# 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

### fourier jr

I've only ever seen "equivalent" used with categories; what does it mean in analysis?

3. Oct 13, 2004

### matt grime

I think it just means show there is a bijection from one to the other.
Equivalent isn't particularly standard here. Have the same cardinality, yes, which means lie in the same isomorphism class of sets in the category SET. Also used is equipollent.

Take the obvious bijection from the set {0,1,2....} to {1,2,...}?

Both are in bijection with the rationals inside the interval [0,1]. Can you now see a bijection between [0,1] and (0,1]? (hint map irrationals to themselves identically)
and rinse and repeat.

(I've no idea what 'under the irrationals' means, by the way)

4. Oct 13, 2004

### HallsofIvy

Since the set of rationals in (0,1) is countable, they can be ordered:{r1,r2,...}

Define f(x) for x in [0, 1] by: f(0)= r1, f(1)= r2, f(rn)= rn+2 and f(x)= x if x is irrational.

5. Oct 13, 2004

### Fredrik

Staff Emeritus
It's not a great idea to post the same question in several forums.

This is what I answered in the homework forum:

This is of course the same thing that HallsofIvy suggested, but he also mentioned the obvious extension to the irrationals:

$$f(x)=x$$

Last edited: Oct 13, 2004
6. Oct 13, 2004

### tink

Thank you guys sooo much! I really appreciate the help... and sorry about posting this twice! I figured if I posted in two places that I'd be more likely to get a response! This forum is awesome, my new home away from home!