Clear explanation about uniform continuity. Can anyone explain the uniform continuity clearly with picture if possible?? I have read the section on this topic in my text book but I am still not clear on this. Thanks.
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In my text book, it gives two definitions of a continuous function. One with sequence in the domain of a function and $\varepsilon$-$\delta$ definition. I think I am good with a continuous function at a point. The definition of the uniform continuity in my text book is this.
Let $f$ be a real-valued function defined on a set $S \subseteq R.$ Then $f$ is uniformly continuous on $S$ if $\forall \varepsilon >0,\exists \delta > 0\text{ s.t } x,y \in S \text{ and } |x-y| < \delta \Rightarrow \left|f(x)-f(y)\right|<\varepsilon.$
From this definiton the points $x,y$ are not fixed points in the domain. So for given $\varepsilon >0$ we find $\delta >0$ such that the distance between $x$ and $y$ in the domain is less than $\delta$ implies the distance between $f(x)$ and $f(y)$ less than given $\varepsilon.$ Then the function is uniformly continuous. This is as far as I know about the uniform continuity.
 A: First, uniform continuity is about an interval, not just a point.  You can say f(x) is continuous on the interval $(a,b)$ means: $\lim_{x \rightarrow c} f(x) = f(c)$ for all $c$ in $(a,b)$.  Then you have to say what you mean by a limit, and you get the old $\varepsilon$-$\delta$ statement:  for every $\varepsilon$ there exists a $\delta$ such that $|x-c| < \delta \Rightarrow |f(x) - f(c)| < \varepsilon$.
Now what they never tell you is that $\delta$ depends on $x$.  For some functions $\delta$ has to get smaller and smaller to get $|f(x) -f(c)| < \varepsilon$. An example of this is the function $f(x) = 1/x$ on $(0,1)$.  It's continuous there everywhere, but as $x$ approaches $0$ the function gets steeper and steeper.  That means you have to take your $x$ and $c$ closer and closer together -- i.e $\delta$ smaller and smaller,  to get $|f(x) -f(c)| < \varepsilon$.
What uniform continuity means is that the $\delta$ does NOT depend on x.  You can find a single $\delta$ that depends only on $\epsilon$ for the entire interval.
Intuitively uniform continuity means your function can't get infinitely steeper on $(a,b)$ the way $1/x$ does on $(0,1)$.  Steepness doesn't only mean that $f$ may go off to infinity in your interval.  It could also oscillate around in some unpleasant way.  Look for example at $f(x) = \sin(1/x)$ on $(0,1)$.
A: One intuitive way to think about it: if a function is uniformly continuous, as long as the "distance" between "$a$" and "$b$" is within some number $\delta$, you are guaranteed that the "distance" between $f(a)$ and $f(b)$ will be within some number $\varepsilon$. The rigorous correct way to think about it is the other way around. You first fix $\varepsilon$ and try to find whether you can find such $\delta$.
A: Uniform continuity gives you a continuous function for all $x$ in your domain, while pointwise continuity does not guarantee this. For example, prove that the sequence of functions $f_n(x) = x^n$ for $x \in [0,1]$ is pointwise continuous, but not uniformly continuous. This picture should give you an idea of the behavior for each $f_n$  
