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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.

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    $\begingroup$ Welcome to the site. It’s much more the site’s etiquette to show one’s own effort to solve the stated problems and/or to give a little context. $\endgroup$ – k.stm Mar 18 '15 at 7:20
  • $\begingroup$ @k.stm Hi! This is an exercise from Chapter 2.3 (Basis for a given topology) from Moriss' Topology without tears. 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 from the button up. The other question I posted includes my attempt for the first part of that question. $\endgroup$ – user224530 Mar 18 '15 at 7:56
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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!

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  • $\begingroup$ Ok, so here is where the dumbing down begins. So lets take it step by step: Firstly, I completely agree that the first thing we need to do is to verify that it covers the space. So, as you said we need to show that $C[0,1]$ is covered by $M$. Now you seem to do this by fixing $\epsilon$ and then showing that the space $C[0,1]$ is a subset of $\bigcup_{f \in C [0,1]} M(f,\epsilon )$ ... Please elaborate on how this first part works. Thanks. $\endgroup$ – user224530 Mar 18 '15 at 15:14
  • $\begingroup$ Well, just fix some $\epsilon$. Then we have that $f\in M(f,\epsilon)$ since $\int_0^1 |f-f|=0<\epsilon$. So for every $f\in c[0,1]$, $f\in M(f,\epsilon)\in M$. Thus $C[0,1]$ is covered by $M$. $\endgroup$ – Moya Mar 18 '15 at 16:49
  • $\begingroup$ Ok, I think I got that. But another question, I might already know this but I would feel more grounded if you explained it. Why is it necessary that we fix an epsilon? I mean since we know that epsilon is positive wont the statment $\int_0^1 |f-f|=0 <\epsilon$ hold no matter what the epsilon is? Please bear with me... $\endgroup$ – user224530 Mar 19 '15 at 7:10
  • $\begingroup$ Sure. You can choose just any $\epsilon$, and the inclusion holds. You could choose a different $\epsilon$ for each $f$, but since any $\epsilon$ will work, just choose one fixed $\epsilon$ for this case. In the other case you have to vary the positive constants. $\endgroup$ – Moya Mar 19 '15 at 8:09
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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...]

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