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Find the cardinality of the set $\{ X \subset \mathbb{N} : |X| = \aleph_0\} $.

First off all I could easily prove that the set given above, say $S$, is uncountable.

Because $P(\mathbb{N})$ is uncountable and $|P(\mathbb{N})|=c$, and $P(\mathbb{N}) = S \cup F$, where $F$ = $ \{$family of all finite subsets of $\mathbb{N}$ $\}$. So, $S$ is countable $\implies$ $P(\mathbb{N})$ is countable as $F$ is clearly countable. Which is a contradiction.

But I can't find the cardinality of $S$.

Now, $$S \subset P(\mathbb{N})$$ $$\implies |S| \le |P(\mathbb{N})|=c$$

And, $$|S| \gt \aleph_0$$ Again, $$\aleph_0 \lt c$$

But in order to say, $$|S|=c$$ I have to use Continuum Hypothesis.

But can I do that because I know it's a hypothesis and also with current mathematical tools it cannot be proved that this is false(I got to know this from internet)?

Asaf Karagila
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Itachi
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2 Answers2

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Let $A$ be the set of even naturals and $B$ the odds and consider $$ \{A \cup C : C \in \mathcal{P}(B)\}. $$ How big is this set?

Jonathan Schilhan
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  • This set is equivalent to $\mathcal{P}(B)$. Now this implies that $|S| \gt c$ $[ |\mathcal{P}(B)|= 2^{\aleph_0}=c]$ because your set is a proper subset of $S$. Thus $|S|=c$. Am I correct? – Itachi May 18 '21 at 11:00
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Your set $S$ is the power set of $\aleph_0$ (cardinality $2^{\aleph_0}=\mathfrak{c}$) minus the set of finite subests of $\aleph_0$ (cardinality $\aleph_0$). So the answer is $2^{\aleph_0}-\aleph_0$, which is...? (This doesn't require the Continuum Hypothesis.)

Edited to add: The OP asks in a comment why the Axiom of Choice is not required to show that $2^{\aleph_0}-\aleph_0=2^{\aleph_0}$. If we can exhibit injections in both directions between $S$ and $\mathcal{P}(\Bbb N)$, then we can invoke the Schröder–Bernstein theorem.

The injection from $S$ to $\mathcal{P}(\Bbb N)$ is just the identity.

To construct an injection from all subsets of $\Bbb N$ to infinite subsets of $\Bbb N$, let $A$ be the set of infinite subsets of $\Bbb N$ that don't contain $1$, and let $B$ be the set of infinite subsets of $\Bbb N$ that do contain $1$ (I am assuming the convention $\Bbb N=\{1,2,3,\ldots\}$). Now we we can map the infinite subsets to $A$ by adding $1$ to each element; and we can map the finite subsets to $B$ by first mapping them to $A$ in the same way, and then taking their complement.

TonyK
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