How do I prove that the Cardinality of the set of functions from Natural Numbers to Real Numbers is equal to the power set of Real Numbers?
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7Does this answer your question? cardinality of all real sequences – Alejandro Tolcachier Jan 25 '20 at 20:22
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I didn't see this earlier. Thanks for suggesting this. However, I still don't understand the language and notations. Could you please simplify it for me? – Severus Jan 25 '20 at 20:38
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The first thing is that a function from $\mathbb{N}\to\mathbb{R}$ is a sequence of real numbers. So your question is about the cardinality of all real sequences. The notation $A^\mathbb{N}$ is just the set of functions from $\mathbb{N}$ to $A$. – Alejandro Tolcachier Jan 25 '20 at 21:00
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There are $2^{\aleph_0}$ reals, so there are $\left(2^{\aleph_0}\right)^{\aleph_0}=2^{\aleph_0^2}=2^{\aleph_0}$ functions from $\Bbb N$ to $\Bbb R$.
J.G.
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