sequences and series - Show that $S$ has the same cardinality with the set of real numbers.

Suppose that $a_n$ is a sequence with values in $\mathbb R$ which is not ultimately constant.

Let $S$ be the set of the subsequences of $a_n$.

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Question: Show that $S$ has the same cardinality with the set of real numbers.

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Attempt: First, I tried to write $S$ in my logic as $S=\{a_{f_n}\;|\; f_n \text{is monotone increasing}, f_n:\mathbb N\to\mathbb N \}$

If two sets have the same cardinality, there must be a bijective mapping between them. If I am able to show there are uncountable many increasing $f_n$, then I am done, but I couldnot. Besides, how can one prove this by using my logic or in other method?

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Comment:
Moreover, I can understand intuitively that there cannot be countable many such subsequences since $a_n$ is not convergent, that is not going to the any number, changes everytime.

Answer

The $\{0,1\}$ sequences are not ultimately constant (if we forget countably many being constant from some term), hence
$\mathfrak{c}=2^{\aleph_0}\leq|{S}|\leq\mathfrak{c}^{\aleph_0}=\mathfrak{c}$