The absolute value is compatible with limits

Let $(x_k)$ be a sequence of reals converging to $x\in \mathbb{R}$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences, Limits and Cauchy sequences of reals). Then $(\mathrm{\mid }{x}_k\mathrm{\mid })$ converges to $\mathrm{\mid }x\mathrm{\mid }$.

In the single case $x=0$ the implication reverses: $\mathrm{\mid }{x}_k\mathrm{\mid }\to 0$ if and only if $x_k\to 0$. Whether the implication can be reversed for $x\mathrm{\ne }0$ is taken up in the remarks below; it is no part of what the proof establishes.

• The converse fails at every nonzero limit. This is not established by the proof above, which proves only the forward implication and the equivalence at $x=0$; the witness is exhibited here instead. Fix a real $c\mathrm{\ne }0$, let $(s_k)$ be the alternating sequence of $1$ and $-1$ constructed in FALSE: every bounded sequence converges, which is shown there not to converge, and put $w_k:=s_{k}c$. Then $\mathrm{\mid }{w}_k\mathrm{\mid }=\mathrm{\mid }{s}_k\mathrm{\mid }\mathrm{\mid }c\mathrm{\mid }=\mathrm{\mid }c\mathrm{\mid }$ for every $k$ (Basic properties of the absolute value), so $(\mathrm{\mid }{w}_k\mathrm{\mid })$ is the constant sequence $\mathrm{\mid }c\mathrm{\mid }$ and converges to $\mathrm{\mid }c\mathrm{\mid }$ (Sequences of reals: bounded, eventually, frequently, tails, subsequences). But $(w_k)$ does not converge: if it converged to some $\mathrm{\ell }$, then $(s_k)=(c^{-1}{w}_k)$ would converge to $c^{-1}\mathrm{\ell }$ by the scalar-multiple rule (Algebra of limits: sums, scalar multiples, products and quotients), which it does not. Passing to absolute values destroys sign information, and only at $0$ is there no sign information to destroy.

• Combined with Algebra of limits: sums, scalar multiples, products and quotients this gives the usual companions: the identities $\max \{u,v\}=\frac{1}{2}(u+v+\mathrm{\mid }u-v\mathrm{\mid })$ and $\min \{u,v\}=\frac{1}{2}(u+v-\mathrm{\mid }u-v\mathrm{\mid })$ (Maximum and minimum of a set), each a two-case check on the sign of $u-v$, exhibit $\max \{x_k,y_k\}$ and $\min \{x_k,y_k\}$ as sums of convergent sequences, so they converge to $\max \{x,y\}$ and $\min \{x,y\}$.

• The lemma is the sequential form of the statement that $u\mapsto \mathrm{\mid }u\mathrm{\mid }$ is continuous, but continuity is not available yet and is not needed: the reverse triangle inequality does the work directly.