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I was wondering if anyone could show me how to express the cardinality of all n-element subsets of real numbers.

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you think that number is countable –  Adi Dani Jan 11 '13 at 14:18
What if $n=1$ ? –  Ilya Jan 11 '13 at 14:24
Then the cardinality is continuum. –  Bilbo Jan 11 '13 at 14:24
@Anna: ok, so it is at least continuum for the general case. Now, can you write any $n$-element subset of $\Bbb R$ through its coordinates? –  Ilya Jan 11 '13 at 14:32
What do you mean by subset's coordinates? –  Bilbo Jan 11 '13 at 14:57

1 Answer 1

up vote 4 down vote accepted

Let $[{\bf R}]^n$ be the set of $n$-element subsets of $\mathbf R$. Assumming $n > 0$, we have $|[{\bf R}]^n| \ge |{\bf R}| = 2^{\aleph_0}$. Conversely the map $f \colon [\mathbf R]^n \to \mathbf R^n$ defined by $f(\{x_1, ..., x_n\}) = (x_1, ..., x_n)$ where $x_1 < ... < x_n$ is injective. So $|[\mathbf R]^n| \le |\mathbf R^n| = |\mathbf R|^n = |\mathbf R| = 2^{\aleph_0}$. Thus $|[\mathbf R]^n| = 2^{\aleph_0}$.

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