Real Analysis Question

1. Oct 13, 2004

tink

Show that the intervals (0,1) and [0,1] are equivalent. (Hint: consider rationals and irrationals separately).

I'm able to find a function that shows a bijection between (0,1) an [0,1] under the irrationals, but i can't figure out the rationals. Also... the next step (i believe) would be to find a bijection between these two functions. If anybody can help me figure out an answer to this problem that would be so awesome! Thanks so much, guys!

2. Oct 13, 2004

stunner5000pt

i'm not really good at this but i think this would be relevant to the proof

In both cases the supremum (least upper bound) is 1 and the greatest lower bound is 0.

Also for all x in (0,1) 0<x<1
and for for all x in [0,1], 0<=x<=1, and it follows that 0<x<1 for this interval as well.

Thus (0,1) is a subinterval of [0,1]

If you are familiar with delta-epsilon proofs you might want to prove that

0 + delta < x < 1 + delta is equivalent to 0<=x<=1 for a suitable delta > 0.

3. Oct 13, 2004

Fredrik

Staff Emeritus
The set of rational numbers is countable. That means that the set of rationals in (0,1) can be arranged in a sequence, like this:

$$r_2, r_3, r_4,\dots$$

If you define

$$r_0=0$$

and

$$r_1=1$$

The function f defined by

$$f(r_n)=r_{n+2}$$

maps the rationals in [0,1] bijectively onto the rationals in (0,1).