-1
$\begingroup$

A local book problem: A function $f:\mathbb{R}\to\mathbb{R}$ is continuous on $\mathbb{R}$ and $f(x)=0$ for all $x\in\mathbb{Q}$. Prove that $f(x)=0$ for all $x\in\mathbb{R}$. There was a hint in the book:Let $c\in\mathbb{R}$. Consider a sequence of rational point ${c_n}$ converging to c. Use sequential criterion for continuity. I could not understand how to use the sequential criterion of continuity. I also have a question about the sequential criteria. Sequential criterion states that: Let $D\subset\mathbb{R}$ and $f:D\to\mathbb{R}$ be a function. Let $c\in D\cap D^c$. $f$ is continuous at $c$ if and only if for every sequence ${x_n}$ in $D$ converging to $c$, the sequence ${f(x_n)}$ converges to $f(c)$. I could not understand how can there be a common point between a set and it's complement set. Please help. Thanks in advance.

$\endgroup$
  • $\begingroup$ Please mention the reason too for down voting so I could improve. $\endgroup$ – Mansi Mar 16 at 3:45
1
$\begingroup$

Let $c\notin \mathbb Q$. There is a sequence of rational $(c_n)$ s.t. $c_n\to c$ when $n\to \infty $. Now, $$f(c_n)=0.$$ By continuity, $$\lim_{n\to \infty }f(c_n)=f(c).$$ Since $f(c_n)=0$ for all $n$, we get, $f(c)=0$. Since $c\in \mathbb R\setminus \mathbb Q$ is unspecified, we get $f(x)=0$ for all $x\in\mathbb R$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks! But can a sequence of rationals converge to an irrational? $\endgroup$ – Mansi Mar 16 at 3:43
  • $\begingroup$ Oh sorry! Got it. Sequence of rationals can converge to any number (rational or irrational) depending upon the sequence. Sorry and thanks again!! $\endgroup$ – Mansi Mar 16 at 3:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.