Ultra filters are countably complete?

I am reading friends lecture notes. It says "Let $F$ be an ultra filter on a set $X$ such that $F$ is not countably complete. Then there are $Y_n \in F, n \in \omega$ such that $\bigcap_n Y_n = \varnothing$"

But: from it seems to follow that "every ultra filter is countably complete". And, here is counter example to above statement: let $\mathbb R$ have standard topology, $x \in \mathbb R$, $F$ be the nbhood base of $x$ consisting of all nbhoods $N_x$ of $x$. Then for all $(Y_n)_{n}$ with $Y_n \in F$, $\bigcap Y_n \neq \varnothing$.

I asked the friend but he doesn't remember and we would like to know what statement was intended. Can someone help us and tell us what statement probably was intended by lecturer? Many thank you.

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 Wait, the assumption is: $F$ is not coutbly complete, while you said that "every ultra-filter is countbly comlete." I am quite confused what you meant. Thanks for any explanation then. – awllower Jan 31 at 11:00

Note that filters can be countably complete, even if every ultrafilter extending them is not. The collection of open neighborhoods of $x$ is not an ultrafilter, so this is not a counterexample.
The fact is that every infinite set has a free ultrafilter which is not countably complete. The proof is simple, let $X$ be an infinite set, and let $A$ be a countably infinite subset. Let $\cal F$ be an ultrafilter on $A$ which extends the co-finite filter. It's not hard to see that $\cal F$ is not countably complete. Now define $$\mathcal F^*=\{Y\subseteq X\mid A\cap Y\in\cal F\}$$
You can show that this is a filter which includes $\cal F$, and in particular the sequence witnessing its incompleteness.
It should be pointed out that the existence of free ultrafilters which are closed under countable intersections is not provable from $\mathsf{ZFC}$.
You probably mean "let $\mathcal{F}$ be an ultrafilter on $A$...." I tried to correct this myself so as not to bug you about it, but someone rejected my edit :-( – Trevor Wilson Jan 31 at 20:59