Suppose that $\kappa$ is an uncountable cardinal. Let $\kappa^{<\omega}$ denote the family of all finite subsets of $\kappa$. Does there exist a family $S\subset\kappa^{<\omega}$ such that for every $i\in\kappa$, there are only finitely many elements of $S$ that do not contain $i$?
Note that for $\kappa=\omega$, we can just take $S=\{\{0\},\{0,1\},\{0,1,2\},\cdots\}$. Each element of $S$ is a finite subset of $\omega$, and every $i\in\omega$ does not belong to finitely many elements of $S$.
I came up with this question by suspecting that every ordinal $\alpha$ is $\kappa$-compact with $\kappa<\operatorname{cf}\alpha$, so that every open cover of $\alpha$ by at most $\kappa$ sets has a finite subcover. The well-known fact is that $\omega_1$ is countably compact, and if there is such a family $S$ for $\kappa$ then I can imitate the proof given here: Suppose that there is an open cover $(O_i)_{i\in\kappa}$ of $\alpha$ that does have a finite subcover, then for each $E\in S$, we can pick $x_E\in\alpha$ that is not in $\displaystyle\bigcup_{i\in E}O_i$. Since $\{x_E:E\in S\}$ has cardinality at most $\kappa$, it has a least upper bound $x$ in $\alpha$. We must have $x\in O_i$ for some $i\in\kappa$, but the neighborhood $O_i$ of $x$ can only contain finitely many $x_E$, a contradiction (if $i\in E$ then $x_E\notin O_i$).
So does such a family really exist after all? Thank you in advance.
Edit. As the comment pointed out, the question was not meaningful. See the answer of myself. However, I would still be happy with a proof or a counterproof that every ordinal $\alpha$ is $\kappa$-compact with $\kappa<\operatorname{cf}\alpha$.
Best Answer
At least we can prove the following result: If $\operatorname{cf}\kappa>\omega$, then there is no $S\subset\kappa^{<\omega}$ with $|S|=\kappa$ such that for every $i\in\kappa$, there are less than $\kappa$ elements in $S$ that do not contain $i$.
Suppose that such $S$ exists. Let $f:\kappa\to\omega$ be a surjection. For each $E\in S$, pick $x_E\in\omega$ that is not in $f(E)$ (which is possible since $f(E)$ is finite). For each $j\in\omega$, pick $i\in f^{-1}(j)\subset\kappa$, then the set $\{E\in S:x_E=j\}$ is included in $\{E\in S:i\notin E\}$ ($i\in E$ implies that $f(i)\neq x_E$, so $x_E\neq j$). By assumption we have $|\{E\in S:x_E=j\}|<\kappa$, then $S = \displaystyle\bigcup_{j\in\omega}\{E\in S:x_E=j\}$ is the union of $\omega$ subsets with cardinality $<\kappa$, which contradicts $\operatorname{cf}\kappa>\omega$.