Why is every continuous function on the reals determined by its value on rationals?

Question

Why is every continuous function on the reals determined by its values on the rationals?

Answer 1

Since the rationals are dense in the reals, we can construct a sequence $(q_n)_1^\infty$ of rational numbers approaching any real number $x \in \mathbb{R}$.

Since $f$ is continuous, we know that can achieve the second equality in the statement below (we can pass limits under application of continuous functions): $$f(x) = f\left( \lim_{n \to \infty} q_n \right) = \lim_{n \to \infty} f(q_n).$$

The expression on the right hand side is only in terms of the value of the function at rational numbers. Thus, since we know what a continuous function $f$ does with rational numbers that are arbitrarily close to a real number $x$, we know what it will map that real number $x$ to.

Answer 2

Since the rationals are dense in $\mathbb{R}$.

If we choose a $x_i \in \mathbb{Q}^{\mathcal{c}}$, we can construct a sequence of rational numbers $(x_n)_{n\in\mathbb{N}}$ such that $\lim_{n\to\infty}x_n = x_i$.

Recall, a characterization of continuity is that $\lim_{x\to x_0} f(x) = f(x_0)$. Hence,

$$ \lim_{n\to\infty}f(x_n) = f(x_i) $$

So, if we know $f|_{\mathbb{Q}}$, we actually can extend this to $f$.

Answer 3

Suppose I have two continuous functions $f,g:\mathbb{R}\to\mathbb{R}$ that agree at every rational number. You want to conclude that $f(x)=g(x)$ for every real number $x$. Alternatively, you can show that $f(x)-g(x)=0$ for every real number $x$.

$f-g$ is a continuous function on $\mathbb{R}$, and $(f-g)(q)=0$ for every rational number $q$. Let $x$ be an arbitrary real number. Since the rationals are dense in the reals, we choose a sequence of rational numbers converging to $x$. On this sequence $f-g$ is identically zero, and passing to the limit by continuity, we conclude that $(f-g)(x)=0$. Since $x$ was arbitrary $f-g$ is identically zero on $\mathbb{R}$. So a continuous function on $\mathbb{R}$ is uniquely determined by its values on $\mathbb{Q}$.

Answer 4

Hint: You need to use the property of continuous functions that says continuous functions map convergent sequences to convergent sequences, that is if $f$ is continuous and $ x_n \to a $, then $ f(x_n) \to f(a) $. The other fact you need is that if $a$ is irrational, then there exists a Cauchy sequence of rational numbers such that $x_n \to_{n\to \infty} a$.