Proof of Properties of the Product of Two Real Inner Product Spaces

Throughout we use that the operations of $E_{1}\times E_{2}$ are componentwise, so that $x-y=(x_{1}-y_{1},x_{2}-y_{2})$, and that two pairs are equal exactly when their components are.

Claim 1. Each axiom of Vector Space over a Field asserts an equality between two elements of $E_{1}\times E_{2}$. Since the addition and scalar multiplication of the product are componentwise, the components of the two sides are the two sides of the corresponding axiom in $E_{1}$ and in $E_{2}$; as two pairs are equal exactly when their components are, each axiom holds. The same reasoning identifies $(0_{E_{1}},0_{E_{2}})$ as the zero vector and $(-x_{1},-x_{2})$ as the additive inverse of $(x_{1},x_{2})$, these being the componentwise assertions of the identity and inverse axioms.

We check that the product pairing satisfies the four conditions of Real Inner Product Space §inner-product. Let $u=(u_{1},u_{2})$ be a further element of $E_{1}\times E_{2}$. Symmetry holds because, by condition (a) in $E_{1}$ and in $E_{2}$,

Additivity in the first argument holds because, by condition (b) in $E_{1}$ and in $E_{2}$ and the commutativity and associativity of the addition of $\mathbb{R}$,

Homogeneity in the first argument holds because, by condition (c) in $E_{1}$ and in $E_{2}$ and distributivity in $\mathbb{R}$,

For positive definiteness, put $a=\langle x_{1},x_{1}\rangle_{E_{1}}$ and $b=\langle x_{2},x_{2}\rangle_{E_{2}}$, so that $\langle x,x\rangle=a+b$ with $0\le a$ and $0\le b$ by condition (d) in $E_{1}$ and in $E_{2}$; then $0\le a+b$ by claim 2 of Elementary Arithmetic in an Ordered Field. If moreover $a+b=0$, then $a\le a+b=0$ by claim 3 of Elementary Arithmetic in an Ordered Field applied to $0\le b$, so $a=0$ by antisymmetry of the order, and $b=0$ in the same way; condition (d) in $E_{1}$ and in $E_{2}$ then gives $x_{1}=0_{E_{1}}$ and $x_{2}=0_{E_{2}}$, that is, $x=0_{E_{1}\times E_{2}}$.

Hence $E_{1}\times E_{2}$, with the product pairing, is a real inner product space, and its norm and distance are those of Real Inner Product Space §norm and Real Inner Product Space §distance.

Claim 2. By the definition of the norm and that of the product pairing,

Since $0\le|x_{2}|_{E_{2}}^{2}$, claim 3 of Elementary Arithmetic in an Ordered Field gives $|x_{1}|_{E_{1}}^{2}\le|x|^{2}$. Both $|x_{1}|_{E_{1}}$ and $|x|$ are nonnegative, so $|x_{1}|_{E_{1}}\le|x|$: otherwise $|x|<|x_{1}|_{E_{1}}$ by trichotomy, and claim 1 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field would give $|x|^{2}<|x_{1}|_{E_{1}}^{2}$, a contradiction. The bound for $|x_{2}|_{E_{2}}$ is obtained in the same way.

For the last inequality, write $p=|x_{1}|_{E_{1}}$ and $q=|x_{2}|_{E_{2}}$. Since $0\le pq$ by claim 5 of Elementary Arithmetic in an Ordered Field, and $0\le pq+pq$ by claim 2 of that lemma, we get

and $p+q$ is nonnegative by claim 2 of Elementary Arithmetic in an Ordered Field, so $|x|\le p+q$ by the same trichotomy argument as above.

Claim 3. By the definition of the distance, $d(x,y)=|x-y|$, and $x-y=(x_{1}-y_{1},x_{2}-y_{2})$, so claim 2 gives

Write $p=d_{E_{1}}(x_{1},y_{1})$, $q=d_{E_{2}}(x_{2},y_{2})$ and $\rho=\rho(x,y)=\max\{p,q\}$, which is one of $p$ and $q$ and satisfies $p\le\rho$ and $q\le\rho$. All three are nonnegative. From $\rho^{2}\in\{p^{2},q^{2}\}$ and claim 3 of Elementary Arithmetic in an Ordered Field we get $\rho^{2}\le p^{2}+q^{2}=d(x,y)^{2}$, hence $\rho\le d(x,y)$ by the trichotomy argument of claim 2. Also, by claim 5 of Elementary Arithmetic in an Ordered Field, $p^{2}\le\rho^{2}$ and $q^{2}\le\rho^{2}$, so

since $(2\rho)^{2}=4\rho^{2}=\rho^{2}+\rho^{2}+\rho^{2}+\rho^{2}$ and $0\le\rho^{2}$; hence $d(x,y)\le2\rho$.

