Show for $f$ defined on $\mathbb Q \cap [0,1]$ uniformly continuous, exists extension $g$ defined on $[0,1]$ that is continuous Suppose that $f$ is a uniformly continuous function defined on $\mathbb Q \cap [0,1]$. Show that there is a unique continuous function $g$ defined on $[0,1]$ so that $$g(x)=f(x) \text{ for every } x \in \mathbb Q \cap [0,1].$$

Not sure what I am supposed to do here. I've seen answers where a sequence was defined to converge to a point $x \in \mathbb Q \cap [0,1]$ but I'm not understanding why we are doing this.
I know that continuity preserves converges of sequences and uniform continuity preserves Cauchy sequences.
Any help on getting started with this exercise?
 A: If such a continuous $g$ exists, then, as you stated, it must preserve convergence of sequences. At this point we don't know the value of $g$ at irrational points $x$, but we do know it must satisfy
$$f(x_n) = g(x_n) \to g(x)$$
for any sequence of rational points $x_n$ that converge to the irrational point $x$.
Thus, an obvious candidate for the definition of $g(x)$ for irrational $x$ is to find a sequence of rationals $x_n$ converging to $x$, and define $g(x)$ to be the limit of $f(x_n)$.
The only thing we have left to check is that the choice of sequence of rationals $x_n$ does not change the definition of $g(x)$. That is, we must show that if $x_n$ and $x'_n$ are two sequences of rationals that both converge to $x$, then
$$\lim_{n \to \infty} f(x_n) =\lim_{n \to \infty} f(x'_n).$$
Why does each limit exist?

 The existence of each limit is due to uniform continuity of $f$ implying each sequence is Cauchy.

Why are the limits the same? (Hint: show that the sequences $x_n$ and $x'_n$ get arbitrarily close for large $n$, and then apply uniform continuity.)

 If you fix $\delta > 0$, there will exist an $N$ such that $|x_n - x| < \delta/2$ and $|x'_n - x|<\delta/2$ for all $n \ge N$, so the triangle inequality implies $|x_n - x'_n| < \delta$ for all $n \ge N$. Combining this with the definition of uniform continuity, we see that for any $\epsilon > 0$ there exists some $N$ such that $|f(x_n) - f(x'_n)| < \epsilon$ for any $n \ge N$.

A: Sketch of proof: With the facts that you know, you can take any real number $r$ and think of it is an equivalence class of Cauchy sequences of rationals converging to that $r$. By uniform continuity, this give a Cauchy sequence which must be convergent, since $\mathbb{R}$ is complete. You only need to check that this is independent of the choice of representative of $r$ as a Cauchy sequence. To do this, just observe that if you had two different representatives of this class, both being arbitrarily close to $r$ implies they are arbitrarily close to one another. Continuity and the triangle inequality should finish it.
A: Another approach is to consider the graph
$$\Gamma_f=\{(a,f(a)):a\in\Bbb Q\cap[0,1]\}$$
of $f$, and consider its closure $\overline{\Gamma_f}$ in $\Bbb R^2$.
First, prove that $\overline{\Gamma_f}$ is the graph $\Gamma_g$
of a function from $[0,1]$ to $\Bbb R$. (This involves the same
sort of Cauchy series manipulation seen in the other solutions.)
Then one needs to prove that $g$ is continuous. But the projection
map $\pi_1:\Gamma_g\to[0,1]$ will be a continuous bijection from a compact
space to a Hausdorff space, and so a homeomorphism. Composing
its inverse with the other projection function gives $g$, which
therefore is continuous.
