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.

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

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 $x\in\mathbb R$.

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  • $\begingroup$ 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

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