Wordings of definition of an accumulation point of a net

Question

Let $O(x)$ denote the set of open sets containing a point $x$. I read in a definition,

A point $x$ in a topological space $(M,\tau)$ is called an accumulation point of the net $(x_i)_{i\in I}$ in $M$ if $$ \forall U\in O(x)\forall i\in I\exists j\in I(j\geq i\wedge x_j\in U).\qquad (*) $$

How do you read (*) in words? "For every open set $U$ containing $x$ and every $i\in I$, there exists $j\in I$ with $j\geq i$ such that $x_j\in U$? I think it sounds weird to say "every" twice before and after "and". Is there another way to say it?

Answer 1

There’s nothing at all weird about your wording, but it is possible to avoid the repetition while sticking close to the structure of $(*)$. For instance, one can say ‘for each open nbhd of $x$ and $i\in I$ there is a $j\ge i$ such that $x_j\in U$’. But one can also paraphrase the statement, for instance as ‘the net is in each open nbhd of $x$ cofinally often’ or ‘every open nbhd of $x$ intersects every tail of the net’.

Answer 2

You might understand this statement as a generalization of the statement "a sequence $(x_n)_{n \in \mathbb N}$ is in any neighborhood $U$ of $x$ infinitely often." When the net is $\mathbb N$ with its usual order, the last two parts of your statement $(*)$ become $$ \forall n \in \mathbb N \quad \exists m \in \mathbb N \quad (m \geq n \land x_m \in U),$$ Meaning that if we fix any large index $n$, there is always some index after $n$ for which the sequence re-enters $U$. For example, the sequence given by $x_n = (-1)^n$ "accumulates at" $1$ without converging to it. For any neighborhood $U$ of $1$ and any $n \in \mathbb N$, the next even number $m$ has $x_m = 1$, which of course is in $U$. In other words, this sequence is in $U$ infinitely often. This is different from convergence to $1$, which requires that the sequence be "eventually" in any neighborhood of $1$ (try to formulate this difference!).