# 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).