Prove two bases are not equivalent on $C[0,1]$ This is starred question #2.3.4 from S. Morris's Topology without Tears.
Let $C[0,1]$ be the set of all continuous real - valued functions on $[0,1]$
(i) Show that the collection  $M$ where $M=\{M(f,\epsilon) : f \in C[0,1]$ and $\epsilon$ is a positive real number$\}$ and $M(f,\epsilon) = \{g : g \in C[0,1]$ and $\int_0^1 \left| f-g \right| < \epsilon \} $ is a basis for a topology $\tau_1$ on $C[0,1]$.
(ii) Show that the collection $U$, where $U=\{U(f,\epsilon) : f \in C[0,1] $ and $\epsilon$ is a positive real number$\}$ and $U(f,\epsilon) = \{g : g \in C[0,1]$ and $\sup_{x\in[0,1]} \left|f(x) - g(x)\right| < \epsilon \}$, is a basis for a topology $\tau_2$ on $C[0,1]$.
(iii) Prove that $\tau_1 \ne \tau_2$.
I wish I could present an attempt, but I can't since I don't even know what the mentioned sets are. I really need help on this one. The other question I posted includes my attempt for the first part of that question.
Please simplify your answers as much as possible so that a beginner might understand. Thank you.
 A: Here's a hint:
(i) What are the characteristics that define a basis for a topology? 


*

*First, it must cover your space. 


This is easy, since $\forall f\in C[0,1]$, we have  $f\in M(f,\epsilon)$ for every $\epsilon>0$. So fix an $\epsilon$. 
Then, $C[0,1]\subset \bigcup_{f\in C[0,1]}M(f,\epsilon)$. 


*

*Next, you have to show that if $h\in M(f,\epsilon)\cap M(g,\delta)$, there is another basis element $N$ such that $h\in N\subset M(f,\epsilon)\cap M(g,\delta)$. 


So suppose $h\in M(f,\epsilon)\cap M(g,\delta)$. You have to think of another continuous function $k\in C[0,1]$ and a positive number $\gamma$ so that $h\in M(k,\gamma)\subset M(f,\epsilon)\cap M(g,\epsilon)$. 
This is where you need to fiddle around with the problem a bit on paper, but the idea would be to use the functions and constants you have, namely $h,f,g$ and $\delta,\epsilon$. Think of linear combinations of these and try to make them work.
(ii) The first part is the same as before, and the second part is just finding a basis element for $T_1$ that is not in $T_2$ and vice versa.
Hope that helps somewhat!
A: I will first outline some strategies for thinking about the question.  Then, I will provide my answer to part (i):
Strategies:
1) What is $M(f,\epsilon)$?  Since $f, g\in C[0,1]$, focus on what $\int_{0}^{1}|f-g|$ means. Since $|f-g|$ is the positive difference between the two functions, then the integral of that is total positive area between $f$ and $g$.
2) What is $U(f,\epsilon)$?  Since $f, g\in C[0,1]$, focus on what $\sup_{x\in[0,1]}|f-g|$ means. Since $|f-g|$ is the positive difference between the two functions, then the supremum of that is the least upper bound of the positive distance between $f$ and $g$.
3) In other words, $M$ is looking at total area between $f$ and $g$ over the whole interval $[0,1]$, whereas $U$ is looking at the one slice of $[0,1]$ where the difference between $f$ and $g$ is biggest.
4) I think the key to this question is to think about the many different ways that two functions can be close to or far from each other over an interval, and how that affects the area between them as well as the maximum distance between them.
Answer:
i) Moya answered the first part about covering and outlined the goal for the second part. 
 But I think I have a slightly different way of going about it:


*

*You want to show that there is a basis element $N$ such that $h\in N\subset M(f,\epsilon)\cap M(g,\delta)$.


Consider $N = M(h, \alpha_3)$.  We want to choose $\alpha_3$ so that $N$ is inside the given intersection.
Now, since we already know that $h\in M(f,\epsilon)\cap M(g,\delta)$, we have:
(*)$\int_0^1 |f-h|=k_1 <\epsilon$ 
and   
(**) $\int_0^1 |g-h|=k_2 <\delta$
Now, $\forall p\in M(h, \alpha_3)$, we know:  $\int_0^1 |h-p|<\alpha_3$
Combining this with (*), we have:
$\int_0^1 |f-h| + \int_0^1 |h-p| < k_1 +\alpha_3$
But note that on the left we may combine the integrals and apply the triangle inequality to get the following:
$\int_0^1 |f-p| < k_1 +\alpha_3$
So in order for $p\in M(f,\epsilon)$, we need to choose: $\alpha_3 < \epsilon - k_1$
Repeat this argument with (**) in order to get the condition:  $\alpha_3 < \delta- k_2$
Let $\alpha_3 < \min\{ \epsilon - k_1, \delta- k_2\}$
Comments on intuition:  In retrospect, I wasted a lot of time trying to find a specific function that would work.  I feel like I finally made a breakthrough when I started to realize that since $h$ was in both basis sets, any other function in the intersection couldn't stray too far away from $h$ without potentially opening up too much area between $f$ or $g$.  This led me to the realization that each function in my set had an integral with a (unknown) fixed value, but that I could  still use the $\alpha_3$ to squeeze that set of functions as close to $h$ as needed in order not to violate the $\epsilon$ or $\delta$ boundaries.
[part 2 is similar, perhaps in a later edit...]
