Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 $n$.

share|cite|improve this question
up vote 7 down vote accepted

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 $U\cap K_n=\varnothing$.

share|cite|improve this answer

Exists an open set containing $x$ such that $U\cap K_n$ is empty for infinitely many $n$.

share|cite|improve this answer

$\neg$(for all $U$ open containing $x$, $U \cap K_n \ne \emptyset$ for all but finitely many $n$).

share|cite|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.