Proving discontinuity using $\epsilon-\delta$ definition of continuity My definition is a function f(x) is discontinuous at a point $x_{0}$ means that
$$
\exists \space \epsilon \space \forall \space \delta \space \exists \space x \space where \space |x-x_{0}| < \delta \space \land \space |f(x)-f(x_{0}| \ge \epsilon
$$
My function
$$
\begin{cases}
x^2+1 & if \space x \space is \space irrational \\
2x & if \space x \space is \space rational
\end{cases}
$$
If I’m understanding this correctly then I need to pick an $\epsilon$ and an x. The question asks to show discontinuity at $x_{0} = \sqrt{2}$.
My work -
Let $\epsilon$ = 1, if $\delta \gt 1$ let $x = \pi$ and if $\delta \le 1$, take x = $\frac{\delta}{2}$.
In the first case I get;
$$
|x-x_0| < \delta \land |f(x)-f(x_0)| = |\pi^2 - 2 | < |16-2| \ge 1 = \epsilon 
$$
In the second case I get;
$$
|x-x_0| \lt \delta \land |f(x) -f(x_0)| = | 2(\frac{\delta}{2}) - 3 | = | \delta - 3| \le|1/2 - 3| = 5/2 \ge 1 = \epsilon 
$$
Therefore I would write that $ x = min(\pi, \frac{\delta}{2})$.
Am I understanding this correctly, or have I made a mistake here? Appreciate any help.
 A: At $x_0 = \sqrt{2}$, for $x$ near $x_0$ and $x$ rational, you have that  
$f(x) = 2x \approx 2x_0 = 2\sqrt{2}.$
At $x_0 = \sqrt{2}$, for $x$ near $x_0$ and $x$ irrational, you have that 
$f(x) = x^2 + 1 \approx (x_0)^2 + 1 = 3.$

This is where things get very tricky.  This is a Math problem that actually requires sophisticated intuition.
Suppose that you have that
$$|x_1 - x_2| = A.$$
How could you then show that either
$$|x_1 - x_0| \geq \frac{A}{2}$$
or
$$|x_2 - x_0| \geq \frac{A}{2}?$$
Answer:
By invoking the triangle inequality.
$$A = |x_1 - x_2| \leq |x_1 - x_0| + |x_0 - x_2|. \tag1 $$
So, if both of the terms on the RHS of (1) above were $~< \dfrac{A}{2},~$ this would force the LHS of (1) above, $|x_1 - x_2|$ to be $< A$, which would contradict the premise that $|x_1 - x_2| = A.$

Based on the above analysis, the first thing to do is set
$\epsilon$ to be something like $~\dfrac{1}{40},~$ which is well below
$$[\sqrt{2}^2 + 1] - [2\sqrt{2}] = 3 - 2\sqrt{2}.$$

Once this is done, the problem would then reduce to showing that no how small $\delta$ is taken, there will be an $x_1$ and an $x_2,$ both within a neighborhood of radius $\delta$ around $x_0 = \sqrt{2}$ such that $x_0, x_1, x_2$ are all distinct from each other, and
$$|f(x_1) - f(x_2)| > \frac{1}{20}.$$
This will establish, per the previous analysis, that at least one of the elements in the neighborhood of radius $\delta$ around $x_0$, either $x_1$, or $x_2$ will be such that either
$$|f(x_1) - f(x_0)| < \frac{1}{40} ~~~\text{or}~~~ |f(x_2) - f(x_0)| < \frac{1}{40}.$$

This can be done by establishing the following, some or all of which might already be established:

*

*The function $g(x) = 2x$ is a continuous function at $x_0 = \sqrt{2}.$


*The function $h(x) = x^2 + 1$ is a continuous function at $x_0 = \sqrt{2}.$


*The rational numbers are dense in $\Bbb{R}.$ 
That is, given any $x_0 \in \Bbb{R},$ and given any neighborhood of radius $\delta$ around $x_0$, 
no matter how small (positive) $\delta$ is, 
there exists a rational number not equal to $x_0$ within the neighborhood of radius $\delta$ around $x_0$.


*The irrational numbers are dense in $\Bbb{R}.$ 
That is, given any $x_0 \in \Bbb{R},$ and given any neighborhood of radius $\delta$ around $x_0$, 
no matter how small (positive) $\delta$ is, 
there exists an irrational number not equal to $x_0$ within the neighborhood of radius $\delta$ around $x_0$.
