Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a homework problem, so I'll tag it as such, but I'm having a bit of trouble in my Real Analysis class. The problem I have is this:
Prove that the function $f$ is continuous at the point $a$ if and only if, for every $\epsilon>0$, there is a $\delta>0$ such that $|f(x)-f(y)|<\epsilon$ whenever $x$ and $y$ are both in the interval $(a-\delta, a+\delta)$.

I feel like this should be rather simple, but I just need a starting place - some of the logic is still new to me and I don't really know where to begin. If anyone would be so kind as to give me a hand I'd really appreciate it!


EDIT: I know as much to say that if $f$ is continuous at $a$ then there exists an $\epsilon$-neighborhood about $a$ in the domain, but why does $x$ and $y$ being in the $\delta$-neighborhood of $a$ imply that they are in the $\epsilon$-neighborhood? If $x$ and $y$ were both in the $\epsilon$-neighborhood of $a$, $|f(x)-f(y)|$ would always be less than $\epsilon$, correct? And thus the conclusion, but how does the $\delta$-neighborhood come into play?

share|cite|improve this question
Do you mean $|f(x)-f(y)|<\epsilon$? – Brian M. Scott Oct 22 '12 at 1:22
I do. It's been fixed. – roboguy12 Oct 22 '12 at 1:23
The epsilon-delta criterion for continuity can be expressed as such: A function $f(x)$ is continuous at $x_o$ if $\forall$ e>0, $\exists$ d>0 such that 0 < |$x$ - $x_o$| < d $\rightarrow$ |$f(x)$-f($x_o$)| < e – Arthur Collé Oct 22 '12 at 1:26
up vote 2 down vote accepted

I’ll give you a hand with one direction to get you started. Suppose that $f$ is continuous at $a$, and let $\epsilon>0$ be arbitrary; we must find a $\delta>0$ such that $$|f(x)-f(y)|<\epsilon\quad\text{whenever}\quad x,y\in(a-\delta,a+\delta)\;.$$ Since $f$ is continuous, we know that there is a $\delta_\epsilon>0$ such that $$|f(x)-f(a)|<\epsilon\quad\text{whenever}\quad x\in(a-\delta_\epsilon,a+\delta_\epsilon)\;,$$ but that’s not quite good enough: it tells us that $|f(x)-f(a)|<\epsilon$ and $|f(y)-f(a)|<\epsilon$ whenever $x,y\in(a-\delta_\epsilon,a+\delta_\epsilon)$, but a little experimentation should convince you that $|f(x)-f(y)|$ can still be larger than $\epsilon$.

However, we have a tool that directly relates $|f(x)-f(y)|$ to $|f(x)-f(a)|$ and $|f(y)-f(a)|$: the triangle inequality, which tells us that


Thus, we can conclude that $$|f(x)-f(y)|<2\epsilon\quad\text{whenever}\quad x,y\in(a-\delta_\epsilon,a+\delta_\epsilon)\;.$$

If we’d chosen a $\delta_{\epsilon/2}$ so that $|f(x)-f(a)|<\frac{\epsilon}2$ whenever $x\in(a-\delta_{\epsilon/2},a+\delta_{\epsilon/2})$, instead of the $\delta_\epsilon$ that we actually did use, then ... ?

share|cite|improve this answer
Then we'd have $|f(x)-f(y)|<\epsilon$! Brilliant, I'd never think to use the Triangle Inequality here, but what value of $\delta_\epsilon$ can we pick to make this work? I'm a little shaky on how the values of $\delta$ can affect the value of $\epsilon$. – roboguy12 Oct 22 '12 at 1:42
@roboguy12: Since you’re proving a general result rather than working with a specific function at a specific point $a$, you don’t have to worry about such computational details. Exactly how $\delta$ and $\epsilon$ interact isn’t important; all that matters is that the hypothesis that $f$ is continuous ensures that a $\delta_\epsilon$ with the stated properties actually exists. – Brian M. Scott Oct 22 '12 at 1:45
Okay, so because $f$ is continuous, there exists a specific $\delta_{\frac{\epsilon}{2}}$ such that, if $x\in(a-\delta_{\frac{\epsilon}{2}},a+\delta_{\frac{\epsilon}{2}})$ then $|f(x)-f(a)|<\frac{\epsilon}{2}$, just because of the definition of continuity? – roboguy12 Oct 22 '12 at 1:49
@roboguy12: Yes, that’s right. The definition of continuity of $f$ at $a$ is that for each $\epsilon>0$ there is a $\delta_\epsilon>0$ such that $|f(x)-f(a)|<\epsilon$ whenever $x\in(a-\delta_\epsilon,a+\delta_\epsilon)$. – Brian M. Scott Oct 22 '12 at 2:08
Okay, got it. Thank you very much!! – roboguy12 Oct 22 '12 at 2:09

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.