Proving two sets have the same cardinality

[quick bridge]

| (0,1) | = | [0,1] |
Proving these are equal using Schroder-Bernstein Theorem

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[quick bridge]

f: (0,1) -> [0,1] be given by f(x)=x
g: [0,1] -> (0,1) be given by g(x) = 2^(x-2)
f((0,1)) = (0,1) ⊆ [0,1]
g([0,1]) = [1/4,1/2] ⊆ (0,1)
so are both these functions fine

[untold skiff]

Schroder-Bernstein says that if you have an injective function A->B and B->A then they have the same cardinality. So, you're on the right track.

Your f looks right

As does your g, though maybe you should spend a couple sentences explaining why g([0,1]) is what it is (exponentials are increasing thus injective, then compute g(0) and g(1))

Basically - you're very much on the right track! Just give a bit more of an explanation as to what's going on!

[keen torrent]

This is incidental to your question, but you may find it interesting or useful to note how you can explicitly construct a bijection for these sets.

[sharp warren]

Invariance

[keen torrent]

that is, we map every x in (0, 1) to itself, except the powers of 1/2, a sequence which we we "shift over" by 2

   x: 1/2  1/4  1/8  1/16 1/32 1/64 ...
f(x): 1/8  1/16 1/32 1/64 ...

which frees up two spots (1/2 and 1/4) to map 0 and 1 into

[keen torrent]

and while that g(x) is perfectly fine, it bears mentioning that there's a simpler way to map [0, 1] to another closed interval like [1/4, 1/2]

[quick bridge]

Ty this helps with another part of the question where I have to create an explicit example of a bijection but couldn’t figure out how to really do it

[quick bridge]

+close

[primal girder]

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[primal girder]

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