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Let $C[0; 1]$ be the set of all continuous real-valued functions on $[0; 1]$.

(i) Show that the collection $M$, where $M = \{M(f,\varepsilon ) : \text{$f\in C\left[0; 1\right ]$ and $\varepsilon $ is a positive real number}\}$ and $M(f,\varepsilon) =\{g : \text{$g\in C\left[0; 1\right ]$ and $\int_{0}^{1}\left|f-g\right| < \varepsilon $}\}$, is a basis for a topology $\mathcal{T}_{1}$ on $C[0; 1]$.

(ii) Show that the collection $U$, where $U = \{U(f,\varepsilon ) : \text{$f\in C\left[0; 1\right ]$ and $\varepsilon $ is a positive real number}\}$ and $U(f,\varepsilon ) =\{g : \text{$g\in C\left[0; 1\right ]$ and $\sup_{x\in \left[0,1\right]}$$\left|f-g\right|<\varepsilon $}\}$, is a basis for a topology $\mathcal{T}_{2}$ on $C[0; 1]$.

(iii) Prove that $\mathcal{T}_{1}\neq \mathcal{T}_{2}$.

(i)and (ii) are similar by using the property of absolute value $\left|f-g\right|\leq\left|f\right|+\left|g\right|$ for (i) let $M_{1}$ and $M_{2}\in M$ where $M_{1}(f_{1},\varepsilon) =\{g : \text{$g\in C\left[0; 1\right ]$ and $\int_{0}^{1}\left|f_{1}-g\right|<\varepsilon $}\}$, $M_{2}(f_{2},\varepsilon) =\{g : \text{$g\in C\left[0; 1\right ]$ and $\int_{0}^{1}\left|f_{2}-g\right|<\varepsilon $}\}$ then $M_{1}\cap M_{2}=M(\dfrac{f_{1}+f_{2}}{2},\varepsilon )$ so $M$ is a base for $C[0; 1]$.

But I am not sure for (iii) by using the mean value theorem of integrals if $g$ is in some $m\in M$ then there may has no $u\in U$ since $\int_{0}^{1}\left|f-g\right|=\left|(f-g)\right| \left|(\xi)\right|<\varepsilon$ ($\xi \in [0,1]$) but if $\left|(f-g)\right| \left|(\xi)\right|<\sup_{x\in \left[0,1\right]}\left|f-g\right|$ so $g$ is not in some $u$ in $U$. I have no ideal about what to do next.

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$\left|(f-g)\right| \left|(\xi)\right|$ <$\sup_{x\in \left[0,1\right]}$$\left|f-g\right|$ so g is not in some u in U. I have no ideal about what to do next. –  Jebei Aug 8 '13 at 9:44
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Keep in mind a strategy, to prove that some set is open in one topology but not in the other. In this case one topology "refines" the other, i.e. all the open sets of one are open in the other, but not conversely. So first figure out which topology is "more refined", i.e. has the smaller open sets. –  hardmath Aug 8 '13 at 10:15
    
@hardmath U(f,ε) may not open in M(f,ε) by using mean value theorem. So U is smaller.This way? –  Jebei Aug 8 '13 at 10:19
    
Right, it's enough to look at open ngbhds of the origin ($f=0$) as both topologies are translation invariant. $U(0,\epsilon) \subseteq M(0,\epsilon)$, so any open ngbhd of the origin in $T_1$ is also open in $T_2$. –  hardmath Aug 8 '13 at 10:34
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This image shows (the beginning of) a sequence of functions all within the same $M$-neighbourhood of $f(x) = 0$, but not the same $U$-neighbourhood. It was made for a different purpose, and is therefore centered around the origin, but it should give you an idea of what counter-example to put forward. –  Arthur Aug 8 '13 at 10:50

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up vote 2 down vote accepted

Parts (i) and (ii) are routine, as we are asked to show the families $M$ and $U$ are open bases for topologies, hereafter called $\mathcal{T}_1$ and $\mathcal{T}_2$. What needs to be shown for each family is that (1) it covers the space $C[0,1]$, continuous real functions on the unit interval, and (2) that for any "point" $f$ in the intersection $S_1 \cap S_2$ of two members of such a family, there exists a member $S_3$ of that family s.t. $f \in S_3 \subseteq S_1 \cap S_2$.

The covering property is evident for both families from the definitions of the $M$ and $U$ families. The intersection property is only slightly more work, as we illustrate for the family $M$ of sets $M(f,\varepsilon)$, using triangle inequality arguments.

Suppose $f \in M(f_1,\varepsilon_1) \cap M(f_2,\varepsilon_2)$. Since $\int_0^1 |f_1 - f| dx \lt \varepsilon_1$ and $\int_0^1 |f_2 - f| dx \lt \varepsilon_2$, we pick $\varepsilon_3 > 0$ to be the minimum of $\varepsilon_1 - \int_0^1 |f_1 - f| dx $ and $\varepsilon_2 - \int_0^1 |f_2 - f| dx $. Then for any $g \in M(f,\varepsilon_3)$ we would have (from a triangle inequality) respectively:

$$ \int_0^1 |f_1 - g| dx \le \int_0^1 |f_1 - f| dx + \int_0^1 |f - g| dx \lt \varepsilon_1 $$ $$ \int_0^1 |f_2 - g| dx \le \int_0^1 |f_2 - f| dx + \int_0^1 |f - g| dx \lt \varepsilon_2 $$

and the interesection property is established: $f \in M(f,\varepsilon_3) \subseteq M(f_1,\varepsilon_1) \cap M(f_2,\varepsilon_2)$.

The argument for family $U$ is similar except that the sup-norm rather than the $L^1$-norm is involved.

The more difficult part of the exercise seems to be showing the resulting topologies $\mathcal{T}_1$ and $\mathcal{T}_2$ are different. The image linked by @Arthur gives the idea; a function can be close to the origin in the $L^1$-norm, but arbitrarily far away in the sup-norm.

It suffices, per comments under the Question, to show that $U(0,1/2)$, an open set in $\mathcal{T}_2$, is not open in $\mathcal{T}_1$. In particular there does not exist any $\varepsilon \gt 0$ such that $0 \in M(0,\varepsilon) \subseteq U(0,1/2)$, which would be required for $U(0,1/2)$ to be an open neighborhood of the origin in $\mathcal{T}_1$.

To prove this, consider the elementary sequence of real functions $f_n(x) = x^n$ on the unit interval, $n = 1,2,3,\ldots$:

$$ \int_0^1 |0 - x^n| dx = 1/n $$

Since all but finitely many $f_n$ belong to $M(0,\varepsilon)$ but none of them belong to $U(0,1/2)$ (because each *sup*$|f_n(x)| = 1$ on $[0,1]$), clearly $M(0,\varepsilon) \not \subseteq U(0,1/2)$. Open sets in $\mathcal{T}_2$ are not necessarily open in $\mathcal{T}_1$, so these topologies are not the same.

As it happens, the converse holds however, that open sets in $\mathcal{T}_1$ are always open in $\mathcal{T}_2$. In a case like this we say $\mathcal{T}_2$ is strictly finer than $\mathcal{T}_1$.

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