Proof of The Complex Coordinate Space is a Complex Hilbert Space

Components and componentwise operations are those of The Complex Coordinate Space, with additive inverses as in The Complex Coordinate Space is a Complex Vector Space; in particular the $k$-th component of $u-v$ is $u_{k}-v_{k}$. By claim 2 of The Standard Inner Product Makes the Complex Coordinate Space an Inner Product Space, $\lVert w\rVert^{2}=\sum_{k=1}^{n}|w_{k}|^{2}$, the finite sum being taken in $\mathbb{R}$, with $|\cdot|$ the modulus. We use the ordered field properties of $\mathbb{R}$ and the following two facts, the second proved as in the proof of Properties of Complex Conjugation and Modulus: if $p\le q$ and $0\le r$ then $pr\le qr$; and if $p,q$ are nonnegative reals with $p^{2}\le q^{2}$, then $p\le q$.

Step 0 (convergence as a real null sequence). Let $(X,\rho)$ be a metric space, $(x_{m})$ a sequence in $X$ and $x\in X$. Since $0\le\rho(x_{m},x)$, the absolute value of $\rho(x_{m},x)-0$ is $\rho(x_{m},x)$ itself, so $(x_{m})$ converges to $x$ if and only if the real sequence $\bigl(\rho(x_{m},x)\bigr)_{m}$ converges to $0$. We use this for $(\mathbb{C}^{n},d)$ and for $(\mathbb{C},d_{\mathbb{C}})$, where $d_{\mathbb{C}}(z,w)=|z-w|$ is the metric of claim 9 of Properties of Complex Conjugation and Modulus.

Step 1 (each component is dominated by the norm). Let $w\in\mathbb{C}^{n}$ and let $k$ satisfy $1\le k\le n$. Every term $|w_{j}|^{2}$ is nonnegative, and each term of a finite sum of nonnegative real numbers is at most the sum: this follows by applying the principle of induction to $n$, the case $n=1$ being trivial, and the inductive step using that $\sum_{j=1}^{S(m)}a_{j}=\sum_{j=1}^{m}a_{j}+a_{S(m)}$ together with claim 4 of Properties of Finite Sums, which gives $0\le\sum_{j=1}^{m}a_{j}$, and the nonnegativity of $a_{S(m)}$. Hence

and since $|w_{k}|$ and $\lVert w\rVert$ are nonnegative, $|w_{k}|\le\lVert w\rVert$.

Step 2 (the component sequences converge). Let $(u^{(m)})$ be a Cauchy sequence in $(\mathbb{C}^{n},d)$ and fix $k$ with $1\le k\le n$. By Step 1 applied to $w=u^{(m)}-u^{(l)}$,

Given a positive real $\varepsilon$, choosing $N$ with $d(u^{(m)},u^{(l)})<\varepsilon$ for all $m,l\ge N$ therefore gives $d_{\mathbb{C}}(u^{(m)}_{k},u^{(l)}_{k})<\varepsilon$ for all $m,l\ge N$. So $(u^{(m)}_{k})_{m}$ is a Cauchy sequence in $(\mathbb{C},d_{\mathbb{C}})$, and by The Complex Numbers are Complete in the Modulus Metric there is a complex number $u_{k}$ to which it converges; by Step 0 the real sequence $\bigl(|u^{(m)}_{k}-u_{k}|\bigr)_{m}$ converges to $0$.

Step 3 (the squared norms tend to zero). Put $u=(u_{1},\dots,u_{n})\in\mathbb{C}^{n}$. For each $k$, applying claim 2 of Arithmetic of Limits of Real Sequences to the sequence $\bigl(|u^{(m)}_{k}-u_{k}|\bigr)_{m}$ and itself shows that $\bigl(|u^{(m)}_{k}-u_{k}|^{2}\bigr)_{m}$ converges to $0\cdot0=0$. Applying claim 1 of Arithmetic of Limits of Real Sequences and the principle of induction on the number of summands, the real sequence

converges to $\sum_{k=1}^{n}0$, which is $0$ (again by induction, using the recursion of Finite Sum Notation in a Field and $0+0=0$). Since the $k$-th component of $u^{(m)}-u$ is $u^{(m)}_{k}-u_{k}$, this sequence is exactly $\bigl(\lVert u^{(m)}-u\rVert^{2}\bigr)_{m}$.

Step 4 (from squared norms to norms). Let $\varepsilon$ be a positive real number; then $\varepsilon^{2}$ is positive. By Step 3 there is $N$ such that $\lVert u^{(m)}-u\rVert^{2}<\varepsilon^{2}$ for all $m\ge N$. For such $m$ we must have $\lVert u^{(m)}-u\rVert<\varepsilon$: otherwise $\varepsilon\le\lVert u^{(m)}-u\rVert$, and multiplying this inequality first by the nonnegative number $\varepsilon$ and then by the nonnegative number $\lVert u^{(m)}-u\rVert$ gives $\varepsilon^{2}\le\lVert u^{(m)}-u\rVert^{2}$, contradicting the previous line. Hence $d(u^{(m)},u)<\varepsilon$ for all $m\ge N$, so $(u^{(m)})$ converges to $u$ in $(\mathbb{C}^{n},d)$.

Thus every Cauchy sequence in $(\mathbb{C}^{n},d)$ converges to a point of $\mathbb{C}^{n}$, which proves claim 1 by Complete Metric Space.

Claim 2. By claim 1 the metric induced by the norm is complete, so $\mathbb{C}^{n}$ with the standard inner product is a complex Hilbert space. The qubit state space is by definition the complex inner product space $\mathbb{C}^{2}$ with the standard inner product, that is the case $n=2$, and is therefore a complex Hilbert space.