Induction of topological space by an $L^2$-space Let $(L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R}),\langle\cdot,\cdot\rangle)$ be a Hilbert space, where $L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ denotes the set of all (equivalence classes of) $\boldsymbol{\Xi}$-measurable $\mathbb{R}$-valued functions that are square $\mu$-integrable on $\Xi$, and $\langle\cdot,\cdot\rangle:L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\times L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\to\mathbb{R}$ denotes its standard inner product given by
$$\langle f,g\rangle=\int fg\,\mathrm{d}\mu.$$
I know that this Hilbert space induces a topological space whose induced standard topology $\mathcal{O}$ is the one generated by all open balls whose radii are measured using the induced standard metric $d:L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\times L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\to\mathbb{R}_0^+$ given by
$$d(f,g)=\langle f-g,f-g\rangle^{\tfrac{1}{2}}.$$
My question is how to write this standard topology symbolically.
At the moment I have defined the topological basis:
$$\mathscr{B}=\{B_r(f)\subset L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\,\mid\, r\in\mathbb{R}^+,f\in L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\},$$
where $B_r(f):=\{g\in L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\mid d(f,g)<r\}$ is the open ball of radius $r$ centered at $f$.
Then, we define $\mathcal{O}=\varphi(\mathscr{B})$. Here $\varphi$ represents a map which takes a topological basis as an input and returns a topology as an output. In measure theory, we have the map $\sigma$ which generates a $\sigma$-algebra from a given collection of subsets. I thought that the same could be done here. However, I haven't found the actual symbol for performing such an operation on a topological basis yet.
So, is this the correct way to write the induced standard topology symbolically? Also, is there a better way to write the same thing?
Thank you, Frederick.
 A: Given a set $X$, a collection $\mathscr{B}$ of subsets of $X$ must satisfy two conditions to be a topological basis:
$1$. For any $x \in X$, there is $B\in \mathscr{B}$ such that $x \in B$.
$2$. For any $B_1, B_2 \in \mathscr{B}$ and for any $x \in B_1 \cap B_2$, there is a $B_3 \in  \mathscr{B}$ such that $x \in B_3$.
The topology generated by $\mathscr{B}$ is usually indicated by $\tau(\mathscr{B})$ (but this notation is not as standard as the $\sigma$ for the generated $\sigma$-algebra). Using this notation, we have:
$\tau(\mathscr{B})$ is the collection of arbitrary union of sets is $\mathscr{B}$.
So, while any collection of subsets can be used to generate a $\sigma$-algebra, it is not true that any collection of subsets is a basis for a topology.
If you want to mimic the concept of "generated $\sigma$-algebra" closer, you should use the concept of topological subbase:
Given a set $X$ and any collection $\mathscr{S}$ of subsets of $X$, the generated topology, $\tau(\mathscr{S})$, is defined as:
$\tau(\mathscr{S})$ is the collection of arbitrary unions of finite intersections of sets in $\mathscr{S}$.
Note that a partiular case of  finite intersection of sets in $\mathscr{S}$ is the nullary intersection who result is $X$.  So $X \in \mathscr{S}$.
In your specific example, $X= L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})$ and we have that
$$\mathscr{B}=\{B_r(f)\subset L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\,\mid\, r\in\mathbb{R}^+,f\in L^2(\Xi,\boldsymbol{\Xi},\mu;\mathbb{R})\}$$
is actually a basis for the topology (the standard Hilbert space topology).
You could write $\tau(\mathscr{B})$ to indicate the generated topology.
