Open sets and annihilator of functions A topological space $X$ is said to be completely regular provided
that it is a Hausdorff space such that, whenever $F$ is a closed
set and $x$ is a point in its complement, there exists a function
$f\in C(X)$, the ring of all continuous real function over $X$,
such that $f(x) = 1$ and $f(F) = 0$, in short, $F$ and ${x}$ are
completely separated by a continuous function. And let
$ann(f)=\{g\in C(X)\mid fg=0\}$, where $fg$ is the pointwais multiplication. \
Now if $u_1\subsetneq u_2\subsetneq u_3 \subsetneq...$ is a proper
chain of open subsets of $X$, how can we construct $f_i\in C(X)$
for $i=1, 2, 3,...$, such that $ann(f_1)\subsetneq
ann(f_2)\subsetneq ann(f_3)\subsetneq...$?
 A: Assuming that the members of ann$(f)$ are continuous : It is not possible in every case.
Let each $u_i$ be a dense open subset of $\mathbb R.$  E.g.  $u_i=\mathbb R \backslash \{n\in \mathbb N: n>i\}$. For each $i , $ let $f_i:\mathbb R\to [0,1]$ be continuous such that $f_i(x)\ne 0\iff x\in u_i.$ If $g$ is continuous and $g(x)f(x)=0$ for all $x$ then the closed set $g^{-1}\{0\}$ contains the dense subset $u_i , $  so $g=0.$ 
A: A natural idea to construct the sequence $\{f_i\}$ is to find for each $i$ a function $f_i$ such that the supporter $\operatorname{supp} f_i=\{x\in X: f_i(x)\ne 0\}=u_i$. But on this way we encounter two problems.  The first is that supporters are  $F_\sigma$ sets, which may fails for $u_i$ for non-perfect spaces $X$. The second is if the set $u_i$ is dense (as in Daniel Wainfleet’s answer) then $\operatorname{ann}(f_i)=\{0\}$. 
So the construction of the required sequence $\{f_i\}$ should be  more specific. Given a completely regular space $X$ we shall proceed as follows. 
If the space $X$ is finite then $C(X)$ is finitely dimensional, so each proper chain of the annulators (which are linear subspaces of $C(X)$) is finite. Thus from now we assume that the space $X$ is infinite.
If the space $X$ is discrete then we can pick any infinite sequence $\{x_i\}$ of points of the space $X$ and for each $i$ put $f_i(x)=0$ iff $x\in \{x_1,\dots, x_i\}$, otherwise $f_i(x)=1$. Then $\operatorname{ann}(f_i)=\{f\in C(X): f|X\setminus\{x_1,\dots, x_i\}=0\}$, so the sequence $\{f_i\}$ is the required.
From now we assume that the space $X$ is non-discrete. Let $y$ be an arbitrary non-isolated point of the space $X$. Starting from the empty set $F_0$ we inductively construct the sequence $\{f_i\}$ as follows. Given a closed subset $F_{i-1}\not\ni y$ of the space $X$, pick any point $x_i\in X\setminus F_{i-1}$ distinct from $y$. There exists a continuous function $f_i:X\to [0,1]$ such that $f_i(y)=1$ and $f_i(\{x_i\}\cup F_{i-1})=\{0\}$. Put $F_i=f_i^{-1}([0,1/2])$. Then $F_i\not\ni y$ is a closed subset of the space $X$. Since the sequence of the sets of zeros of the functions $f_i$ is non-decreasing, the sequence $\{\operatorname{ann} f_i\}$ of its annulators is non-decreasing too. Now for each $i$ pick an arbitrary function $g_i\in C(X)$ such that $g_i(x_{i+1})=1$ and $g_i(f_{i+1}^{-1}([1/2,1])=\{0\}$. Then $g_i\in \operatorname{ann} f_{i+2}\setminus \operatorname{ann} f_i$.
