TITLE: How exactly can't $\delta$ depend on $x$ in the definition of uniform continuity?
QUESTION [20 upvotes]: I'm told that a function defined on an interval $[a,b]$ or $(a,b)$ is uniformly continuous if for each $\epsilon\gt 0$ there exists a $\delta\gt 0$ such that $|x-t|\lt \delta$ implies that $|f(x)-f(t)|\lt \epsilon$. Then it gives a little note saying that $\delta$ cannot depend on $x$, it can only depend on $\epsilon$. 
With ordinary continuity, the $\delta$ can depend on both $x$ and $\epsilon$. I'm just a little lost on why $|x-t|\lt \delta$ implies $|f(x)-f(t)|\lt \epsilon$, and how $\delta$ can't depend on $x$ but only $\epsilon$.

REPLY [2 votes]: The following might help: $f:[a,b]\to \mathbb R$ is continuous means: $$\forall \epsilon >0, \forall x\in [a,b];\exists \delta >0,\forall y\in [a,b]:|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$$
while $f:[a,b]\to\mathbb R$ is uniformly continuous means  $$\forall \epsilon >0;\exists \delta >0,  \forall x\in [a,b],\forall y\in [a,b]:|x-y|<\delta \implies |f(x)-f(y)|<\epsilon$$
so typographically, the only difference is in the order of the second and third quantifiers. The semantic meaning though is that for that first you first choose $\epsilon $ and $x$ and only then find $\delta $, while for the second one you first choose $\epsilon$, then find $\delta$. In the former you have more information and the $\delta$ may depend on $x$. But in the latter, all the information you get is $\epsilon$ and out of that alone you must find the corresponding $\delta$.