# Set Theory – Family of ?^

cardinalscompactnessset-theory

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 $$\kappa<\operatorname{cf}\alpha$$.

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 $$\operatorname{cf}\kappa>\omega$$.