Convergence of a z-filtre to an outer point

Question

Let $X$ be a completely regular topological space and let the set of all continuous functions from the topological space $X$ into the topological space $\mathbb{R}$ is denoted by $C(X)$. Let $Z(X)=\{Z(f) \mid f\in C(X)\} $, where $Z(f) = \{x\in X: f(x) = 0\}$.

A nonempty subfamily $F$ of $Z(X)$ is called a $z$-filter on $X$ provided that

(i) $\{\}\not\in F$;

(ii) if $Z_1, Z_2\in F$, then $Z_2 \cap Z_2\in F$; and

(iii) if $Z \in F$, $Z' \in Z(X)$, and $Z \subseteq Z'$, then $Z' \in F$.

A $z$-filter $F$ on $X$ is said to converge to the limit $p\in X$ if every neighborhood of $p$ contains a member of $F$.

Now if $X\subseteq T$ are two completely regular topological spaces and $p\in T$ and $F$ is $z$-filter on $X$. What is the definition of convergence of $F$ to $p$?(The topology on $X $ is induced)

Answer 1

I think that answering your question we may consider $F$ as a filter (not necessary a $z$-filter) on $T$ and then apply to it the standard definition of a convergence of a filter, which is the same as for $z$-filters, that is a filter $F$ on $T$ is said to converge to the limit $p\in T$ if every neighborhood of $p$ contains a member of $F$.