From: Marc LeBrun <mlb@well.com> I'm curious how this applies to another sort of edit, the "rebasing" operations, for instance

G(X) := replace 2^n in the binary expansion of X with 3^n.

G maps rationals into rationals, the unit interval into itself, preserves ordering and so on.

...

(5) Is F(X) continuous? (I think yes. Certainly if X and Y are both rational and |X-Y|-->0, then |F(X)-F(Y)|-->0.)

Also true for G, I guess? However there are an awful lot of "holes" in the range -- every number that contains a 2 anywhere in its ternary expansion.

--umm. Actually in view of the intermediate value theorem, neither F nor G can be continuous, because of these holes. Oops.