Real Analysis question

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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!

[fourier jr]

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

[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)

[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.

[Fredrik]

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

This is what I answered in the homework forum:

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).

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

f(x)=x

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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!