I am working on this proof, and wanted someone to check it and to help me understand what is happening in case (ii). The proof:

Let $f(x) = 1$ for rational numbers $x$, and $f(x)=0$ for irrational numbers. Show $f$ is discontinuous at every $x$ in $\mathbb{R}$.

We will consider two cases.

(i) $x \in \mathbb{Q}$

Consider the sequence $x_n = x + \frac{\sqrt{2}}{n}$. We have that $(x_n) \rightarrow x$, yet $x_n$ is irrational $ \forall \ n$. $\Rightarrow x_n \ \in \ \mathbb{Q}$

$\Rightarrow x_n - x \in \ \mathbb{Q}$.

$\Rightarrow n(x_n-x) = \sqrt{2} \in \mathbb{Q}$.$\ \rightarrow \leftarrow$

This is a contradiction, therefore we have that $f(x_n) = 0 \ \forall \ n \Rightarrow \lim f(x_n) =0 \neq 1 =f(x)$.

Therefore, $f(x)$ is not continuous at any $x \in \mathbb{Q}$.

(ii) $x \not\in \mathbb{Q}$

Given the density of the rationals, there exists a subsequence of rational numbers that must converge to $x$. Call this subsequence $(x_{n_r})$. Therefore, we have that:

$f(x_{n_r})=1 \ \forall \ n \Rightarrow \lim f(x_{n_r}) = 1 \neq 0 = f(x).$

$\therefore \ f(x)$ is not continuous at any $x \in \mathbb{Q}$.

By cases (i) and (ii), we have that $f(x)$ is not continuous at any $x \in \mathbb{R}$.

For case (i), I used the hint in the back of my text book to generate a sequence I could work with. Part (ii) is almost identical to the solution in my text- I don't fully understand why we are using the denseness of the rationals, rather than just working with the sequence made in case (i).

  • $\begingroup$ It is pretty easy to prove using the topological definition of continuity. Are you familiar? $\endgroup$ Oct 11, 2015 at 22:20
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    $\begingroup$ @uniquesolution That's a pretty harsh thing to say to aomeone trying to learn and showing effort. $\endgroup$ Oct 12, 2015 at 2:00
  • $\begingroup$ @TylerHG- I am not familiar. I'll look into for my own studies, though I expect my prof will want us to stick with our sequential definitions and/or the epsilons and deltas... $\endgroup$ Oct 12, 2015 at 2:07
  • $\begingroup$ @YoTengoUnLCD I certainly thought so... my question was not far removed from the questions s/he has answered in the past. Ah well. $\endgroup$ Oct 12, 2015 at 2:19

2 Answers 2


When $x$ is irrational, the sequence defined in case (i) might not consist of only rationals. For example, if $x = \sqrt 2$, then $x + \frac {\sqrt 2} n = \frac {n+1} n {\sqrt 2} $ is irrational. (It will converge to $x$, but it doesn't accomplish what's needed.)

For (ii), you need a sequence of rationals converging to the irrational $x$. In theory, we already know one: consider the decimal expansion of $x$. When $x$ is irrational, the sequence is necessarily infinite, doesn't eventually repeat itself forever. Suppose $$x = m + 0.d_1 d_2 \dotso$$ where $m$ is an integer. Let $$x_n = m + 0. d_1 \dotso d_n$$ In other words, $x_n$ is the decimal representation of $x$ cut off at the $n^{th}$ digit after the decimal point ($x_0 = m$). Then every $x_n$ is rational, and: $$lim_{n \to \infty} x_n = x$$

  • $\begingroup$ Ok. My professor actually suggested $x + \frac{\pi}{n}$ instead, though that would yield the same problem, wouldn't it? $\endgroup$ Oct 12, 2015 at 2:05
  • $\begingroup$ Using $\pi$, some of the terms of the resulting sequence might be rational (suppose $x = k\pi)$. A little extra verbiage is needed to show that the sequence is eventually all rationals. I emended my answer to give a more concrete example of a sequence that works. $\endgroup$
    – BrianO
    Oct 12, 2015 at 2:40
  • $\begingroup$ Thank you- that was helpful for part (ii). So, this would suffice to serve as a better way to show the second case (instead of trying to use the denseness argument), correct? $\endgroup$ Oct 12, 2015 at 2:48
  • $\begingroup$ Actually yes, I thnk so. The argument using density of the rationals is certainly true, but it's nonconstructive, in that it just plucks a sequence $x_n$ out of thin air, The sequence I gave is more constructive in flavor: if you can say anything explicit about the irrational $x$ (e.g. how to compute it, to within any degree of precision), then we can figure out how to convert that to our standard sequence of approximations given by the decimal representation. $\endgroup$
    – BrianO
    Oct 12, 2015 at 2:59
  • $\begingroup$ Thanks, BrianO. I often struggle with when we really need to construct something vs when we can just use pick something that will suffice for the proof. $\endgroup$ Oct 12, 2015 at 3:43

Rationals and irrationals are both dense in the real numbers. If we pick $x\in\mathbb{Q}$ then there is a sequence of irrationals converging to $x$, thus proving discontinuity of $f$ at $x$. If $x\notin\mathbb{Q}$ then there is a sequence of rationals...(I'm sure you can now formally make your complete proof)

  • $\begingroup$ Ah, gotcha. Thanks, that makes sense. $\endgroup$ Oct 12, 2015 at 2:06

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