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I have to solve the following:

For every function $f\in C[0,1]$, $\varepsilon>0$ and every finite set $A$, set $U(f,A,\varepsilon)$ is defined by $U(f,A,\varepsilon)=\{g\in C[0,1]:\forall x\in A\; |f(x)-g(x)|<\varepsilon\}$ and family of those sets forms a base of some topology $\mathcal{T}$ on $C[0,1]$. If $U$ is topology induced by metric $d_\infty$ and $M$ topology induced by $d_1$, compare topologies $\mathcal{T}$, $\mathcal{U}$, $\mathcal{M}$.

(By $d_\infty$ and $d_1$ we denote $d_\infty(f,g)=\max_{x\in[0,1]}|f(x)-g(x)|$ and $d_1(f,g)=\int_0^1 |f(x)-g(x)|dx$). So, set $U(f,A,\varepsilon)$ contains continuous functions whose graphs pass through the finite set of $\varepsilon$-balls (whose centers are points from $A$). We have to check two properties of base for topology $\mathcal{T}$ (first, sets $U(f,A,\varepsilon)$ cover $C[0,1]$ and the second one, with intersection of two base sets, that I'm having problem with). Also, I have got that $\mathcal{M}\subset\mathcal{U}$ (it is easy to check that $d_1(f,g)\leq d_\infty(f,g)$ ) and those topologies are not equivalent, and $\mathcal{U}\subset\mathcal{T}$. But, what happens with $\mathcal{M}$ and $\mathcal{T}$? Also, are topologies $\mathcal{T}$ and $\mathcal{U}$ equivalent?

Detailed explanations are welcome. Thanks in advance.

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  • $\begingroup$ Do you mean $$U(f,A,\epsilon) = \{g\in C[0,1] : |f(x) - g(x)| < \epsilon \quad\forall x \in A\}$$ $\endgroup$ – Prahlad Vaidyanathan Nov 12 '13 at 9:35
  • $\begingroup$ Yes, that was typo. $\endgroup$ – alans Nov 12 '13 at 9:36
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    $\begingroup$ What does compare precisely mean -- decide whether or not equivalent? $\endgroup$ – Rasmus Nov 12 '13 at 9:52
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A partial answer that should help you a little - Comparing $\mathcal{T}$ and $\mathcal{M}$ :

a) Consider $V:= U(0,\{1\}, 1/2) \in \mathcal{T}$, I claim that $V \notin \mathcal{M}$ :

Define $f \in V$ to be the constant function $f \equiv 1/4$. Now I claim that for any $\delta > 0$, there is a $g \in C[0,1]$ such that $$ d_1(f,g) < \delta \text{ and } g \notin V $$ For this you can think of a picture of a function $g$ such that $$ g(1) = 1, \text{ and } g(x) = 1/4\quad\forall x < 1-\delta $$ and $g$ describes a thin triangle between $x = 1-\delta$ and $x=1$. Then, $$ d_1(f,g) = \int_{1-\delta}^1 |g(x) - 1/4|dx \leq \int_{1-\delta}^1 (1-1/4)dx = \frac{3\delta}{4} < \delta $$ Hence, $d_1(f,g) < \delta$. However, $$ |g(1) - 0| = 1 > 1/2 \Rightarrow g \notin V $$

b) Consider $W := \{g \in C[0,1] : \int |g(x)|dx < 1\} \in \mathcal{M}$, I claim that $W \notin \mathcal{T}$ :

Choose $f \equiv 1/2$. For any finite subset $A \subset [0,1]$ and any $\delta > 0$, we can construct a function $g \in C[0,1]$ such that $$ g(x) = 1/2 \quad\forall x \in A, \text{ but } \int |g(x)|dx > 1 $$ In fact, just take the smallest $x_0 \in A$, and build a really large triangle from $(0,2)$ to $(x_0,1/2)$. Now let $g$ be the hypotenuse of that triangle for $x \leq x_0$ and $g(x) = 1/2$ for all $x > x_0$.

Hence, $\mathcal{T}$ and $\mathcal{M}$ are not related by inclusion.

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