Prove that the given collection of sets is a basis for a topology $\tau_1$ on $C[0,1]$ Question (from Topology without tears exercise 2.3 #Q4-i)

Let $C[0,1]$ be the set of all continuous real-valued functions on [0,1].
(i) Show that the collection $\mathcal{M}$  , where $\mathcal{M}=\left\{M(f,\varepsilon):f \in C[0,1] \;\text{and} \;\varepsilon\;\text{is a positive real number}\right\}$ and $M(f,\varepsilon)=\left\{g:g\in C[0,1]\;\text{and}\int_0^1|f-g|\lt\varepsilon\right\}$, is a basis for a topology $\tau_1$ on $C[0,1]$.  

Proof: Here's my attempt at a proof.
To prove that it is a basis, first I fixed an $\varepsilon$ and then I attempt to prove that:  


*

*$\bigcup_{f\in C[0,1]}M(f,\varepsilon)=C[0,1]$

*Let there be some index set $J$ which is probably uncountable. If I pick arbitrary $i,j\in J$ such that $i\neq j$. Then I have to show that, $M(f_i,\varepsilon)\bigcap M(f_j,\varepsilon)$ can be expressed as a union of members of $\mathcal{M}$.



1). This is trivial, by assuming some $f_k \in \bigcup_{f\in C[0,1]}M(f,\varepsilon)$ , it obviously belong to $C[0,1]$.
And for the reverse part I can assume $f_k\in C[0,1]$ then I have to show that $\exists$ a $f\in C[0,1]$ such that $f_k\in M(f,\varepsilon)$. I can choose $f=f_k+\frac\varepsilon2 $ so that $\int_0^1|f-f_k|=\frac\varepsilon2\lt\varepsilon$. Hence the result.
2). If an arbitrary element $f_k$ for some $k \in J$ belongs to $M(f_i,\varepsilon)\bigcap M(f_j,\varepsilon)$ then $f_k$ must be such that $f_k\in C[0,1]$ and $\int_0^1|f_i-f_k|\lt \varepsilon$ and $\int_0^1|f_j-f_k|\lt \varepsilon$. But I am not able to find a single function $g$ or some sequence of functions $g_n$ such that $\bigcup_n M(g_n,\varepsilon)=M(f_i,\varepsilon)\bigcap M(f_j,\varepsilon)$
I tried the function $g=\dfrac{f_i+f_j}{2}$ but i could only prove that $M(f_i,\varepsilon)\bigcap M(f_j,\varepsilon)\subseteq M(g,\varepsilon)$ with such choice of $g$.
 A: First, note you cannot fix $\varepsilon$, $\mathcal M$ ranges over all $f \in C[0,1]$ and all $\varepsilon >0$. Now to show $\mathcal M$ is a basis we must verify two properties.


*

*For all $f \in C[0,1]$ there is a $M \in \mathcal M$ such that $f \in M$.

*If $M_1,M_2 \in \mathcal M$ and $f \in M_1 \cap M_2$ then there exists an $M_3$ containing $f$ such that $M_3 \subseteq M_1 \cap M_2$.
1.) As you noted this is rather trivial as the set $M(f,\varepsilon)$ contains $f$ and is in $\mathcal M$ by definition.
2.) Let $M(g_1,\varepsilon_1),M(g_2,\varepsilon_2) \in \mathcal M$ be given and suppose $f \in M(g_1,\varepsilon_1)\cap M(g_2,\varepsilon_2)$. Consider the set $M(f,\varepsilon)$ with  $$\varepsilon = \min_{i}\{\varepsilon_i-\int_0^1|g_i-f |\}$$
Let $h \in M(f, \varepsilon)$. Then 
$$\int_0^1 |g_i-h|\leq \int_0^1|g_i -f |  \, + \int_0^1|f-h| <\int_0^1|g_i-f| +\varepsilon <\varepsilon_i$$
Thus $h \in M(g_i,\varepsilon_i)\, \forall i$ implying $h \in M(g_1,\varepsilon_1) \cap M(g_2,\varepsilon_2)$. Hence $M(f,\varepsilon)\subseteq M(g_1,\varepsilon_1) \cap M(g_2,\varepsilon_2)$ as desired.
We can now conclude by 1.) and 2.) that $\mathcal M$ is a basis for a topology on $C[0,1]$.
