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Consider $f(x)=\begin{cases} x, & \text{if $x$ is rational} \\ \frac{1}{x}, & \text{if $x$ is irrational} \end{cases}$

I proved that $f$ is continuous only at $1$ and $-1$ as follows:

Case $1$: $c=0$.

Case $2$: $c$ is rational number that is different than $0,1$ and $-1$.

Case $3$: $c$ is irrational number.

And in each case I assumed that $f$ is continuous and manged to arrive to a contradiction by constructing a sequence say $\{x_n\}$ that converges to $c$, but $\{f(x_n) \}$ either diverges or it converges to a number different than $f(c)$.

And for the case of $1,-1$, I proved that f is continuous by the definition of continuity.

Is my result correct or did I miss something ? And is there a shortcut or faster method than what I did ?

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The approach is sound. Without checking the individual steps, it's difficult to verify the proof beyond this point, or to recommend shortcuts.

I don't see to much to shortcut here anyway. You could find most of the points of discontinuity by using the density of the rationals and the irrationals. For any point $x \in \Bbb{R}$, you know that there is a sequence of rationals $a_n \to x$ and a sequence of irrationals $b_n \to x$. Since $x$ is continuous, $f(a_n) = a_n \to x$. Since $\frac{1}{x}$ is continuous (for $x \neq 0$), $f(b_n) \to \frac{1}{x}$.

If $x \neq \frac{1}{x}$ (and $x \neq 0$), then you have two sequences converging to $x$, that map to sequences converging to different limits. This would contradict $f$ being continuous at $x$.

This doesn't establish that $f$ is continuous at $x = \pm 1$, so more individual verification is needed. The $x = 0$ case needs to be dealt with separately too.

So, I wouldn't consider it much of a shortcut! But it might cut down your argument a little.

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  • $\begingroup$ Why would you check $x=0$ separately ? $x=\frac1x$ cannot be fulfilled. $\endgroup$ – Yves Daoust Jun 10 at 16:17
  • $\begingroup$ @YvesDaoust That is what I'm worried about ... do I have to check that case since your argument dose not account for a separate case fo x=0 $\endgroup$ – yousef magableh Jun 10 at 16:19
  • $\begingroup$ @YvesDaoust Because it doesn't work with the argument used for $x \neq 0$. The argument used the continuity of $\frac{1}{x}$ at the given $x$. This is not true (or meaningful) at $x = 0$. $\endgroup$ – Theo Bendit Jun 10 at 16:20
  • $\begingroup$ @TheoBendit: ok. Anyway it is pretty immediate, as $1/x$ diverges. $\endgroup$ – Yves Daoust Jun 10 at 16:21
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The limit of $f(x)$ by the rationals is $x$. The limit of $f(x)$ by the irrationals is $\dfrac1x$. So for the limit of $f(x)$ to exist, one must fulfill

$$x=\frac1x.$$


In a neighborhood of $x=1$, $|x-1|$ and $|\frac1x-1|=\frac{|x-1|}x$ are virtually equal, so that a $\delta-\epsilon$ argument is easy. Same around $-1$.

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  • $\begingroup$ How to make this argument precise ( I mean using the definition ) ? $\endgroup$ – yousef magableh Jun 10 at 16:01
  • $\begingroup$ @yousefmagableh: $x=1/x$ is an obvious necessary condition (convergent subsequences). You can show that it is sufficient by computing the limits at $\pm1$. $\endgroup$ – Yves Daoust Jun 10 at 16:04
  • $\begingroup$ You mean using Bolzano theorem ? $\endgroup$ – yousef magableh Jun 10 at 16:07
  • $\begingroup$ @yousefmagableh: nope. If two subsequences converge to different limits, the limit does not exist. $\endgroup$ – Yves Daoust Jun 10 at 16:09
  • $\begingroup$ @yousefmagableh: no, it doesn't require BW. $\endgroup$ – Yves Daoust Jun 10 at 16:19
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You method (if you did it correctly) is sound and solid.

An alternative that is perhaps more direct but not shorter or easier would be an $\epsilon$ delta proof.

Rough argument For any $c$ not equal to $\pm 1$ and any $\delta$ you can find rational $x < c$ and irrational $y> c$ so that $x,y \in (c-\delta, c+\delta)$ so that also; either $1<x<c < y$ if $c > 1$; or $0 < x < c < y < 1$ if $0<c < 1$; or $1 < x < 0 < y < 1$ if $c = 0$; or $-1 < x < c <y$ if $-1 < c < 0$; or $x < c < y < 1$ if $c< -1$.

If we compare the values of $f(x)$ to $f(y)$ to $f(c)$... well, if $c =0$ then $|f(y)-f(c)| = |\frac 1y - 0| = \frac 1y > 1$. Otherwise $f(c)$ will be on "one side" of $\pm 1$ whereas one or the other of $f(x)$ or $f(y)$ will be on the other so that will be a $|f(x|y) - f(c)| > |c -(\pm 1)|$.

So for any $\epsilon$ less than $|c-(\pm 1)|$ there is no $\delta$ that can guarantee that $|w - c|< \delta \implies |f(w) - f(c)|< \epsilon$.

So $f$ is not continuous at $c$.

Of course there are cases to be made. ($c$ rational vs. $c$ irrational and $c= 0$ vs. $0< |c| < 1$ vs. $|c| > 1$ ) But the argument is the same: you can always find an irrational or rational $w$ so that $f(w) >:< \pm 1$ while rational or irrational $c$ is such that $f(c) <:> \pm 1$.

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