metric Space (i), where $\mathcal{T}_{d_{Y}}$ is the set of subsets of $Y$ open in $(Y,d_{Y})$; both are $\sigma$-algebras, on $X$ and on $Y$, containing their generating sets, by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iii).

Open sets. Let $U$ be open in $(X,d)$. Then $U\in\mathcal{T}_{d}$ by Metric Spaces: Balls, the Metric Topology, and the Real and Complex Numbers as Metric Spaces (iii), and $\mathcal{T}_{d}\subseteq\sigma(\mathcal{T}_{d})=\mathcal{B}(X)$, so $U\in\mathcal{B}(X)$. This proves (i).

Closed sets. Let $A\subseteq X$ be closed in $(X,\mathcal{T}_{d})$. Then $X\setminus A\in\mathcal{T}_{d}$ by Closed Subset of a Topological Space (i), so $X\setminus A\in\mathcal{B}(X)$ by (i). Since $\mathcal{B}(X)$ is closed under complements in $X$ by σ-Algebra and Measurable Space (i), $A=X\setminus(X\setminus A)\in\mathcal{B}(X)$. This proves (ii).

Points and closed balls. Let $x\in X$, let $r\ge0$ be real, and put $\bar{B}=\bar{B}_{d}(x,r)$. We show that $X\setminus\bar{B}$ is open in $(X,d)$. Let $y\in X\setminus\bar{B}$. By Closed Ball in a Metric Space (ii), $d(x,y)\le r$ fails, so $r<d(x,y)$ by totality of $\le$ (The Real Numbers as a Type (v)), and $s=d(x,y)-r$ is a real number (Differences, Quotients and Absolute Values of Real Numbers, Read in the Type Real (i)) with $0<s$, by adding $-r$ to both sides of $r<d(x,y)$ (The Real Numbers as a Type (vi), The Real Numbers as a Type (iv)). Let $z\in B_{d}(y,s)$; then $z\in X$ and $d(y,z)<s$ by Open Ball in a Metric Space (ii). By Metric Space (v) and Metric Space (iv),

$$d(x,y)\le d(x,z)+d(z,y)=d(x,z)+d(y,z)<d(x,z)+s,$$

the last step adding $d(x,z)$ to both sides of $d(y,z)<s$ (The Real Numbers as a Type (vi), The Real Numbers as a Type (iv)); so $d(x,y)<d(x,z)+s$ by transitivity (The Real Numbers as a Type (v)), and $d(x,z)>d(x,y)-s=r$ by adding $-s$ to both sides (The Real Numbers as a Type (vi), The Real Numbers as a Type (iv)); hence $d(x,z)\le r$ fails (The Real Numbers as a Type (v)), and $z\notin\bar{B}$ by Closed Ball in a Metric Space (ii); thus $z\in X\setminus\bar{B}$. Hence $B_{d}(y,s)\subseteq X\setminus\bar{B}$ for every $y\in X\setminus\bar{B}$, and $X\setminus\bar{B}$ is open in $(X,d)$ by Open Subset of a Metric Space (i). So $X\setminus\bar{B}\in\mathcal{T}_{d}$, $\bar{B}$ is closed in $(X,\mathcal{T}_{d})$ by Closed Subset of a Topological Space (i), and $\bar{B}\in\mathcal{B}(X)$ by (ii). This proves (iv).

For the singleton, let $y$ be any set. By Closed Ball in a Metric Space (ii), $y\in\bar{B}_{d}(x,0)$ if and only if $y\in X$ and $d(x,y)\le0$; by Metric Space (ii) and antisymmetry of $\le$ (The Real Numbers as a Type (v)) this holds if and only if $y\in X$ and $d(x,y)=0$, and by Metric Space (iii) if and only if $y=x$. So $\{x\}=\bar{B}_{d}(x,0)$, which belongs to $\mathcal{B}(X)$ by (iv). This proves (iii).

