2

Consider the set of dyadic rationals

$$D= \{a/2^n: a,n \in \mathbb{N}\}$$

This is dense in $[0, \infty[$.

I have a function $f: D\to \mathbb{R}$ such that for every $t > 0$, the function $f\vert_{D \cap [0,t]}$ is uniformly continuous. Is is true that $f$ has a continuous extension on $[0,\infty[$?

If yes, what theorems do you invoke to prove this?

(This question comes from a proof in Billingsley that shows that Brownian motion exists with continuous paths)

Dave L. Renfro
  • 36,843
  • 1
    Is $F(x)=\lim_n f( 2^{-n} \lfloor x 2^n \rfloor)$ well-defined and continuous ? – reuns Feb 26 '20 at 17:57
  • Thanks @reuns. I'll try to prove that $F$ is a continuous extension. –  Feb 26 '20 at 19:05
  • If you have proved it, you should give an answer to your own question. – Paul Frost Feb 26 '20 at 22:53
  • FYI, your original title was not correct, since you're not dealing with one or more dense subsets of the dyadic rationals, but rather with intervals of dyadic rationals (none of which is a dense subset of the dyadic rationals). – Dave L. Renfro Mar 02 '20 at 08:20
  • Yes I know but I can't make my titles arbitrarily long. –  Mar 02 '20 at 09:47

1 Answers1

2

The required claim follows from the fact that for each $t > 0$ the function $f\vert_{D \cap [0,t]}$ admits a continuous extension and that a family $\{[0,t):t>0 \}$ is an open cover of the space $[0,\infty)$. Also it directly follows from Lemma 2 from this my answer.

Alex Ravsky
  • 90,434