Suppose that $B$ is a disjoint set of subsets of $\Bbb N$: if $A,A' \in B$ and $A \neq A'$, then $A \cap A' = \emptyset$. Show that B is countable

Question

Suppose that $B$ is a disjoint set of subsets of $\Bbb N$: if $A,A' \in B$ and $A \neq A'$, then $A \cap A' = \emptyset$. Show that B is countable.

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Does my proof look fine or contain logical gaps and flaws? Thank you so much for your verification!

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My attempt:

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Lemma: Let $\mathfrak P$ be a partition of $\Bbb N$. Then $\mathfrak P$ is countable.

Proof

If not, then $\mathfrak P$ is uncountable. It's clear that $\{\min X\mid X\in \mathfrak P\}$ and $\mathfrak P$ are equinumerous. Thus $\{\min X\mid X\in \mathfrak P\}$ is uncountable. On the other hand, $\{\min X\mid X\in \mathfrak P\}\subseteq \Bbb N$, then $\{\min X\mid X\in \mathfrak P\}$ is countable. This leads to a contradiction. Hence $\mathfrak P$ is countable.

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We proceed to prove our main theorem.

It's clear that $B\subseteq \mathfrak P$ and $\mathfrak P$ is countable by Lemma. Then $B$ is countable. This completes the proof.

Answer 1

From the statement I deduce that you're using countable in the sense of

either finite or equinumerous with $\mathbb{N}$

In this case, the statement that a partition of $\mathbb{N}$ is countable is correct. If $\mathfrak{B}$ is a partition of $\mathbb{N}$, then you can define $$ f\colon\mathfrak{B}\to\mathbb{N} \qquad f(A)=\min A $$ which is an injective map. This is what you did.

There is a small detail: a partition, by definition, doesn't contain the empty set, but your set $B$ might.

Moreover you are failing to produce a suitable partition which $B$ is a subset of.

Let's call quasipartition of $X$ a set $\mathfrak{B}$ of subsets of $X$ such that $\mathfrak{B}\setminus\{\emptyset\}$ is a partition of $X$. Then every quasipartition of $\mathbb{N}$ is countable, because adding one element to a countable set produces a countable set.

Now consider your set $B$. If $B_0=\bigcup_{A\in B}A$, then $$ \mathfrak{B}=B\cup\{\mathbb{N}\setminus B_0\} $$ is a quasipartition of $\mathbb{N}$ (prove it).

Answer 2

Each set of B, discarding a possible empty set, has at least one point. As the sets of B are pairwise disjoint, |B| <= |$\cup$B| <= |N|. Thus B is countable.