# Uniform continuity

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- Don't forget a uniformly continuous map is continuous and co - Alec (talk) 21:14, 8 April 2017 (UTC)

## Contents

## Definition

Let [ilmath](X,d_1)[/ilmath] and [ilmath](Y,d_2)[/ilmath] be metric spaces and let [ilmath]f:X\rightarrow Y[/ilmath] be a map between them. We say [ilmath]f[/ilmath] is *uniformly continuous* if^{[1]}:

- [ilmath]\forall\epsilon>0\exists\delta>0\forall x,y\in X\big[d_1(x,y)<\delta\implies d_2(f(x),f(y))<\epsilon\big][/ilmath]
- For comparison: continuity at [ilmath]x\in X[/ilmath] (in a map between metric spaces) is [ilmath]\forall\epsilon>0\exists\delta>0\forall y\in X\big[d_1(x,y)<\delta\implies d_2(f(x),f(y))<\epsilon\big] [/ilmath] - uniform continuity differs by supposing given an [ilmath]\epsilon >0[/ilmath] there is some [ilmath]\delta>0[/ilmath] that'll "work" for all [ilmath]x,y\in X[/ilmath], not just for a fixed-before-[ilmath]\epsilon[/ilmath] [ilmath]x[/ilmath].