## [0,1] ~ [0,1) (Cardinality)

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Govalant
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### [0,1] ~ [0,1) (Cardinality)

I've been trying to prove that [0,1] is equipotent to [0,1) by trying to find a bijection between them and had no luck. I then asked my teacher and he told me that I should find an injection from each to the other, as that guarantees they are equipotent.

Now I have two questions:
Where can I find a proof of that theorem?
and
Can I still find a bijection which is not horribly complicated?

Thanks!

Edit: The injections were f:[0,1) -> [0,1] f(x) = x and g:[0,1] -> [0,1) g(x) = x/2
Now these points of data make a beautiful line.

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-Entropy is winning.

the tree
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### Re: [0,1] ~ [0,1) (Cardinality)

Govalant wrote:I asked my teacher and he told me that I should find an injection from each to the other, as that guarantees they are equipotent.

Where can I find a proof of that theorem?
It's called the Cantor–Bernstein–Schroeder Theorem, the Wikipedia article is okay.

Govalant wrote:Can I still find a bijection which is not horribly complicated?
You wont find anything particularly pretty, no.

Torn Apart By Dingos
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### Re: [0,1] ~ [0,1) (Cardinality)

It's easy to find an explicit bijection. All you want to do is to remove a single point from your set and show that your new set has the same cardinality. This should be familiar to you: how do you show that {0,1,2,3,...} has the same size cardinality as {1,2,3,...}? A bijection in this case is f(n)=n+1. You just "shift" your set so that you lose one point. The same idea works for [0,1]: choose a countable sequence inside [0,1], and shift it. To be precise, f:[0,1]->[0,1) is a bijection, where f(1/n)=1/(n+1) for all integers n>=1, and f(x)=x otherwise.

There exists no continuous bijection, because if f:[0,1]->[0,1) is continuous, then the image of f is compact (because [0,1] is), which [0,1) is not.

Nic
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### Re: [0,1] ~ [0,1) (Cardinality)

Pretty is in the eye of the beholder.
Spoiler:
Halve everything that's a power of 1/2, and fix everything that isn't.

phlip
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### Re: [0,1] ~ [0,1) (Cardinality)

Nic wrote:Pretty is in the eye of the beholder.
Spoiler:
Halve everything that's a power of 1/2, and fix everything that isn't.

Spoiler:
Which, as a bonus, is what you get if you apply the aforementioned Cantor-Bernstein-Schroeder Theorem (or, at least, the procedure used in the first proof on the Wikipedia page) to the pair of injections mentioned in the OP...

Code: Select all

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[he/him/his]

mike-l
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### Re: [0,1] ~ [0,1) (Cardinality)

phlip wrote:
Nic wrote:Pretty is in the eye of the beholder.
Spoiler:
Halve everything that's a power of 1/2, and fix everything that isn't.

Spoiler:
Which, as a bonus, is what you get if you apply the aforementioned Cantor-Bernstein-Schroeder Theorem (or, at least, the procedure used in the first proof on the Wikipedia page) to the pair of injections mentioned in the OP...

Wow.. I've talked about that example probably 100 times, and used both methods, and never noticed!
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