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Given that $f(x) = -1$ if $f$ is irrational and $f(x)=1$ if $f$ is rational, show that $f$ is not continuous anywhere.

  1. Let's show that between any 2 rationals there is always an irrational number

Let's consider $a,b \in Q$ such that $a<b$, there is an infinite number of rational $r$ such that $a<r<b$. Where $r=\frac{a+b}{n}$ where $n \in Z$

Also $$a<r<b$$, $$ a +\sqrt{2}<r<b+\sqrt{2}$$ , $$a<r- \sqrt{2}<b$$

It follows that we have always an irrational between any two rationals

  1. Between any two irrationals, we always have a rational.

Here I am not sure how to proceed.

  1. One can find a rational in between two irrationals and vice versa an irrational between two rationals.

Therefore as the value of $x$ approaches a value from the left or right of $r$ , $x$ will oscillates between a rational and irrational infinitely. Therefore limits will oscillate between $1$ and $-1$ infinitely. it shows that $$f(r^-) \neq f(r^+) \neq f(r)$$

it follows that there no continuity anywhere on $D_f$

Is this correct? Is there a more efficient (clean) way to show the discontinuity?

Any input is much appreciated

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    $\begingroup$ For another approach: Take two sequences approaching to $c$: one sequence has only rationals, the other only irrationals. Now your function converges to two different limits as $x \to c$, hence discontinuous at $x=c$. $\endgroup$
    – Anurag A
    Jul 11, 2017 at 22:20
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    $\begingroup$ As written, $f$ is actually continuous. Assume $f(x)=-1$, then $f$ is rational ($-1$ is a rational number), a contradiction, since $f(x)=-1$ only if $f$ is irrational. Thus, $f(x)=1$ for all $x$. $\endgroup$ Jul 11, 2017 at 23:51

3 Answers 3

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Pick a real number $x$. We will show that $f$ is discontinuous at $x$.

Side Remark: Discontinuity of $f$ at $x$ can be shown by exhibiting at least once sequence $y_{n}$ that converges to $x$, yet violates the requirement $$ \lim_{n \rightarrow \infty} f(y_{n}) = f(\; \lim_{n \rightarrow \infty} y_{n} \;). $$ (Essentially, $f$ is continuous at $x$ if and only if $f$ commutes with the limit operation of convergence to $x$.)

Now, the proof:

Case 1: $x$ is rational. Then there exists a sequence $y_{n}$ of irrational numbers that converges to $x$. Thus, $$ \lim_{n \rightarrow \infty} f(y_{n}) = -1 \neq 1 = f(\; \lim_{n \rightarrow \infty} y_{n} \;). $$

Case 2: $x$ is irrational. Then there exists a sequence $y_{n}$ of rational numbers that converges to $x$. By an argument analogous to that in Case 1, $$ \lim_{n \rightarrow \infty} f(y_{n}) \neq f(\; \lim_{n \rightarrow \infty} y_{n} \;). $$

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Take a rational $x $ and the irrational sequence $x_n=x+\frac {\pi}{n}$ such that $$\lim_{\infty}x_n=x$$ then

$$\lim_{\infty}f (x_n)=-1\ne f (x) $$ hence, $f $ is not continuous at $x $.

Take an irrational $y $ and the rational sequence $y_n=\frac {\lfloor 10^ny \rfloor}{10^n} $ such that $$\lim_{\infty}y_n=y $$ then $$\lim_{\infty}f (y_n)=1\ne f (y) .$$ You can conclude.

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  • $\begingroup$ If $y$ is irrational then how can $y_n+\frac{1}{n}$ be a "rational" sequence converging to $y$?? $\endgroup$
    – Anurag A
    Jul 12, 2017 at 3:06
  • $\begingroup$ I think you are missing the point. I am "not" saying that a rational sequence cannot converge to an irrational number, in fact that is the very thing I have written myself in the comments above. What I am saying is that if $y$ is an irrational number then $y+1/n$ will be irrational for all $n \geq 1$, hence it cannot be the sequence of "rational" terms converging to $y$. $\endgroup$
    – Anurag A
    Jul 12, 2017 at 16:17
  • $\begingroup$ @AnuragA Hello, You're right. i edited.Thanks. $\endgroup$ Jul 12, 2017 at 21:55
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The key is exactly that between two real numbers there are a rational number and an irrational number.

Your final argument is a bit vague. You can make it rigorous by recalling that, when a function is continuous and positive at a point, it assumes only positive values in a neighborhood of the point. Let's state it more precisely.

Suppose $f$ is continuous at $c\in\mathbb{Q}$; then there exists $\delta>0$ such that, for $|x-c|<\delta$, $f(x)>0$.

There exists $x$ irrational such that $c-\delta<x<c+\delta$: contradiction.

Similarly for $c$ irrational.


More generally, the function $$ g(x)=\begin{cases} a & x\in\mathbb{Q} \\[4px] b & x\in\mathbb{R}\setminus\mathbb{Q} \end{cases} $$ is nowhere continuous when $a\ne b$, because $$ f(x)=\frac{2(g(x)-b)}{a-b}-1 $$

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