We now read off the topological statements. If $(z_{m})$ converges to $z$ in $(E_{1}\times E_{2},d)$ and $\varepsilon$ is positive, then $\rho(z_{m},z)\le d(z_{m},z)<\varepsilon$ for large $m$, so $(z_{m})$ converges to $z$ in $(E_{1}\times E_{2},\rho)$; conversely, if $\rho(z_{m},z)<\tfrac{\varepsilon}{2}$ for large $m$, then $d(z_{m},z)\le2\rho(z_{m},z)<\varepsilon$. The same two estimates prove the statement about Cauchy sequences. If $U$ is open in $(E_{1}\times E_{2},\rho)$ and $z\in U$, choose a positive $\delta$ such that every $w$ with $\rho(z,w)<\delta$ lies in $U$; then every $w$ with $d(z,w)<\delta$ satisfies $\rho(z,w)\le d(z,w)<\delta$ and so lies in $U$, whence $U$ is open in $(E_{1}\times E_{2},d)$. Conversely, if $U$ is open in $(E_{1}\times E_{2},d)$ and $z\in U$ with $\delta$ as above for $d$, then every $w$ with $\rho(z,w)<\tfrac{\delta}{2}$ satisfies $d(z,w)\le2\rho(z,w)<\delta$ and lies in $U$. So the two metrics have the same open sets, and therefore the same closed sets, these being the complements of the open sets by Closed Subset of a Topological Space.

Claim 4. By claim 2 applied to $z_{m}-z$,

for $i\in[2]$. The left inequality shows that convergence of $(z_{m})$ to $z$ implies convergence of both component sequences; the right one, applied with $\tfrac{\varepsilon}{2}$ for each component, shows the converse. Replacing $z$ by $z_{\ell}$ throughout gives the corresponding statement for Cauchy sequences.

Claim 5. Linearity of $\pi_{1}$ and $\pi_{2}$ is immediate from the componentwise operations, and $|\pi_{i}x|_{E_{i}}\le|x|$ is claim 2; hence $\pi_{i}$ is bounded, with $1$ as an admissible constant. Linearity of $j_{1}$ and $j_{2}$ is immediate as well, and by claim 2 and Elementary Identities in a Real Inner Product Space §zero, $|j_{1}u|^{2}=|u|_{E_{1}}^{2}+|0_{E_{2}}|_{E_{2}}^{2}=|u|_{E_{1}}^{2}$, so $|j_{1}u|=|u|_{E_{1}}$, both being nonnegative; the same holds for $j_{2}$. Finally $j_{1}\pi_{1}x+j_{2}\pi_{2}x=(x_{1},0_{E_{2}})+(0_{E_{1}},x_{2})=(x_{1},x_{2})=x$.

Claim 6. Let $(z_{m})$ be a Cauchy sequence in $(E_{1}\times E_{2},d)$. By claim 4 the sequences $(\pi_{1}z_{m})$ and $(\pi_{2}z_{m})$ are Cauchy in $(E_{1},d_{E_{1}})$ and $(E_{2},d_{E_{2}})$, which are complete because $E_{1}$ and $E_{2}$ are real Hilbert spaces; let $z^{1}$ and $z^{2}$ be their limits. By claim 4 again, $(z_{m})$ converges to $(z^{1},z^{2})$. Hence $(E_{1}\times E_{2},d)$ is complete, and $E_{1}\times E_{2}$ is a real Hilbert space.

Claim 7. Let $D_{1}\subseteq E_{1}$ and $D_{2}\subseteq E_{2}$ be countable dense subsets, which exist because $(E_{1},d_{E_{1}})$ and $(E_{2},d_{E_{2}})$ are separable. The set $D_{1}\times D_{2}$ is countable by claim 1 of Products and Powers of Countable Sets, and is a subset of $E_{1}\times E_{2}$.

Let $z\in E_{1}\times E_{2}$ and let $\varepsilon$ be positive. By Characterization of the Closure in a Metric Space by Open Balls there are $p_{1}\in D_{1}$ and $p_{2}\in D_{2}$ with $d_{E_{1}}(\pi_{1}z,p_{1})<\tfrac{\varepsilon}{2}$ and $d_{E_{2}}(\pi_{2}z,p_{2})<\tfrac{\varepsilon}{2}$. By claim 2,

Thus every open ball of $(E_{1}\times E_{2},d)$ about $z$ meets $D_{1}\times D_{2}$, so $z$ lies in the closure of $D_{1}\times D_{2}$ by Characterization of the Closure in a Metric Space by Open Balls. As $z$ was arbitrary, $D_{1}\times D_{2}$ is dense, and $(E_{1}\times E_{2},d)$ is separable.