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 $n$.
Let us write the statement formally:
$$\forall U(x\in U\rightarrow\exists n\forall k(k&gt;n\rightarrow U\cap K_k\neq\varnothing))$$
For every $U$ (open of course), if $x\in U$ then there is some $n$ that for all $k&gt;n$ 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:

$$\exists U(x\in U\land\forall n\exists k(k&gt;n\land U\cap K_k=\varnothing))$$
Or in words, there exists an open set $U$ such that $x\in U$ but for every $n$ there is some $k&gt;n$ 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 $U\cap K_n=\varnothing$.
$\neg$(for all $U$ open containing $x$, $U \cap K_n \ne \emptyset$ for all but finitely many $n$).