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I saw somewhere that there are no sets between $\mathbb Q$ and $\mathbb R$ in the sense that there are no set $S\subset \mathbb R$ s.t. $|\mathbb Q|<|S|$ but $|S|<|\mathbb R|$, i.e. all set $S\subset \mathbb R$ s.t. $|\mathbb Q|<|S|$ should have the cardinality of the continuum. Now, what about the Cantor set ? It's a set of measure $0$, but it's uncountable. Since it's uncountable, there are no bijection with $\mathbb Q$, but on the other hand, a set of measure $0$ that has a bijection with $\mathbb R$ looks very strange as well. So, what do you think ? Is the Cantor set having the cardinality of the continuum ?

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    $\begingroup$ (1) The statement that "there are no sets between $\mathbb{Q}$ and $\mathbb{R}$" in the sense you write is called the "Continuum Hypothesis". It is an statement that is independent from regular set theory (can neither be proven nor disproven), just like the parallel postulate is independent from the remaining geometric axioms. You can work in theories where it is true that no such sets exist, and you can work in theories where it is false that no such sets exist. (2) As for the Cantor set, it definitely has the cardinality of $\mathbb{R}$.(cont) $\endgroup$ – Arturo Magidin Mar 21 at 16:56
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    $\begingroup$ Elements of the Cantor set are precisely those that have a ternary (base 3) expansion that does not contain any 1s. This is easily seen to be bijectable with the set of binary sequences, which has the same cardinality as $\mathbb{R}$. Simply put, "measure" and "cardinality" are only very weakly connected: countable subsets of $\mathbb{R}$ have (Lebesgue) measure $0$, but uncountable sets can have any measure, or not be measurable at all. $\endgroup$ – Arturo Magidin Mar 21 at 16:57
  • $\begingroup$ @ArturoMagidin: I thought every set with cardinality less that $\mathbb R$ had Lebesgue measure $0$? Do I misremember? $\endgroup$ – celtschk Mar 21 at 19:00
  • $\begingroup$ @celtschk that is precisely what Arturo Magidin said. If a set has cardinality less than $\mathbb R$ it has Lebesue measure $0$. If the set has cardinality of $\mathbb R$ it doesn't have to have measure $0$.... but it could. That Cantor set is uncountable with measure $0$. $[0,1]$ is uncountable with measure $1$. $\mathbb R$ is uncountable with infinite measure. and so on. $\endgroup$ – fleablood Mar 21 at 19:19
  • $\begingroup$ @celtschk: Assuming the Continuum Hypothesis, any set with cardinality less than $|\mathbb{R}|$ is countable, hence has Lebesgue measure zero. $\endgroup$ – Arturo Magidin Mar 21 at 19:21
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Yes. Cantor set has cardinality of the reals (continuum).

As Cantor Set $\subset \mathbb R$ it's cardinality is at most $|\mathbb R|$ and as it is uncountable it's reasonable that we can't have found a contradiction to "Continuum Hypothesis" and found a cardinality between $|\mathbb Q|$ and $|\mathbb R|$ so it reasonable that Cantor set has the cardinality of the reals.

But to seal the deal we need a bijection between Cantor set and $\mathbb R$.

Following a comment by Arturo Magidin:

If $x \in [0,1]$ then $x = \sum\limits_{i=0}^{\infty} b_i 3^{-i}$ for some sequence of $b_i$ where each $b_i=0,1,2$. If we disallow infinite tailing $0$s then this sequence is unique. This is just writing $x$ is decimal in base $3$. But where all terminating decimals are replaced with tailing $2$s.

Likewise if $y \in [0,1]$ then $y = \sum\limits_{i=0}^{\infty} c_i 2^{-1}$ for some sequence of $c_i = 0,1$. And if we disallow infinite tailing $0$s 0 this sequence is unique. This is just the base $2$ decimal.

If $x = \sum b_i 3^{-i}$ is in the Cantor set then none of the $b_i = 1$. That is because we removed the middle third of all segments and $b_k = 1$ means $\sum\limits_{i=0}^{k-1} b_i 3^{-i} < x < \sum\limits_{i=0}^{k-1} b_i 3^{-i} + 2*3^{-k}$ would mean $x$ is in some middle third.

So let $f(\sum b_i 3^{-i}) = \sum c_i 2^{-i}$ where if $b_i = 0$ then $c_i = 0$ and if $b_i = 2$ then $c_i = 1$. $f$ is a bijection between the Cantor set and $[0,1]$.

but on the other hand, a set of measure 0 that has a bijection with R looks very strange as well.

Ah.... not really. It seems counterintuitive because ... to have measure $0$ no two points can be connected in the set so $\color{red}{\text{for any point there must be a measurable distance before the "next" one}}$ and there can only be countably many such points. But that clause in $\color{red}{\text{red}}$ is completely erroneous and is based on a naive concept of numbers must "follow each other". Uncountable numbers don't.

And the Cantor set exists merely to be a simple counter example.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Pedro Tamaroff Mar 23 at 13:00

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