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

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

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

Now these points of data make a beautiful line.

How's things?

-Entropy is winning.

How's things?

-Entropy is winning.

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

It's called the Cantor–Bernstein–Schroeder Theorem, the Wikipedia article is okay.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?

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

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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.

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.

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

Pretty is in the eye of the beholder.

**Spoiler:**

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

Nic wrote:Pretty is in the eye of the beholder.Spoiler:

**Spoiler:**

Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

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

phlip wrote:Nic wrote:Pretty is in the eye of the beholder.Spoiler:Spoiler:

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

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