1

Let $(A_m^{(n)})$ be a sequence of sets indexed by $m$ and $n$. Let $$\Delta \triangleq \bigcap_{n=1}^N \bigcup_{m=1}^\infty A_m^{(n)}.$$

From the identity $(a+b)(c+d) = ac+ad+bc+cd$, I guess $\Delta$ can be written as a countable union, i.e., $$\Lambda = \bigcup_{k \in \mathbb N^N} \bigcap_{n=1}^N A_{k_n}^{(n)},$$ where $k_n$ is the $n$-th coordinate of the multi-index $k \in \mathbb N^N$.

I would like to ask if my proof is correct and if below statement is correct, i.e.,

Countable intersection of countable unions can be written as countable union of countable intersections.

Of course, if $N = \mathbb N$, then $\mathbb N^N$ is uncountable.

Proof: For $x \in \Delta$, $x \in \bigcup_{m=1}^\infty A_m^{(n)}$ for all $n$. Then for each $n$, there is some $k_n$ such that $x \in A_{k_n}^{(n)}$. Hence $x \in \bigcap_{n=1}^N A_{k_n}^{(n)}$ and thus $x \in \Lambda$.

For $x \in \Lambda$, $x \in \bigcap_{n=1}^N A_{k_n}^{(n)}$ for some $k = (k_1, \ldots,k_N) \in \mathbb N^N$. Then $x \in A_{k_n}^{(n)}$ for all $n$. Hence $ x \in \bigcup_{m=1}^\infty A_m^{(n)}$ for all $n$ and thus $x \in \Delta$.

Akira
  • 17,367
  • 2
    It is correct and it does generalize to infinite intersections of infinite unions. – Kavi Rama Murthy Oct 23 '21 at 08:35
  • 1
    @KaviRamaMurthy If $N =\mathbb N$, then $\mathbb N^N$ is uncountable. – Akira Oct 23 '21 at 08:36
  • Yes it is. @KaviRamaMurthy is pointing out that in general, for any cardinal $\kappa$, there's an analogous identity between a $\kappa$-ary intersection of (say) countable unions, and countable unions of $\kappa$-ary intersections. And yes, you can change "countable" to "$\lambda$-ary" for $\lambda$ a cardinal other than $\aleph_0$. – BrianO Oct 23 '21 at 09:05
  • @BrianO So if $\kappa = \aleph_0$, then the $\kappa$-array is no more countable? – Akira Oct 23 '21 at 09:10
  • 2
    That's right. In that case, the union of intersections is taken over $\mathbb{N}^{\mathbb{N}}$ the set of all functions $k \colon \mathbb{N} \to \mathbb{N}$, which has cardinality $2^{\aleph_0}$. The identity is $$\Lambda = \bigcup_{k \in \mathbb{N}^{\mathbb{N}}} \bigcap_{n\in \mathbb{N}} A_{k(n)}^{(n)}$$. – BrianO Oct 23 '21 at 09:24
  • 1
    Thank you so much @BrianO. – Akira Oct 23 '21 at 09:25

0 Answers0