Which topologies on $X$ can be generated by picking $F$, a set of functions into $\mathbb{R}$, and making them continuous? Let $\varepsilon$ denote the empty set.
Let $X$ be a topological space. We will be considering multiple possible topologies on $X$.
Let $\langle B \rangle$ denote the closure under arbitrary unions, including unions of zero elements, and finite intersections of a set of sets, $B$.
Let $T_{\mathbb{R}}$ denote the standard topology on the reals.
Which topologies can be generated by picking $F$, a set of maps from $X$ to $\mathbb{R}$, arbitrarily declaring them to be continuous, and then computing the inverse images of elements of $T_{\mathbb{R}}$.
For instance, I can get the standard topology on $\mathbb{R}\!\times\!\mathbb{R}$ by picking the two functions $f_1$ and $f_2$ below
$$ f_1(x, y) = x $$
$$ f_2(x, y) = y $$
Computing $\langle f^{-1}_1(T_{\mathbb{R}}), f^{-1}_2(T_{\mathbb{R}}) \rangle$ gives me the topology I want. I can construct any epsilon ball in $\mathbb{R}\!\times\!\mathbb{R}$ that I want by unioning together a countable set of open squares. The epsilon balls form a basis of the standard topology on $\mathbb{R}\!\times\!\mathbb{R}$, so I'm done.
However, we can't generate the Sierpiński topology, $\{ \varepsilon, \{0\}, \{0, 1\}\}$.
Any potential function $f$ must send $0$ to a single real number and $1$ to a single real number and every real number is in some open set.
Which topologies can we generate by picking some set of functions to $\mathbb{R}$ to be continuous and then leveraging the existing topological structure of $\mathbb{R}$?
 A: The topologies you can generate that way are just the completely regular topologies. A topological space $X$ is completely regular if, for every closed set $A$ and every point $p\notin A$, there is a continuous function $f:X\to\mathbb R$ such that $f(p)\gt0$ and $f(x)=0$ for all $x\in A$. (Note that this definition does not require a completely regular space to be $T_1$.)
It's clear from the definition that a completely regular topology is generated by its continuous real-valued functions. The converse requires a bit of argument. Let $X$ be a set and let $F$ be a set of functions $f:X\to\mathbb R$. Then $\mathcal S=\{f^{-1}(V):f\in F,\ V\text{ open in }\mathbb R\}$ is a subbase for a topology $\tau$ on $X$. I claim that $(X,\tau)$ is completely regular.
Let $A$ be a closed set in $(X,\tau)$, and let $p\in U=X\setminus A$. Then there are sets $S_1,\dots,S_n\in\mathcal S$ such that $p\in S_1\cap\cdots\cap S_n\subseteq U$, and there are functions $f_1,\dots,f_n\in F$ and open sets $V_1,\dots,V_n\subseteq\mathbb R$ such that $S_i=f_i^{-1}(V_i)$; moreover we may assume that $V_i\ne\mathbb R$. Let $A_i=\mathbb R\setminus V_i\ne\emptyset$, and define a continuous function $f:X\to\mathbb R$ by setting
$$f(x)=\prod_{i=1}^nd(f_i(x),A_i).$$
Then $f(p)\gt0$, while $f(x)=0$ for all $x\in A$.
A: First of all, it looks like the topology $T$ on $X$ doesn't play any role in the question -- can't we ignore/eliminate it from consideration? In fact, why not just call $X$ a set?
I'll add the observation that for any $A \subset X$, $\chi_A^{-1}( (1/2, 3/2) ) = A$, so we can get any arbitrary subset into our sub-basis, but at the cost of getting $X \setminus A$ in there as well. Initially I had thought that it completely answered the question, but now I think it just means we can generate topologies that "have many open sets", but like the Sierpinksi example shows, some "small" topologies are not achievable. I'm leaving this answer up just in case someone finds the observation useful.
