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I am trying to better understand uniform continuity and working with the definition. My knowledge is that Given $\epsilon >0$ your choice for $\delta$ must only depend on $\epsilon$. However, this definition seems very similar to continuity on a set. What exactly is the difference with continuity on $S$ and uniform continuity on $S$?

for each $x_0 \in S$ and $\epsilon>0$ there exists $\delta>0$ s.t $x\in S$ and $|x-x_0|<\delta \implies |f(x) - f(x_0)|<\epsilon$

If I changed $x\in S$ to $x \in dom(f)$ would I get continuity on a set?

However is the following statement correct? $f$ is uniformly continuous on $S$ iff for $x_n \in S$ converging to $x_0 \implies \lim f(x_n) = \lim f(x_0)$ for all $x_0$ in $S$

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I don't understand what change you are proposing when you write, "If I changed $x\in S$...". If $S$ is the domain of $f$, these are literally the same thing. A relevant reference: – Jonas Meyer Nov 28 '12 at 6:23
I am saying $S$ is not the domain of $f$ – CodeKingPlusPlus Nov 28 '12 at 6:27
Your statement is the sequential characterization of continuity and is not equivalent to uniform continuity. – JSchlather Nov 28 '12 at 6:27
@CodeKingPlusPlus: Thank you. I still do not understand what that question means, but maybe it's just me. A function can be defined to be continuous on a set if it is continuous at each point in the set. Your other questions are clear, and I expect good answers to be posted reasonably soon. – Jonas Meyer Nov 28 '12 at 6:32

1 Answer 1

up vote 12 down vote accepted

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:

  1. If $|y - 10| < \delta$ then $|f(y) - 15| < \epsilon$.
  2. If $|y - 20| < \delta$ then $|f(y) - 95| < \epsilon$.
  3. If $|y - 30| < \delta$ then $|f(y) - 10| < \epsilon$.

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:

  1. If $|y - 10| < \delta_1$ then $|f(y) - 15| < \epsilon$.
  2. If $|y - 20| < \delta_2$ then $|f(y) - 95| < \epsilon$.
  3. If $|y - 30| < \delta_3$ then $|f(y) - 10| < \epsilon$.

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.

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+1 great example – Sir Jective Oct 14 at 23:49

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