The relation ~ is defined on P(N): A~B if |A| = |B|.
I need to prove that the cardinality of the equivalence classes is countable.
Any ideas??
Cardinality of equivalence classes
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$\begingroup$
cardinals
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0I think that must use the continous hypotesis... – 2017-01-02
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1Continuum hypothesis is definitely not required here. If $A\subset \mathbb{N}$ is infinite, then it is countably-infinite. The only remaining ones are the finite sets, so define a function $\mathbb{N} \to \mathcal{P}(\mathbb{N})/\sim$ by sending $0$ to $[\mathbb{N}]$, and $n$ to $[\{1,\ldots, n-1\}]$ for $n\geq 1$. – 2017-01-02
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0Is there any more context? What is $P(N)$ here? – 2017-01-02
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0@Hayden I think I got it. Shouldn't the function be defined n to [(1,..,n}]? – 2017-01-02
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0In that case, you'd miss $[\emptyset]$, which was intended to be the image of $1$ (a notation like $\{1,\ldots, n-1\}$ is usually meant to denote the empty set when $n-1 < 1$). – 2017-01-02
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0I see it now, thanks ! – 2017-01-02
1 Answers
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Hint: The cardinalities of the subsets of $\mathbb{N}$ are the natural numbers themselves and $\aleph_0$ ($=|\mathbb{N}|$), so this amounts to proving that the set $\mathbb{N} \cup \{ \aleph_0 \}$ is countable.