continuitylimitsreal-analysis
I was reading from this website,
But as far as I know is not it the same as definition of continuity. For example Kenneth Ross's book has the same definition for continuity
Let $f$ be a real valued function whose domain is a subset of $\mathbb{R}$. Then $f$ is continuous at $x_0\in dom(f)$ iff for each $\epsilon>0$ there exist $\delta>0$ such that $x\in dom(f)$ and $|x-x_0|<\delta$ imply $|f(x)-f(x_0)|<\epsilon$
I am confused.
I think you will see the difference if we explicitly write out the definitions for continuity and uniform continuity.
Let $f: X \rightarrow \mathbb{R}$ be a function from some subset $X$ of $\mathbb{R}$ to $\mathbb{R}$.
Def 1: We say that $f$ is continuous at $x$ if for every $\epsilon > 0$ there exists $\delta > 0$ such for every $y \in X$, if $|y - x| < \delta$ then $|f(y) - f(x)| < \epsilon$.
Def 2: We say that $f$ is continuous if $f$ is continuous at every $x \in X$.
This definition corresponds to what you called "continuity on a set" in your question.
Def 3: We say that $f$ is uniformly continuous if for every $\epsilon > 0$ there exists $\delta > 0$ such that for any two points $x, y \in X$, if $|y - x| < \delta$ then $|f(y) - f(x)| < \epsilon$.
As you have already noted in your question, the difference between uniform continuity and ordinary continuity is that in uniform continuity $\delta$ must depend only on $\epsilon$, whereas in ordinary continuity, $\delta$ can depend on both $x$ and $\epsilon$.
To illustrate this with a concrete example, let's say $f$ is a function from $\mathbb{R}$ to $\mathbb{R}$, and $f(10) = 15, f(20) = 95$ and $f(30) = 10$.
In uniform continuity, if I pick $\epsilon > 0$, you must give me a single $\delta > 0$ such that the following statements all hold:
On the other hand, in ordinary continuity, given the same $\epsilon > 0$ as above, you are free to pick three different deltas, call them $\delta_1, \delta_2, \delta_3$ such that the following statements all hold:
To answer the second part of your question, as Jacob has already pointed out, your statement is the sequential characterisation of ordinary continuity. Since ordinary continuity does not imply uniform continuity the "if" part of your statement is false. However, it is easy to see that uniform continuity implies ordinary continuity, so the "only if" part of your statement is true.
I believe in order to write a proof, one needs to be able to visualize what they are trying to prove mentally.
So here is an illustration I made for definition 2
Let $y=f(x)$ be a function.Let $x=x_o$ be a point of domain of f .The function f is said to be continuous at $x=x_o$ iff given $\epsilon \gt 0$,there exists $\delta \gt 0$ such that if $x\in (x_o−\delta,x_o+\delta )$, then $f(x)\in (f(x_o)−\epsilon ,f(x_o)+\epsilon )$.
And here is an illustration I made for definition 1
$f(x_0)$ exists;
$\lim_{x \to x_o} f(x)$ exists; and
$\lim_{x \to x_o} f(x)$ =$f(x_o)$.
Best Answer
The only difference there is that for the existence of the limit $f(a)$ does not need to be defined.
For continuity $f(a)$ must be defined (and must be equal to the limit).