Use of directed sets in the definition of nets in topology In topology, we use nets instead of sequences. The motivation is quite natural since the sequence is not "long" enough if the neighborhoods of some point "separate" too much. 
What I am confused about is the concept of directed set, is the only reason why we need the set to be directed because we want the definition of convergence of net to be uniform with that of sequence? Or there is another reason, thanks
 A: If $(x_d)_{d\in D}$ is a net in $X$ with $(D,\preceq)$ a directed preorder, $(x_d)$ converges to $x$ iff it's eventually in every neighborhood of $x$, where "eventually" means "from some $\underline{d}$ on": for every neighborhood $U$ of $x$, there is some $\underline{d}\in D$ such that for all $d \succeq \underline{d}$, $x_d \in U$.
The neighborhood system of a point $x$ in a space $X$ is downward directed by $\subseteq$: if $U,V$ are neighborhoods of $x$, then there is a neighborhood $W$ of $x$ such that $W\subseteq U$ and $W\subseteq V$ — namely, $W = U\cap V$.
We need to be able to argue that for any sets $A$ and $B$, if $(x_d)$ is eventually in $A$ and $(x_d)$ is eventually in $B$, then $(x_d)$ is eventually in $A\cap B$. Requiring $(D,\preceq)$ to be directed guarantees that we can.
If nets were defined on non-directed preorders, then "limits" wouldn't necessarily be unique, even in 'nice' (e.g. Hausdorff) spaces. For example, suppose $D = $ the infinite binary tree — all finite sequences of $0$s and $1$s, ordered by inclusion, i.e. extension. $(D, \subseteq)$ is very non-directed: any two incomparable finite sequences in $D$ differ at some index, so they have no common extension.
Suppose $(x_s)_{s\in D}$ is the function $D\to\Bbb R$ such that
$$
x_s = \begin{cases}
3 &\text{if $s_0 = 0$,}\\
5 &\text{if $s_0 = 1$,}\\
0 \text{ ($\text{arbitrarily}$)} &\text{if $s$ is the empty sequence.}\\
\end{cases}$$
Then $(x_s)$ is eventually in every neighborhood of $3$ (for every extension $s$ of $\underline{s} = \langle 0 \rangle$, $x_s$ is in every neighborhood of $3$), and it's eventually in every neighborhood of $5$ (take $\underline{s} = \langle 1 \rangle$).
