$\epsilon{\rm -}\delta$ proof of the discontinuity of Dirichlet's function How can I prove that the function 
$$   f(x) = \left\{\begin{array}{l l}     x &\text{if }x \in \mathbb{Q} \\     -x & \text{if } x \notin \mathbb{Q}   
\end{array} \right. $$
is discontinuous for $x \neq 0$, using $\epsilon$'s and $\delta$'s?
I see that is truth. But I cannot prove using only $\epsilon$'s and $\delta$'s.
 A: Let $x_0\ne 0$ be fixed. Informally, $f$ is continuous at $x_0$ would mean that $f(x)$ can be made as close to $f(x_0)$ as desired by taking $x$ sufficiently close to $x_0$.  With the epsilon business, this means that $\color{maroon}{ \text{for every}}$ $\epsilon>0$, there is a $\delta>0$ such that $$\tag{1}
\underbrace{|f(x_0)-f(x)|<\epsilon}_{f(x)\text{ is close to }f(x_0)}
\quad\text{ whenever }\quad
\underbrace{|x-x_0|<\delta\phantom{f}}_{x\text{ is close to }x_0}.$$
If $f$ were not continuous at $x_0$, this would mean that there  $\color{maroon}{ \text{is some}}$ $\epsilon>0$ such that no matter what $\delta$ you choose, equation $(1)$ is not true. Informally, this means that no  matter how close you take $x$ to $x_0$, you are not guaranteed that  $f(x)$ is close to $f(x_0)$  and, in fact, you can find an $x$ as close to $x_0$ as you wish with $f(x)$ "far" away from $f(x_0)$. 
So you need to show the following is true: 
S: There is some $\epsilon>0$ such that for every $\delta>0$, there is a $x_1$ such that both $|x_1- x_0|<\delta$ and $|f(x_1)-f(x_0)|\ge \epsilon$.
Well, how to accomplish this? Let's consider the graph of our function (as far as we can; of course, it's impossible to actually draw the graph of $f$  but we'll use our imaginations) and try to get some intuition as to what the required $\epsilon$ should be. 
The graph of your function resembles the  graphs of the lines $y=\pm x$; or, if you will, an X shape. For rational values of $x$, the point $\bigl(x,f(x)\bigr)$ is on the graph of  $y=x$; and  for irrational values of $x$, the point $\bigl(x,f(x)\bigr)$ is on the graph of  $y=-x$.
I suggest you draw the picture now (when I have time, I'll provide one though).
Now if you take any open interval $I$ containing $x_0$, you can find both a rational number and an irrational number in $I$. (Why? This is the gist of the whole argument, so it's worthwhile that you convince yourself that you can.) Thus, you can find a point $x_1$ in $I$ that is irrational if $x_0$ is rational and which is rational if $x_0$ is irrational. It will follow then that  $f(x_0)$ and $f(x_1)$ are far apart. 
In fact, they are approximately $2|x_0|$ away from each other when $I$ has small length, since, if you'll allow a sloppy and vague statement, $f(x_0)$ and $f(x_1)$ are on different arms of the X.
This leads us to suspect that $|x_0|$ would serve for the $\epsilon$ needed in statement S.
And
now it's a matter to formalize things. We need to show that statement S holds.
Let $0 < \epsilon < |x_0|$.  
Let $\delta>0$ be arbitrary.
We need to show that there is an $x_1$ with $|x_0-x_1|<\delta$ such that
$|f(x_0)-f(x_1)|\ge\epsilon $.
The preceding discussion points to a way towards finding $x_1$. Choose $x_1$ to be an irrational number with $|x_0-x_1|<\delta$ if $x_0$ is rational. If $x_0$ is irrational, choose $x_1$ to be an  rational number with $|x_0-x_1|<\delta$.
Now we show this choice of $x_1$ "works":
Either 
$\ \ \ \ \ f(x_0)=x_0$ and $f(x_1)=-x_1$ 
or
$\ \ \ \ \ f(x_0)=-x_0$ and $f(x_1)= x_1$. 
In either case $|f(x_0)-f(x_1)| = |x_0|+|x_1|\ge |x_0|>\epsilon. $
And that's it...
A: Let $x>0$, and assume that we can find a $\delta>0$ such that if $|y-x|\leq\delta$ then $|f(x)-f(y)|<x/2$. If $x$ is rational, it means $|x-f(y)|<x/2$ so $x/2<f(y)<3x/2$ for all $y\in (x-\delta,x+\delta)$. In particular, $y\in\mathbb Q$ for all $y\in (x-\delta,x+\delta)$. 
If $x$ i irrational, then $|x+f(y)|<x+2$ so $x/2<-f(y)<3x+2$ and $y$ has to be irrational if it is in $(x-\delta,x+\delta)$.
Do you see why it leads to a contradiction?
