Proof of The Norm Metric of a Real Normed Space

The norm axioms are Norms and Real Normed Spaces (i), Norms and Real Normed Spaces (ii), Norms and Real Normed Spaces (iii) and Norms and Real Normed Spaces (iv).

Norm of the zero vector. By Elementary Identities in a Vector Space (ii), $0_{V}=0\,0_{V}$, so homogeneity gives

$$\lVert0_{V}\rVert=\lVert0\,0_{V}\rVert=|0|\,\lVert0_{V}\rVert=0 .$$

Norm metric. Let $u,v,w\in V$. First, $d_{V}(u,v)=\lVert u-v\rVert\ge0$. If $u=v$, then $u-v=0_{V}$ by Elementary Identities in a Vector Space (vii), so $d_{V}(u,v)=0$ by (i). Conversely, if $d_{V}(u,v)=0$, then $u-v=0_{V}$ by definiteness, and by Elementary Identities in a Vector Space (viii) and The Zero Vector, Negatives and Differences in a Vector Space (i),

$$u=(u-v)+v=0_{V}+v=v+0_{V}=v .$$

For symmetry, $v-u=-(u-v)=(-1)(u-v)$ by Elementary Identities in a Vector Space (xi) and Elementary Identities in a Vector Space (iv), so homogeneity gives $d_{V}(v,u)=|-1|\,\lVert u-v\rVert=d_{V}(u,v)$. Finally, $u-w=(u-v)+(v-w)$ by Elementary Identities in a Vector Space (x), so the triangle inequality for the norm gives $d_{V}(u,w)\le d_{V}(u,v)+d_{V}(v,w)$. These are the conditions Metric Space (ii), Metric Space (iii), Metric Space (iv) and Metric Space (v), so $d_{V}$ is a metric on $V$.

Reverse triangle inequality. By Elementary Identities in a Vector Space (viii) and the triangle inequality,

$$\lVert u\rVert=\lVert(u-v)+v\rVert\le\lVert u-v\rVert+\lVert v\rVert,\qquad\lVert v\rVert\le\lVert v-u\rVert+\lVert u\rVert=\lVert u-v\rVert+\lVert u\rVert,$$

using the symmetry of $d_{V}$ just shown. Thus $\lVert u\rVert-\lVert v\rVert\le\lVert u-v\rVert$ and $-(\lVert u\rVert-\lVert v\rVert)\le\lVert u-v\rVert$; since $\bigl|\,\lVert u\rVert-\lVert v\rVert\,\bigr|$ is one of these two numbers, the inequality follows.

Convergence through norms. For a sequence $(x_{m})$ in $V$ and $x\in V$, the real number $\lVert x_{m}-x\rVert$ is nonnegative, so $\bigl|\lVert x_{m}-x\rVert-0\bigr|=d_{V}(x_{m},x)$. Comparing Convergent Sequence in a Metric Space (i) with Convergent Sequences of Real Numbers (i), $x_{m}\to x$ in $(V,d_{V})$ if and only if the sequence $(\lVert x_{m}-x\rVert)$ in $\mathbb{R}$ converges to $0$. Hence, by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (xii), it suffices for $x_{m}\to x$ to find a sequence $(b_{m})$ in $\mathbb{R}$ converging to $0$ with $\lVert x_{m}-x\rVert\le b_{m}$ for every $m\in\mathbb{N}$. In particular, $\lVert u_{m}-u\rVert\to0$ and $\lVert v_{m}-v\rVert\to0$.

Limit of a sum. By Elementary Identities in a Vector Space (vi), $-(u+v)=(-u)+(-v)$, so rearranging by associativity and commutativity,

$$\lVert(u_{m}+v_{m})-(u+v)\rVert=\lVert(u_{m}-u)+(v_{m}-v)\rVert\le\lVert u_{m}-u\rVert+\lVert v_{m}-v\rVert ,$$

and the right side converges to $0+0=0$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi). Hence $u_{m}+v_{m}\to u+v$.

Limit of a scalar multiple. By Elementary Identities in a Vector Space (ix), Vector Space over a Field (viii) and Elementary Identities in a Vector Space (iv),

$$\lambda_{m}(u_{m}-u)=\lambda_{m}u_{m}-\lambda_{m}u,\qquad(\lambda_{m}-\lambda)u=\lambda_{m}u+(-\lambda)u=\lambda_{m}u-\lambda u,$$

and adding these by Elementary Identities in a Vector Space (x) gives $\lambda_{m}u_{m}-\lambda u=\lambda_{m}(u_{m}-u)+(\lambda_{m}-\lambda)u$. The sequence $(\lambda_{m})$ is bounded by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (iii): there is $M\in\mathbb{R}$ with $|\lambda_{m}|\le M$ for every $m$. The triangle inequality and homogeneity give

$$\lVert\lambda_{m}u_{m}-\lambda u\rVert\le|\lambda_{m}|\,\lVert u_{m}-u\rVert+|\lambda_{m}-\lambda|\,\lVert u\rVert\le M\lVert u_{m}-u\rVert+\lVert u\rVert\,|\lambda_{m}-\lambda| .$$

The right side converges to $M\cdot0+\lVert u\rVert\cdot|\lambda-\lambda|=0$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (i), Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (vi) and Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (viii). Hence $\lambda_{m}u_{m}\to\lambda u$.

Limit of a difference. The constant sequence $(-1)_{m\in\mathbb{N}}$ converges to $-1$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (i), so $(-1)v_{m}\to(-1)v$ by (vi). Since $u_{m}-v_{m}=u_{m}+(-1)v_{m}$ and $u-v=u+(-1)v$ by Elementary Identities in a Vector Space (iv), (iv) gives $u_{m}-v_{m}\to u-v$.

Limit of the norms. By (iii), $\bigl|\lVert u_{m}\rVert-\lVert u\rVert\bigr|\le\lVert u_{m}-u\rVert$, and $\lVert u_{m}-u\rVert\to0$, so $\lVert u_{m}\rVert\to\lVert u\rVert$ by Limits of Sequences of Real Numbers: Uniqueness, Boundedness, Constants, Tails, Arithmetic, Quotients, Absolute Values, Finite Sums, Order, Squeezing, Domination and Subsequences (xii).