Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

$f:X\rightarrow Y $ is continuous iff every open ball in $Y$ has open pre-image. It is between topological spaces.

$f:X\rightarrow Y$ is uniformly continuous if $\forall \epsilon >0,\forall x \in X, |x-y|<\delta \implies |f(x)-f(y)|<\epsilon$.

My question is what is the difference between uniform continuity and continuity of a map.

share|improve this question
1  
The differenc is that for uniform continuity, you compare the preimages of different balls: Especially, every $\epsilon$-ball has a $\delta$-ball in its preimage, where $\delta$ depends only on $\epsilon$ not on the (center of the) ball itself. You may also notice that it is not as easy to get rid of the metric (i.e. the numbers $\epsilon$ and $\delta$) and formulate this purely in the langugae of topology. –  Hagen von Eitzen Feb 3 '13 at 11:53
    
thank you. .... –  user45099 Feb 3 '13 at 12:11

2 Answers 2

up vote 5 down vote accepted

Uniform continuity is a stronger property. To see why, let's write down the $\epsilon-\delta$ definition of continuity:

$$ \forall \epsilon > 0, \forall x \in X, \exists \delta > 0 : |x - y| < \delta \implies |f(x) - f(y)| < \epsilon $$

Compare this with the definition if uniform continuity (and pay extra attention to the order of quantifiers):

$$ \forall \epsilon > 0, \exists \delta > 0, \forall x \in X : |x - y| < \delta \implies |f(x) - f(y)| < \epsilon $$

In the definition of continuity, we find $\delta$ for a particular $\epsilon$ and $x$. $\delta$ depends on both $\epsilon$ and $x$. Each $x$ has its own $\delta$ for a fixed $\epsilon$.

In uniform continuity, $\delta$ depends only on $\epsilon$ and one value of $\delta$ must work for all $x \in X$.

share|improve this answer
    
Thank you. This is perfect. But:Daily vote limit reached; vote again in XX hours. :) –  user45099 Feb 3 '13 at 12:09
    
@user1709828 No problem. Happy to help! –  Ayman Hourieh Feb 3 '13 at 12:21

The difference is that in the uniformly continuity the value of $\delta$ depends only on $\epsilon$ and not on a point in the domain. For example we take $1/x^{2}$ where $x \in (0,1)$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.