# Show for $f$ defined on $\mathbb Q \cap [0,1]$ uniformly continuous, exists extension $g$ defined on $[0,1]$ that is continuous [duplicate]

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?

• See Q# 2497728 and its answers. Commented Oct 31, 2017 at 7:34

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 $n \ge N$.

• If two Cauchy sequences converge to the same number, then they are equivalent Cauchy sequences. This is shown when constructing $\mathbb R$. Is it necessary to do this again? Commented Oct 31, 2017 at 4:56

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.

• How would we go about showing this $g$ is unique? Commented Oct 31, 2017 at 5:05
• Pointwise limits of functions are unique whenever the underlying topological space is Hausdorff. Commented Oct 31, 2017 at 5:07

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.