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Let $X$ be a countably infinite (or larger) set with the cofinite topology. for every $x\in X$ is there exists a family $\xi\subset\tau$ such that $\lbrace x\rbrace=\bigcap\xi $ ? If the answer is yes, then what is the cardinality of $\xi$ ?

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  • $\begingroup$ Due to the way MO parses the MathJax, \{ is ignored. You can use \\{ or \lbrace\rbrace instead. $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 23, 2013 at 17:36

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You should mean $\{x\}=\bigcap\xi$, and the answer is clearly yes, since we can take $\xi$ equal to the set of all open sets containing $x$. Any point $y$ other than $x$ is excluded in this intersection by the open set $X-\{y\}$. The cardinality of this $\xi$ is the same as the number of finite subsets of $X$, which is equinumerous with $X$. And the same will be true of any $\xi$ having your property, since the points other than $x$ must get excluded by elements of $\xi$, but only finitely many at a time, and so the cardinality of $\xi$ must be the same as $X$.

(This argument uses the axiom of choice, in order to know that the collection of finite subsets of an infinite set is equinumerous with that set. In $\neg$AC worlds, the situation is more complicated.)

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    $\begingroup$ Since you only need to intersect co-singletons, your $\xi$ can have cardinality at most that of $X$ even without AC. If you you want $\xi$ to have exactly the cardinality of $X$, even if $X$ is Dedekind-finite, then just throw one more set into $\xi$. An interesting question might be, in the Dedekind-finite case, how much smaller than $X$ you can make $\xi$. $\endgroup$
    – Andreas Blass
    Commented Apr 23, 2013 at 17:34
  • $\begingroup$ Andreas, that is an interesting question. My intuition points to "there is no minimal cardinality". $\endgroup$
    – Asaf Karagila ♦
    Commented Apr 23, 2013 at 17:38
  • $\begingroup$ Yes, I agree with all that. $\endgroup$
    – Joel David Hamkins
    Commented Apr 23, 2013 at 17:42
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    $\begingroup$ In particular, since for Dedekind finite sets one can find $\xi$ of cardinality strictly smaller than $X$, we do know that some AC is involved in the claim that all such $\xi$ have the same size as $X$. $\endgroup$
    – Joel David Hamkins
    Commented Apr 23, 2013 at 19:24

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