What is the negation of this statement? Let $(K_n)$ be a sequence of sets.  
What is the negation of the following statement?
For all $U$ open containing $x$, $U \cap K_n \neq \emptyset$ for all but finitely many $n$.  
 A: Let us write the statement formally:
$$\forall U(x\in U\rightarrow\exists n\forall k(k>n\rightarrow U\cap K_k\neq\varnothing))$$
For every $U$ (open of course), if $x\in U$ then there is some $n$ that for all $k>n$ we have $K_k\cap U\neq\varnothing$.
Now negation flips quantifiers and $\lnot(\alpha\rightarrow\beta)$ is the same as $\lnot\beta\land\alpha$. So we have:
$$\exists U(x\in U\land\forall n\exists k(k>n\land U\cap K_k=\varnothing))$$
Or in words, there exists an open set $U$ such that $x\in U$ but for every $n$ there is some $k>n$ such that $U\cap K_k=\varnothing$. However in the natural numbers to say that something happens unboundedly often is the same as saying it happens infinitely often. So finally we can say:

There exists an open set $U$ such that $x\in U$ and for infinitely many $n$ we have $U\cap K_n=\varnothing$.

A: Exists an open set containing $x$ such that $U\cap K_n$ is empty for infinitely many $n$.
A: $\neg$(for all $U$ open containing $x$, $U \cap K_n \ne \emptyset$ for all but finitely many $n$).
