How to prove (0,1)∼{0,1}N(0,1) \sim \left\{0,1 \right\}^{\mathbb{N}} ?

I know that I need to use binary expansion of a∈(0,1)a \in (0,1), but I am not sure how to formally write the proof.

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shouldn’t that be [0,1][0,1]?

– user300

2 days ago

1

@user300: Why? Do they have different cardinalities?

– Asaf Karagila

2 days ago

Also, for the people who are going to post the “binary expansion” argument, there are two deleted answers that were posted and voluntarily deleted moments later when the answerers realized that the argument is wrong. We don’t need a third deleted answer: the binary expansion is neither unique (so it is not an injection or a well-defined function), nor it is surjective since 00 and 11 (which correspond to the constant 00 and constant 11 sequences) are not in the interval (and also because of the uniqueness issue).

– Asaf Karagila

2 days ago

i think there is some subset relation kind of thing @Asaf Karagila .. but i’m not sure!

– user300

2 days ago

@user300: If you agree that [0,1][0,1] and [12,13][\frac12,\frac13] have the same cardinality, then by the Cantor-Bernstein theorem, [0,1][0,1] and (0,1)(0,1) have the same cardinality; you can also come up with explicit bijections, but this is yet another question that was asked ad nauseum on this site.

– Asaf Karagila

2 days ago

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1 Answer

1

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One way to do this is by showing that there are two injections f:(0,1)→{0,1}N f : (0,1) \to \{0,1\}^\mathbb{N} and g:{0,1}N→(0,1) g : \{0,1\}^\mathbb{N} \to (0,1) then use Cantor-Bernstein-Schrأ¶der.

f(x)=n↦⌊x2n⌋mod2 f(x) = n \mapsto \lfloor x2^n\rfloor \mod 2

g(x)=∑∞n=0x(n)2−2n g(x) = \sum_{n=0}^\infty x(n)2^{-2n}