Vtyjkyuk

An analysis book asks the following question:

Determine where the following function is continuous:

$$ f(x) = left{ beginarrayll x & quad textif x text is
rational \ frac1x & quad textif x text is irrational
endarray right. $$

I claim that $f$ is discontinuous everywhere. For if $(x_n)$ is an irrational sequence converging to some $c in mathbbQ$ then $lim f(x_n) = lim frac1x_n = 1/c ne f(c)$ and similarly if $(x_n)$ is a rational sequence converging to some $c in mathbbR setminus mathbbQ$ then $lim f(x_n) = lim x_n = c ne f(c)$.

Is my answer and justification correct?

Update :
It looks as if $f$ is continuous at $pm 1$. Here's my attempted proof of claim made:

Suppose $(x_n)$ be a real sequence converging to 1. Let $varepsilon > 0$ be given.

Then there exists $N_1 in mathbbN$ such that $|x_n-1|< varepsilon$ for all $nge N_1$.

Also there exists $N_2 in mathbbN$ such that $|x_n-1|< varepsilon / 2 $ for all $nge N_2$.

And finally, there exists $N_3 in mathbbN$ such that $|x_n|ge frac12$ for all $nge N_3$.

Let $N:=max N_1 , N_2, N_3 $. Now for all $n ge N$ and $x_n in mathbbQ$, we have

$|f(x_n) - f(1)|= |x_n -1| < varepsilon $

and for $x_n in mathbbR setminus mathbbQ$, we have

$|f(x_n) - f(1)| = |frac1x_n -1| = fracx_n < 2 cdot fracvarepsilon2 = varepsilon$.

Similarly it can be done for $c=-1$.

Is this okay or is there any better way this can be done?

For $x_0=1$ notice that $$|x-x_0|<epsilonto 1-epsilon<x<1+epsilon$$then we have $$textif xinBbb Q to |x-1|<epsilon\textif xnotinBbb Q to 1-2epsilon<dfrac11+epsilon<dfrac1x<dfrac11-epsilon<1+2epsilon$$where $epsilon$ is small enough therefore $$|dfrac1x-1|<2epsilon$$and the function is continuous $x=1$ similarly $x=-1$