# 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.

• 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. Nov 14, 2022 at 1:32
• 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?) Nov 14, 2022 at 1:36
• Another big flaw is to write things like $|f(x)-f(x_0)| <\dots \ge \epsilon,$ which allow to conclude nothing. Nov 14, 2022 at 1:40
• 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 $x \in \Bbb{R} \setminus \Bbb{Q}$. Nov 14, 2022 at 3:02
• 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 $x=1.41$? Nov 14, 2022 at 4:39

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

$$|x_2 - x_0| \geq \frac{A}{2}?$$

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 $$x_0$$.