Let $f, g : [0, 1] \to [0, 1]$ be continuous functions that agree on the rationals inside $[0, 1]$. Show that $f = g$.

Question

Let $f, g : [0, 1] \to [0, 1]$ be continuous functions that agree on the rationals inside $[0, 1]$. Show that $f = g$. Give an example of a continuous function $f : \mathbb Q \to\mathbb R$ such that there is no continuous function $g : \mathbb R \to\mathbb R$ with the property that $f = g|_\mathbb Q$

I am not really understanding the problem. From my notes, I have " Let $x\in [0,1]$ such that $f(x) \neq f(y)$. Considering a neighborhood of $f(x),$ and $f(y), V$ and $W$ respectively. Then there exists a $q \in V\cup\mathbb Q$ and $r \in W\cup\mathbb Q$. "

How does this prove $f = g$?

Answer 1

Your notes should start "Let $x$ be such that $f(x) \neq g(x)$" Where did $y$ come from? If we want to prove $f(x)=g(x)$ we are interested in the values of the two functions at a common argument. Since the rationals are dense, you can take the limit of $f(x)$ and $g(x)$ approaching $x$. The limits have to agree because $f(x)=g(x)$ on the rationals. You find $f(x)=g(x)$ for all $x$.

Answer 2

To give a failure of continuous extension, here is an example:

Let $f:\mathbb{Q}\rightarrow\mathbb{R}$ be such that $f(p/q)=1/q$, where $p/q$ has its simplest form, assume that there is such a continuous extension $g:\mathbb{R}\rightarrow\mathbb{R}$.

Then consider an irrational point $r\in(0,1)$ and pick a sequence of rational points $(p_{n}/q_{n})\subseteq(0,1)$ such that $p_{n}/q_{n}\rightarrow r$.

The set $\{q_{n}\}$ cannot be of finitely many, for then $r$ will become a rational number.

Therefore a subsequence $(n_{k})$ is such that $q_{n_{k}}\rightarrow\infty$, so with $g(p_{n}/q_{n})\rightarrow g(r)$, we obtain $1/q_{n}\rightarrow r$ and hence $r=0$.

This shows that $g(r)=0$ for all irrational points in $(0,1)$. Now given any rational point $p/q\in(0,1)$ with the simplest form, we can approximate it by irrational points $(r_{n})\subseteq(0,1)$ but then $g(r_{n})\rightarrow g(p/q)$ will entail that $0\rightarrow 1/q$, so $1/q=0$, a contradiction.