Math Help - Quick Uniform Continuity question

1. Quick Uniform Continuity question

If we have a uniformly continuous function, and we know that there exist $\delta$ s.t. $\mid f(x)-f(y)\mid<\epsilon$ $\forall x,y$ s.t. $\mid x-y\mid<\delta$, does this mean that $\forall x,y$ s.t. $\mid x-y\mid\le\delta$ that $\mid f(x)-f(y)\mid\le\epsilon$?
Thanks

2. It looks like you're asking if uniform continuity implies continuity (the answer is yes, because uniform continuity is a stronger condition). However, your quantifiers are not carefully placed. Could you please write out your statements like this:

$(\forall\epsilon>0)(\exists\delta>0)(\forall x)(\forall y)(|x-y|<\delta\to|f(x)-f(y)|<\epsilon)?$

I think that's what you meant for your first statement. Could you correct this if it's wrong, and also write out your second statement that you think follows from the first? Thanks!

3. I know unif continuity implies continuity; I phrased my question poorly, sorry. I know the definition that if f is uniformly continuous that $\forall \epsilon>0$ there exist $\delta>0$ such that if $\mid x-y\mid<\delta$ then $\mid f(x)-f(y)\mid<\epsilon$. Now, does this imply that if $\mid x-y\mid\le\delta$ that $\mid f(x)-f(y)\mid\le\epsilon$? This is a question about how the strict inequalites relate to the non-strict inequalities, a sort of definitional question about uniform continuity. I have the feeling that this is a very trivial question but for some reason I am having conceptual issues with it.

4. Ah, I see what you're getting at. My instinct is that in the definition, with the quantifiers, $\le$ versus $<$ makes no difference. That is, I would claim that the two sets

$\{f\forall x)(\forall y)(\forall\epsilon>0)(\exists\delta>0)(|x-y|<\delta\to|f(x)-f(y)|<\epsilon)\}" alt="\{f\forall x)(\forall y)(\forall\epsilon>0)(\exists\delta>0)(|x-y|<\delta\to|f(x)-f(y)|<\epsilon)\}" /> and

$\{f\forall x)(\forall y)(\forall\epsilon>0)(\exists\delta>0)(|x-y|\le\delta\to|f(x)-f(y)|\le\epsilon)\}" alt="\{f\forall x)(\forall y)(\forall\epsilon>0)(\exists\delta>0)(|x-y|\le\delta\to|f(x)-f(y)|\le\epsilon)\}" />

are the same. I'd need to do some more thinking in order to prove this, but that's what I think.

However, the two statements

$|x-y|<\delta\to|f(x)-f(y)|<\epsilon$ and

$|x-y|\le\delta\to|f(x)-f(y)|\le\epsilon$

are not equivalent on their own, without the quantifiers.