I was wondering what the cardinality of the set of all real sequences is. A random search through this site says that it is equal to the cardinality of the real numbers. This is very surprising to me, since the cardinality of all rational sequences is the same as the cardinality of reals, and it seemed fairly intuitive to me that if cardinality of a set $A$ is strictly greater than the cardinality of the set $B$, then cardinality of $A^{\mathbb{N}}$ should be strictly greater than cardinality of $B^{\mathbb{N}}$. It turns out to be false.
Some technical answers have appeared in this forum elsewhere but I do not understand them. As I am not an expert in this topic, could some one explain me in simple terms why this is happening?
Also is the cardinality of all functions from reals to reals also the same as the cardinality of reals?
Answer
Identify $\mathbb R$ as the set of functions $f : \mathbb N \to \{ 0,1\} $.
Then any sequence $\{ x_n \}$ becomes a sequence $\{f_n \}_n$ where $f_n : \mathbb N \to \{ 0,1\}$. But then, this is simply a function $g : \mathbb N \times \mathbb N \to \{ 0,1\}$:
$$g(m,n) =f_n(m) \,.$$
This way you can construct a bijection from the sequences of real numbers to the set of functions from $\mathbb N \times \mathbb N \to \{ 0,1\}$. Now, since $\mathbb N \times \mathbb N$ and $\mathbb N$ have the same cardinality, you get a bijection from the sequences of real numbers to the set of functions from $\mathbb{N} \times \mathbb{N} \to \{ 0,1\}$, which is just $\mathbb R$.
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