Negation statement a net

Question

Let $X$ be a topological space and $A\subseteq X$. Then $A$ is closed if and only if every net in $A$ that converges to some $x\in X$, we have $x\in A$.

How do I negate the statement

"every net in $A$ that converges to some $x\in X$, we have $x\in A$"?

What troubles me is the wordings "that converges to some" when negating. I would negate this into "that converges to any", which sounds meaningsless. I end up getting

"there exists a net in $A$ that converges to some $x\in X$ and $x\notin A$".

I am still unsure about it. Could anyone explain how to negate a such statement when wordings can be hard to translate into mathematical notations/symbols?

Answer 1

The negation of the statement

$(\forall\,\text{net}\, \Phi$), (If $\lim\Phi$ exists then $\lim\Phi\in F$)

is

$(\exists\, \text{net}\, \Phi$), ($\lim\Phi$ exists, and $\lim\Phi\notin F$)

The last bit is bacause the negation of ($p\Rightarrow q$) is $\big(p\wedge (\sim q)\big)$

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Thus, the negation of

A set $F\subset X$ is closed if and only if $(\forall\,\text{net}\, \Phi$), (If $\lim\Phi$ exists then $\lim\Phi\in F$)

is

A set $F\subset X$ is not closed if and only if $(\exists\, \text{net}\, \Phi$), ($\lim\Phi$ exists, and $\lim\Phi\notin F$)

Answer 2

Just to prove the statement quoted (which is the end goal anyway, I presume):

Suppose $A$ is closed and $\Phi: I \to X$ is a net in $A$ (so $I$ is some directed set) which has limit $x \in X$.

Suppose $x \notin A$, then $X\setminus A$ would be an open neighbourhood of $x$ that contains no points of the net $\Phi$ (as $\Phi(i) \in A$ for all $i \in I$) by assumption) and this contradicts that $x \in \lim \Phi$.

Now suppose $A$ obeys the net condition and we show it’s closed: let $x \in \overline{A}$. Define a net $\Phi$ on $I= \mathcal{N}(x)$, the set of all neighbourhoods of $x$ (ordered by reverse inclusion) which is a directed set in any space, by picking $\Phi(N) \in N \cap A$ for all $N \in \mathcal{N}(x)$. This choice is possible by the fact that $x \in \overline{A}$ so that every neighbourhood of $x$ intersects $A$. This net, basically by definition of convergence of nets, converges to $x$ and so as we have a convergent net in $X$ with limit $x$, $x \in A$ by the assumption on $A$. So $\overline{A} \subseteq A$ so $A$ is closed.

I don’t see why you’d need the negation of the statement on nets in the proof.