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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.

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    $\begingroup$ You can't take $\pi$ because $|\sqrt 2 - \pi| > \delta$ for any $\delta < \pi - \sqrt 2$. To be blunt. No, you don't understand this. This isn't a matter of picking an $x$. It's a matter of picking an $\epsilon$ so that no matter what $\delta$ you pick there will always be $x$ that fails. YOu can't actually pick the $x$ because if you pick a smaller $\delta$ that $x$ won't be in the interval anymore. $\endgroup$
    – fleablood
    Nov 14, 2022 at 1:32
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    $\begingroup$ The intuitive things is that for rational numbers near $\sqrt 2$ the $2x$ are all near $2\sqrt 2$. (Not equal to $2\sqrt 2$ because the $x$s are rational but *near* $2\sqrt 2$). And for the irratioanl numbers nears $\sqrt 2$ the $x^2 + 1$ are near $3$. So you need to turn that informal argument formal. (Hint: Let $\epsilon = 3-2\sqrt 2$ ... do you see what would happen then?) $\endgroup$
    – fleablood
    Nov 14, 2022 at 1:36
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    $\begingroup$ Another big flaw is to write things like $|f(x)-f(x_0)| <\dots \ge \epsilon,$ which allow to conclude nothing. $\endgroup$ Nov 14, 2022 at 1:40
  • $\begingroup$ MathJax tip: Writing if \space x \space is \space irrational is painfully long, and doesn't look great. Instead, consider using the \text{} command. You can write $\text{if $x$ is irrational}$ to get $\text{if $x$ is irrational}$. Alternatively, you can write $x \notin \Bbb{Q}$ to get $x \notin \Bbb{Q}$, or $x \in \Bbb{R} \setminus \Bbb{Q}$ to get $x \in \Bbb{R} \setminus \Bbb{Q}$. $\endgroup$ Nov 14, 2022 at 3:02
  • $\begingroup$ If $\delta>1$, say $\delta=1.1$ and $x_0=\sqrt{2}=1.41\dots$ then your value $x=\pi$ does not satisfy $|x-x_0|<\delta$. So the choice for $x$ is wrong. The choice of $x$ should be very near to $x_0$ so that the desired goal is met. Why not try $x=1.41$? $\endgroup$
    – Paramanand Singh
    Nov 14, 2022 at 4:39

1 Answer 1

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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 $x_0$.

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