Open balls of the subspace. For $y\in Y$ and real $r>0$ write $B_{d_{Y}}(y,r)$ for the open ball of $(Y,d_{Y})$. Since $d_{Y}(y,z)=d(y,z)$ for $z\in Y$, by Domains and Restrictions of Maps (iii) applied to $d$ and $Y\times Y\subseteq X\times X$, Open Ball in a Metric Space (ii), applied in $(Y,d_{Y})$ and in $(X,d)$, gives

$$B_{d_{Y}}(y,r)=B_{d}(y,r)\cap Y. \tag{1}$$

Open sets of the subspace. We prove (v).

Let $V$ be open in $(X,d)$ and $U=V\cap Y$; then $U\subseteq Y$. Let $y\in U$. Since $y\in V$, there is a real $r>0$ with $B_{d}(y,r)\subseteq V$ by Open Subset of a Metric Space (i), and then $B_{d_{Y}}(y,r)=B_{d}(y,r)\cap Y\subseteq V\cap Y=U$ by (1). So $U$ is open in $(Y,d_{Y})$ by Open Subset of a Metric Space (i).

Conversely, let $U$ be open in $(Y,d_{Y})$. Let $\mathcal{S}=\{W\in\mathcal{T}_{d}:W\cap Y\subseteq U\}$, a subset of the set $\mathcal{T}_{d}$ of The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula (ii) given by a predicative condition (Sets and Maps: Ordinary Notation (iii)), and let $V=\bigcup\mathcal{S}$, the union of all open subsets $W$ of $X$ with $W\cap Y\subseteq U$. Since $\mathcal{T}_{d}$ is a topology on $X$ by The Open Subsets of a Metric Space Form a Topology (i), $V\in\mathcal{T}_{d}$ by Topological Space (iii); that is, $V$ is open in $(X,d)$. If $z\in V\cap Y$, then $z\in W$ for some $W\in\mathcal{S}$, so $z\in W\cap Y\subseteq U$. If $x\in U$, then $x\in Y$, and since $U$ is open in $(Y,d_{Y})$ there is a real $r>0$ with $B_{d_{Y}}(x,r)\subseteq U$ by Open Subset of a Metric Space (i). The open ball $B_{d}(x,r)$ is open in $(X,d)$ by Open Ball in a Metric Space is Open (i), so $B_{d}(x,r)\in\mathcal{T}_{d}$ by The Open Sets of a Metric Space Form a Set, Given by a Predicative Formula (ii), and $B_{d}(x,r)\cap Y=B_{d_{Y}}(x,r)\subseteq U$ by (1); hence $B_{d}(x,r)\in\mathcal{S}$. As $d(x,x)=0<r$ by Metric Space (iii), $x\in B_{d}(x,r)$ by Open Ball in a Metric Space (ii), so $x\in V$. Hence $x\in V\cap Y$, and $U=V\cap Y$. This proves (v).

Traces of Borel sets are Borel in the subspace. Let $\mathcal{G}$ be the set of all $B\in\mathcal{P}(X)$ with $B\cap Y\in\mathcal{B}(Y)$. We show that $\mathcal{G}$ is a $\sigma$-algebra on $X$. First, $X\cap Y=Y$, and $Y\in\mathcal{B}(Y)$ by σ-Algebra and Measurable Space (i), $\mathcal{B}(Y)$ being a $\sigma$-algebra on $Y$; so $X\in\mathcal{G}$. If $B\in\mathcal{G}$, then $(X\setminus B)\cap Y=Y\setminus(B\cap Y)\in\mathcal{B}(Y)$ by σ-Algebra and Measurable Space (i), so $X\setminus B\in\mathcal{G}$. If $(B_{n})$ is a sequence in $\mathcal{G}$, then the sequence $(B_{n}\cap Y)$ in $\mathcal{P}(Y)$ of Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii) has all its values in $\mathcal{B}(Y)\subseteq\mathcal{P}(Y)$, so it is a sequence in $\mathcal{B}(Y)$ by Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (v), and

