Can a Continuous Function Map Reals to Rationals?

[mathwonk]

theorem: every continuous function from the reals to the rationals is constant.
proof the image of the map is an interval containing only rationals, hence of form [c,c].

 

[JasonRox]

theorem: every continuous function from the reals to the rationals is constant.
proof the image of the map is an interval containing only rationals, hence of form [c,c].

Yes, but doesn't this theorem only hold for the rationals with the relative topology with respect to the real numbers with the usual topology?

If you give the rationals the indiscrete topology, isn't every function continuous?

 

[DeadWolfe]

Yes, but of course, when asking if there is a continus map between sets X and Y we always need to think of the given topologies on sets, since the question is ill-defined otherwise.

 

[JasonRox]

Yes, but of course, when asking if there is a continus map between sets X and Y we always need to think of the given topologies on sets, since the question is ill-defined otherwise.

Well yeah, but my question a few posts ago asks about a possible metric topology on the rationals such that you can create a continuous function from the reals (usual topology) to the rationals (with new metric topology).

Haven't gave much thought yet. I should write a list of things to think about.

 

[matt grime]

Of course there are continuous functions from the Reals to the rationals, as has been repeatedly pointed out (constant maps). Perhaps you mean continuous *surjections*, or even just non-constant continuous maps.

Yes if f:X\to Y is map either from the discrete or to the indiscrete topology it is trivially continuous.

Mind you, what metric d'ya think gives the indsicrete topology?

 

[JasonRox]

Mind you, what metric d'ya think gives the indsicrete topology?

There is none of course.