choosing maximum number of sets So I am still into this infinite theory setting and I am stuck at a question again. I am currently reading about the topic regarding the power set of a countably infinite set like ($\mathbb{Q},\mathbb{N}$). It is obvious that the power set is continuum. So the initial question was this(this one I solved):
Given a countably infinite set $H$, at most how many subsets can you choose from the power set $P(H)$ such that every countably infinite many have infinite elements in common, however every continuum many have finite elements in common.
Now the answer to the previous question is that you can only choose countably infinite many subsets this way. However there is a problem related to this which I cannot seem to find a solution to. The problem is as follows:
Given a countably infinite set $H$. At most how many subsets can you choose from $P(H)$ such that every $n$ set have infinite elements in common, however every $n+1$ set have finite many in common.($n$ is a positive integer at least 2)
This one seems easier at first read, but I personally find it much harder, and I can't solve it. I am really interested in the answer, and what technique works to prove it.(or if the answer is continuum many then what motivated the construction)
 A: For each $n$ we can certainly get countably infinitely many sets.
Since $H$ is countably infinite, we can index it by the $(n+1)$-element subsets of $\Bbb N$ as
$$H=\{h_F:F\subseteq\Bbb N\text{ and }|F|=n+1\}\,.$$
For each $k\in\Bbb N$ let $H_k=\{h_F\in H:k\in F\}$; clearly $H_k$ is an infinite subset of $H$. If $G\subseteq\Bbb N$, and $|G|=n$, then
$$\begin{align*}
\bigcap_{k\in G}^nH_k&=\{h_F\in H:G\subseteq F\}\\
&=\{h_{G\cup\{k\}}:k\in\Bbb N\setminus G\}
\end{align*}$$
is infinite. If $|G|=n+1$, however, then
$$\bigcap_{k\in G}^nH_k=\{h_F\in H:G\subseteq F\}=\{h_G\}$$
is finite, and if $|G|>n+1$, then
$$\bigcap_{k\in G}^nH_k=\{h_F\in H:G\subseteq F\}=\varnothing\,.$$
Added 5 March 2022: I can now improve this to get a family $\mathscr{H}$ of $\mathfrak{c}=2^\omega$ subsets of $H$ with the desired property.
Let $T$ be the complete binary tree of height $\omega$, and for $k\in\omega$ let $L_k$ be the $k$-th level of $T$. Let $\mathscr{F}$ be the family of all $n$-element sets $F\subseteq T$ such that $F\subseteq L_k$ for some $k\in\omega$; $\mathscr{F}$ is countably infinite, so we can index $H$ as $H=\{h_F:F\in\mathscr{F}\}$.
Let $\mathscr{B}$ be the set of branches of $T$; $|\mathscr{B}|=2^\omega$, and for each finite $\mathscr{C}\subseteq\mathscr{B}$ there is a minimum $m_\mathscr{C}\in\omega$ such the the branches in $\mathscr{C}$ are disjoint at level $m_\mathscr{C}$. For each $B\in\mathscr{B}$ let
$$H_B=\{h_F\in H:F\cap B\ne\varnothing\}\,,$$
and let $\mathscr{H}=\{H_B:B\in\mathscr{B}\}$; clearly $|\mathscr{H}|=2^\omega$. Suppose that $\mathscr{H}_0=\{H_B:B\in\mathscr{B}_0\}$, where $\mathscr{B}_0\subseteq\mathscr{B}$, and $|\mathscr{B}_0|=n$. For $k\ge m_{\mathscr{B}_0}$ let $F_k=L_k\cap\bigcup\mathscr{B}_0$; then $F_k\in\mathscr{F}$, and $F_k\cap B\ne\varnothing$ for each $B\in\mathscr{B}_0$, so $h_{F_k}\in\bigcap\mathscr{H}_0$, and it follows that $\bigcap\mathscr{H}_0$ is infinite.
Now suppose instead that $|\mathscr{B}_0|=n+1$. If $h_F\in\bigcap\mathscr{H}_0$ for some $F\in\mathscr{F}$, then there is a $k\in\omega$ such that $F\subseteq L_k$. Clearly $k<m_{\mathscr{B}_0}$, as otherwise the $n+1$ branches in $\mathscr{B}_0$ are disjoint at level $k$. Thus, $F$ is a subset of the finite set $\bigcup_{k<m_{\mathscr{B}_0}}L_k$, and $\bigcap\mathscr{H}_0$ must be finite.
