# Proof of continuity verification.

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 ?

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.

• Why would you check $x=0$ separately ? $x=\frac1x$ cannot be fulfilled. – Yves Daoust Jun 10 at 16:17
• @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 – yousef magableh Jun 10 at 16:19
• @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 $x = 0$. – Theo Bendit Jun 10 at 16:20
• @TheoBendit: ok. Anyway it is pretty immediate, as $1/x$ diverges. – Yves Daoust Jun 10 at 16:21

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 $$-1$$.

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

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 $$1 if $$c > 1$$; or $$0 < x < c < y < 1$$ if $$0; or $$1 < x < 0 < y < 1$$ if $$c = 0$$; or $$-1 < x < c if $$-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 $$f(c) <:> \pm 1$$.