What is the negation? Sequences on topological spaces Let $X$ be a topological space. We say that $x_n \to x$, if and only if, for every $U \in \mathcal{U}_x,$ a neighborhood system, there is $n_0 \in \mathbb{N}$ such that $x_n \in U$ for all $n > n_0.$
Is this the negation to this as follows?
$x_n$ does not converges to $x$ if, and only if, there is $U \in\mathcal{U}_x$ such that for any $n \in \mathbb{N},$ $x_n$ does not belong to $U$?
The point of my question is that I am trying to prove that if for each subsequence $(x_{n_k})$ of a sequence $(x_n)$ there is a subsequence that converges for $x$, then $(x_n)$ converges to $x.$
My attempt to do this was: 
Suppose that $x_n$ does not converges to $x$. Then there is a $U \in \mathcal{U}_x$ such that $x_n \not\in U$ for every $n$. The none subsequence has a sequence that belongs to $U$.
Thank you!
 A: The exact negation of your statement is:

$x_n$ does not converge to $x$ if there exists a $U \in \mathcal{U}_x$ such that for all $n_0 \in \mathbb{N}$, there exists $n > n_0$ such that $x_n \notin U$.

The statement you proposed is strictly stronger than this. That is, it's possible for $x_n$ not to converge to $x$ while not having a $U$ such that $x_n \notin U$ for all $n$. For example, the sequence $x_n = (-1)^n$ in $\mathbb{R}$ does not converge to $1$, but there is no $U \in \mathcal{U}_1$ such that $x_n \notin U$ for all $n$ because the sequence keeps coming back to $1$.
A: First, write your quantifiers in prenex form: no "trailing quantifiers" the way you wrote "for all $n>n_0$".
The negation of 
$$\forall U \in \mathcal{U}_x \,\, \exists n_0 \in \mathbb{N} \,\, \forall n > n_0 \,\, BLAH
$$
is
$$\exists U \in \mathcal{U}_x \,\, \forall n_0 \in \mathbb{N} \,\, \exists n > n_0 \,\, NOT \, BLAH
$$
The pair of quantifiers "$\forall n_0 \in \mathbb{N} \,\, \exists n > n_0$" is sometimes rewritten in English as "there are arbitrarily large integers $n$" or as "there is a sequence of integers $n_i$ diverging to $+\infty$". So, what you end up with is


*

*There exists a neighborhood $\mathcal{U}_x$ of $x$, and there exists a sequence of integers $n_i$ approaching $+\infty$, such that none of the points $x_{n_i}$ belongs to $\mathcal{U}_x$.

