Cauchy implies bounded

Cauchy implies bounded: Let $(X,d)$ be a metric space and let $(x_n)$ be a Cauchy sequence in $X$. Then $(x_n)$ is bounded : there exist $x\in X$ and $R>0$ such that $d(x_n,x)\le R$ for all $n$.

This is a routine but important tool: Cauchy behavior already forces global control on the sequence.

Proof sketch: Take $\varepsilon=1$ in the Cauchy definition to find $N$ such that $d(x_n,x_N)<1$ for all $n\ge N$. Then all tail terms lie in $B(x_N,1)$. The finitely many initial terms $\{x_1,\dots,x_{N-1}\}$ are bounded (take a maximum distance to $x_N$), so the whole sequence is bounded.