I need help with the following proof:
hence for all thus: i.e.
next we show that is (strictly) increasing: let be two real numbers and then for
some thus: so: i.e. is
now let and choose such that call this since is
increasing, by but thus:
call this now and give us: since is arbitrary,
we must have
Here is another way. If we can show is continous then let be any real number and be a sequence of rationals converging to . Then by continuity . Let us show that is continous WLOG at . Let and choose . Then if we can find a rational so that and then by using the strict property as NonCommAlg shown.
A more interesting question is to find .