Simple question on basis for induced topology Let X be a set and $\mathcal{F}$ be a family of real functions $f:X \rightarrow \mathbb{R}$.
I want to prove that $\mathcal{B}=\{W(x,A, \varepsilon\}|x \in X, A \subset \mathcal{F} \text{ finite },\varepsilon>0 \}$ where $W(x,A,\varepsilon)=\{ y\in X\mid |f(y)-f(x)| < \varepsilon \forall f \in A\}$ is a basis for the induced topology on X (i.e the smallest topology in X that makes the functions from X to $\mathbb{R}$ continuous).
I know that if  $X$ is a set $Y_i, i \in I$ topological spaces and $\mathcal{B}_i $ are bases for $Y_i$ then $\mathcal{B}'=\{ {\cap}_{i=1}^{n} f_i^{-1}(B_i)|B_i \in \mathcal{B}_i, n \in \mathbb{N} \}$ is a basis for X (with the induced topology).
I tried using that with $\mathcal{B}_i=\{(f_i(x)- \varepsilon,f_i(x)+ \varepsilon)| \varepsilon>0,x\in \mathbb{R} \}$ :
$ {\cap}_{i=1}^{n} f_i^{-1}(B_i)= \{y \in X: |f_i(y)-f_i(x_i)|< \varepsilon_i \forall i=1,...,n\} $ since each $B_i=(f_i(x_i)- \varepsilon_i,f_i(x_i)+ \varepsilon_i)$ (trying to show that $\mathcal{B}=\mathcal{B}'$) but it doesn't seem to work
 A: Let $\mathcal{T}$ be any topology that makes all $f \in \mathcal{F}$ continuous. Also, let $\mathcal{T}_w$ be the induced topology on $X$ by $\mathcal{F}$.
Then for each $x \in X$, $\varepsilon>0$ and $f \in \mathcal{F}$ we have that
$$W(x,\{f\},\varepsilon) = f^{-1}[(f(x)-\varepsilon, f(x)+\varepsilon)]$$ by definition, and as open intervals are open in $\Bbb R$ and $f$ is continuous, $W(x,\{f\},\varepsilon) \in \mathcal{T}$.
Conversely, if a topology $\mathcal{T}$ contains all sets of the form $W(x,\{f\},\varepsilon)$, then all $f \in \mathcal{F}$ are continuous: let $O$ be open in $\Bbb R$ and $f \in \mathcal{F}$, then $f^{-1}[O]$ is open in $\mathcal{T}$: let $x \in f^{-1}[O]$, then $f(x) \in O$ and so for some $\varepsilon >0$, $f(x)-\varepsilon, f(x)+\varepsilon) \subseteq O$. But then $W(x,\{f\},\varepsilon) \subseteq f^{-1}[O]$ and so $x$ is a $\mathcal{T}$-interior point of $f^{-1}[O]$.  As $x$ was arbitrary, $f^{-1}[O] \in \mathcal{T}$ for all open $O$ and all $f \in \mathcal{F}$.
Together this shows that $\mathcal{S}= \{W(x, \{f\}, \varepsilon)\mid x \in X, \varepsilon>0, f \in \mathcal{F}\}$ is a subbase (i.e. generating set) for $\mathcal{T}_w$. And as
$$W(x, \mathcal{A},\varepsilon) = \bigcap_{f \in \mathcal{F}} W(x,\{f\},\varepsilon\}$$
the proposed set is the base generated by $\mathcal{S}$ and so a base for $\mathcal{T}_w$. QED

It's also not too hard to check that the proposed base satisfies the two conditions to be a base for some topology. It's clear that every $x$ is covered by $W(x,\{f\},1)$ for any $f \in \mathcal{F}$, so that part is trivial, and if $y \in W(x_1, \mathcal{A}_1, r_1) \cap W(x_2, \mathcal{A}_2, r_2)$ we know that for all $f \in \mathcal{A}_1$, $|f(z) - f(x_1)| < r_1$ and also for all $g \in \mathcal{A}_2$ we have $|f(z) - f(x_2)| < r_2$. Consider the finitely many $f(y) \in (f(x_1)-r, f(x_1)+r)$, where $f$ ranges over $\mathcal{A}_1$; as the interval is open, we can find some $0 < r_1' < r_1$ so that the all intervals around these  $f(y)$ with radius $r'_1$ are still inside $(f(x_1)-r, f(x_1)+r)$ for all $f \in \mathcal{A}_1$; similarly we can find $0 < r'_2 < r_2$ that works for all $f(y), f \in \mathcal{A}_2$; now define $r' = \min(r_1', r_2')>0$ and note that now we have $$W(y, \mathcal{A}_1 \cup \mathcal{A}_2, r') \subseteq W(x_1, \mathcal{A}_1, r_1) \cap W(x_2, \mathcal{A}_2, r_2)$$ so the final axiom to be a base is satisfied.
As $\mathcal{S}$, the subbase for $\mathcal{T}_w$ is already included in this base, as we saw, the generated topology by this base is exactly $\mathcal{T}_w$.
