Thread: continuity

That argument would work fine to *suggest* that there is a discontinuity. However, it would not *prove* that there is a discontinuity. There are many continuous functions such that f(1.41)=1.41, f(1.414...)=0 and f(1.42)=1.42.

The standard argument uses the observation that for any , there are only finitely many points such that .

Spoiler:

So given an irrational point , let . Then if , , so is continuous at the irrational point .

It's straightforward to show is discontinuous at any rational point just using the fact that any interval contains an irrational number.

Spoiler:

If , then , so letting shows that since any small interval around contains an irrational point , .