Bijection between $2^{\Sigma^*}$ and $\mathbb R$

Question

Is there a Bijection in between $2^{\Sigma^*}$ and $\mathbb R$? Here $2^{\Sigma^*}$ denotes the set of all languages over a finite alphabet.

If I have an uncountable set, then its powerset will also be uncountable, but there is no bijection between them.

Of course, sometimes I can find a bijection between two uncountable set: take any uncountable set and itself. And, there is a bijection between $[0,1]$ and $\mathbb R. $
But my question is there is there any possibility of bijection between two uncountable set $2^{\Sigma^*}$ and $\mathbb R$ where $2^{\Sigma^*}$ is the set of all languages. I know that the set $2^{\Sigma^*} $ is uncountable by cantors theorem.

Well, this question shows that $2^{\mathbb N}$ has the same cardinality as $\mathbb R$. So all you need to show is that $\Sigma^*$ is countable. — Rushabh Mehta, Oct 09 '21 at 16:32
$\Sigma^$ is easy to proof countable but there is possible one-to-one correspondence possible between $2^{\Sigma^}$ and $\mathbb R$? — , Oct 09 '21 at 16:42
There's a one to one correspondence between $2^{\mathbb N}$ and $\mathbb R$, so yes. — Rushabh Mehta, Oct 09 '21 at 16:54
$\Sigma^$ is countable, so $2^{\mathbb N}$ is in a one to one correspondence with $2^{\Sigma^}$... — Rushabh Mehta, Oct 09 '21 at 17:04

Vivaan Daga · Accepted Answer · 2021-11-26T07:30:44.973

Say you are working over the alphabet, $A=\{ 0,1 \},$ the set of all finite strings over $A$ obviously has cardinality of the naturals, now in order to get the cardinality of all languages over $A$ you want the cardinality of the power set of the set of all strings which is the cardinality of the the power set of the naturals. It is a well known theorem that the power set of the naturals is equinumoruous with $\mathbb{R}$ whose proof can found in text on real analysis. Note that you can replace $A$ with any finite set.

Bijection between $2^{\Sigma^*}$ and $\mathbb R$

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