$$\Bigl(\bigcup_{n\in\mathbb{N}}B_{n}\Bigr)\cap Y=\bigcup_{n\in\mathbb{N}}(B_{n}\cap Y)\in\mathcal{B}(Y)$$

by σ-Algebra and Measurable Space (i), so $\bigcup_{n}B_{n}\in\mathcal{G}$. Next, for $V\in\mathcal{T}_{d}$ the set $V\cap Y$ is open in $(Y,d_{Y})$ by (v), so $V\cap Y\in\mathcal{B}(Y)$ by (i) applied to the metric space $(Y,d_{Y})$; hence $\mathcal{T}_{d}\subseteq\mathcal{G}$. By Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iv), $\mathcal{B}(X)=\sigma(\mathcal{T}_{d})\subseteq\mathcal{G}$, that is,

$$B\cap Y\in\mathcal{B}(Y)\qquad\text{for every }B\in\mathcal{B}(X). \tag{2}$$

The trace of the Borel sets. Let $\mathcal{C}=\{B\cap Y:B\in\mathcal{B}(X)\}$, the image of $\mathcal{B}(X)$ under the map $B\mapsto B\cap Y$ from $\mathcal{B}(X)$ to $\mathcal{P}(Y)$ of Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii), a set by Sets and Maps: Ordinary Notation (vii) and a subset of $\mathcal{P}(Y)$, and let $\sigma_{Y}(\mathcal{C})$ be the $\sigma$-algebra on $Y$ generated by $\mathcal{C}$. By (2), $\mathcal{C}\subseteq\mathcal{B}(Y)$, so $\sigma_{Y}(\mathcal{C})\subseteq\mathcal{B}(Y)$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iv). If $U\in\mathcal{T}_{d_{Y}}$, then $U=V\cap Y$ for some $V$ open in $(X,d)$ by (v), and $V\in\mathcal{B}(X)$ by (i), so $U\in\mathcal{C}$. Thus $\mathcal{T}_{d_{Y}}\subseteq\mathcal{C}$, and $\mathcal{B}(Y)=\sigma(\mathcal{T}_{d_{Y}})\subseteq\sigma_{Y}(\mathcal{C})$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (v). Hence $\mathcal{B}(Y)=\sigma_{Y}(\mathcal{C})$, which with (2) proves (vi).

A Borel subspace. Suppose $Y\in\mathcal{B}(X)$, and let $\mathcal{A}=\{B\in\mathcal{B}(X):B\subseteq Y\}$. By Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (xi), applied to the $\sigma$-algebra $\mathcal{B}(X)$ on $X$ and the map $B\mapsto B\cap Y$ on $\mathcal{B}(X)$ of Sequences and Maps of Sets: Complements, Traces, Two-Term Sequences, the Disjointing Sequence, and Sequences in a Subset (ii) (a map into $\mathcal{P}(Y)\subseteq\mathcal{P}(X)$, with the same values, hence the same image), $\{B\cap Y:B\in\mathcal{B}(X)\}=\mathcal{A}$ and $\mathcal{A}$ is a $\sigma$-algebra on $Y$. Each $B\in\mathcal{A}$ equals $B\cap Y$, so $\mathcal{A}\subseteq\mathcal{B}(Y)$ by (2). If $U\in\mathcal{T}_{d_{Y}}$, then $U=V\cap Y$ for some $V$ open in $(X,d)$ by (v), $V\in\mathcal{B}(X)$ by (i), so $U\in\mathcal{B}(X)$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (ix), and $U\subseteq Y$; thus $U\in\mathcal{A}$. Hence $\mathcal{T}_{d_{Y}}\subseteq\mathcal{A}$ and $\mathcal{B}(Y)\subseteq\mathcal{A}$ by Closure Properties of σ-Algebras, Intersections and Traces of σ-Algebras, and the Generated σ-Algebra (iv), so $\mathcal{B}(Y)=\mathcal{A}=\{B\cap Y:B\in\mathcal{B}(X)\}$. This proves (vii).