1. modify Cantor function so that all it's growth points are irrational. Then you have a mapping from an interval of irrational numbers to an interval of reals. Then extend to the whole range of numbers.

2. consider a sequence i[n] of irrationals that converges to a rational number. Then if for each member you take the inf of reals that map into it, and consider the limit L of those (it's a bounded sequence so there's a limit), then f(L) is irrational and greater than f(i[n]) for all n. It cannot be rational, so there would be a gap, but since it's an onto mapping there can't be any gaps.

3. if such embedding existed, irrationals and reals would be \infty-equivalent (think infinitely long Ehrenfeucht–Fraïssé game). But they are not. I think. This is your hint :)

1. modify Cantor function so that all it's growth points are irrational. Then you have a mapping from an interval of irrational numbers to an interval of reals. Then extend to the whole range of numbers.

2. consider a sequence i[n] of irrationals that converges to a rational number. Then if for each member you take the inf of reals that map into it, and consider the limit L of those (it's a bounded sequence so there's a limit), then f(L) is irrational and greater than f(i[n]) for all n. It cannot be rational, so there would be a gap, but since it's an onto mapping there can't be any gaps.

3. if such embedding existed, irrationals and reals would be \infty-equivalent (think infinitely long Ehrenfeucht–Fraïssé game). But they are not. I think. (well, it's been shown that such embedding is actually possible, so my assumption was incorrect)

1. modify [Cantor function][1] so that all it's growth points are irrational. Then you have a mapping from an interval of irrational numbers to an interval of reals. Then extend to the whole range of numbers.

2. consider a sequence i[n] of irrationals that converges to a rational number. Then if for each member you take the inf of reals that map into it, and consider the limit L of those (it's a bounded sequence so there's a limit), then f(L) is irrational and greater than f(i[n]) for all n. It cannot be rational, so there would be a gap, but since it's an onto mapping there can't be any gaps.

3. if such embedding existed, irrationals and reals would be \infty-equivalent (think infinitely long [Ehrenfeucht–Fraïssé_game][2]). But they are not. I think. This is your hint :)

(no links for you, because this site sucks and won't allow me to link stuff. Search for them on Wikipedia)

1. modify Cantor function so that all it's growth points are irrational. Then you have a mapping from an interval of irrational numbers to an interval of reals. Then extend to the whole range of numbers.

2. consider a sequence i[n] of irrationals that converges to a rational number. Then if for each member you take the inf of reals that map into it, and consider the limit L of those (it's a bounded sequence so there's a limit), then f(L) is irrational and greater than f(i[n]) for all n. It cannot be rational, so there would be a gap, but since it's an onto mapping there can't be any gaps.

3. if such embedding existed, irrationals and reals would be \infty-equivalent (think infinitely long Ehrenfeucht–Fraïssé game). But they are not. I think. This is your hint :)