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I was thinking about the following problem:

Consider the function $f:\mathbb R \rightarrow \mathbb R$ defined by $$\begin{align}f(x) &=\begin{cases} x^{4} \,\,\,\,\,\,\,\,\,\,\,\,\text{if x is rational} \\ 2x^{2}-1 \,\text{if x is irrational}\end{cases}\end{align}$$ . Let $S$ be the set of points where $f$ is continuous. Then which of the following is correct?

(a) $S=\{1\}$

(b) $S=\{-1\}$

(c) $S=\{-1,1\}$

(d) $S=\phi$.

Can someone point me in the right direction? Please help.Thanks in advance for your time.

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Since your two functions are both continuous on the reals, find the points where they are equal.... – John Dec 14 '12 at 19:17
up vote 2 down vote accepted

First note that $x^4=2x^2-1$ occurs only when $x=\pm 1$.

Fix $x_0\in \mathbb{R}$ and let $(q_n)$ be a sequence of rational numbers and $(i_n)$ be a sequence of irrartional numbers so that $\lim_{n\to +\infty}q_n=\lim_{n\to +\infty}i_n=x_0$ (why do such sequences exist???). Then, $$\lim_{n\to +\infty}f(q_n)=\lim_{n\to +\infty}q_n^4=x_0^4$$ while $$\lim_{n\to +\infty}f(i_n)=\lim_{n\to +\infty}2i_n^2-1=2x_0^2-1$$ What can you say about continuity at $x_0$? What happens when $x_0=\pm 1$?

EDIT: Here is the rest:

If $x_0\neq \pm 1$ then $\lim_{n\to +\infty}f(q_n)\neq \lim_{n\to +\infty}f(i_n)$ which means $f$ is not continuous at $x_0$. If $x_0=\pm 1$ then $\lim_{n\to +\infty}f(q_n)=\lim_{n\to +\infty}f(i_n)$ which means $f$ is continuous at $x_0=\pm 1$.

As a final note, the existence of the two sequences follows from the density of rational and irrational numbers in $\mathbb{R}$

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What can i guess is that the aforementioned function is continuous only at the points $\pm 1.$ So,the set $S=\{-1,1\}.$ Am i right,sir? – user52976 Dec 14 '12 at 19:49
@user52976 You are right. – Nameless Dec 14 '12 at 19:51
Thanks a lot,sir. I have got it. – user52976 Dec 14 '12 at 19:52
@user52976 No problem. I will add some extra details to complete my answer – Nameless Dec 14 '12 at 19:53

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