Prove that if $\mathscr{A}$ is a local base at point $p \in (X, \mathscr{T})$, then $\supset$ (the superset relation) is directed on $\mathscr{A}$.
Here are the definitions I am given for this problem
Call a preorder $R$ on a set $I$ directed if for all $x,y \in I$, there exists $z \in I$ such that $xRz$ and $yRz$.
If $\trianglelefteq$ is a directed preorder on $I$ and $\{x_i\}_{i \in I}$ is an $I-$sequence of points in $(X, \mathscr{T})$, say that $\lim_{i \in I} x_i=y$ if for every open $y \in U,$ there exists $k\in I$ such that $(k \trianglelefteq n \Rightarrow x_n \in U)$.
Given $p \in X, \mathscr{A}$ is a local base at $p$ if $\mathscr{A} \subset \mathscr{T}$ and $\left( \forall A \in \mathscr{A}, p \in A\right)$, and $\left(\forall U \in \mathscr{T} (p \in U \Rightarrow \exists A \in \mathscr{A} : p \in A \subset U) \right)$.
Based on what I have read so far, if a set is directed, then every two elements in $I$ have an upper bound, but not necessarily a least upper bound. We want to show that the preorder $\supset$ has the property that $x \supset z$ and $y \supset z$ for all $x,y \in I$.
I am very confused as to how we proceed in solving this problem. How does one show this result? Any help will be greatly appreciated. Many thanks in advance for your time and patience.
You don’t need the definition of limit along a directed set: it’s not relevant here. This definition of local base at a point is standard; you’ll want to remember it. It just says that $\mathscr{A}$ is a local base at $p$ if $\mathscr{A}$ is a collection of open nbhds of $p$ such that every open nbhd of $p$ is a superset of some member of $\mathscr{A}$. For instance, in $\Bbb R$ the set of intervals of the form $\left(x-\frac1n,x+\frac1n\right)$ for $n\in\Bbb Z^+$ is a local base at $x$.
Now, back to the actual problem. You’re asked to show that if $\mathscr{A}$ is a local base at some point $p\in X$, then $\langle\mathscr{A},\supseteq\rangle$ is a directed set. As you say, this requires showing that if $A,B\in\mathscr{A}$, then $A$ and $B$ have an upper bound in $\mathscr{A}$ relative to the order $\supseteq$. Translate that into simpler terms: you want to show that there is a $C\in\mathscr{A}$ such that $A\supseteq C$ and $B\supseteq C$.
If we knew that $A\cap B$ was a member of $\mathscr{A}$, we could use it; unfortunately, it need not be a member of $\mathscr{A}$. However, $A\cap B$ is guaranteed to be an open nbhd of $p$, and therefore it contains ... ?