Non-empty Intersection of a set with the closure of some other sets implies Non-empty Intersection of a set with the sets

Question

Let $X$ be a topological space.

Let $(x_i)_{i \in I}$ be a net in $X$. Let $I$ be an directed set.

Let $S_i= \{ x_j \in X : j \geq i, \forall i\in I \}$.

Suppose $x\in \cap_{i \in I} \bar{S}_i$, where $\bar{S}_i$ is the closure of $S_i$.

Then we have that any neighbourhood $U_x$ of $x$ intersects with each of $\bar{S}_i$.

From the previous line, why do we have that any neighbourhood $U_x$ of $x$ intersects with each of $S_i$?

Thanks!

Answer 1

Fix $i \in I$ for now, arbitrarily. We know $U_x$ is a neighbourhood of $x$ and we also know that $x \in \overline{S_i}$ by definition (Of the intersection). This means that every neighbourhood of $x$, so also $U_x$ (!), intersects the set $S_i$.

So for every $i \in I$ there is some $j \ge i$ such that $x_j \in U_x$ (the witness to $U_x$ intersecting $S_i$). This says that $x$ is an accumulation (or cluster) point of the net $(x_i)_{i \in I}$.