Uniform Cauchy sequence

A sequence of functions $f_n:X\to Y$ into a metric space $(Y,d)$ is uniform Cauchy on $X$ if, for every $\varepsilon>0$ , there exists $N$ such that for all $m,n\ge N$ and $x\in X$ ,

d\bigl(f_m(x),f_n(x)\bigr)<\varepsilon.

Equivalently,

\sup_{x\in X} d\bigl(f_m(x),f_n(x)\bigr)<\varepsilon \quad \text{for all } m,n\ge N.

Remarks

If $Y$ is complete, a sequence is uniform Cauchy if and only if it converges uniformly to a function $X\to Y$ . Without completeness, a uniform Cauchy sequence need not have a $Y$ -valued limit.

Examples

On $[0,1]$ , $f_n(x)=x/n$ is uniform Cauchy because $\sup_{x\in[0,1]}|f_m(x)-f_n(x)|\le|1/m-1/n|.$
On $[0,1]$ , the sequence $f_n(x)=x^n$ is not uniform Cauchy.