Topology on $C[0,1]$ Let $\mathbb{R}^I$ be the set of all real continous functions in $I=[0,1]$. 
(a) For each $f \in \mathbb{R}^I$, each finite set $F \subset I$ and each $\delta > 0$, consider the set $$U(f,F,\delta)=\{g \in \mathbb{R}^I: |f(x)-g(x)|<\delta, \text{for each } x \in F\}$$
Show that the collection of sets $U(f,F,\delta)$ determines a topology in $\mathbb{R}^I$, such that for each $f \in \mathbb{R}^I$ the collection of the sets of the form $U(f,F,\delta)$ is a local basis for $f$
(b) Determine $\overline{\{f\}}$ for the topology of (a)
I will show that $\{U(f,F,\delta)\}$ is a basis for a topology. To do this I first show that it covers the space, since $g \in U(g,F,\delta)$, for any $g$. Is this right?
Then, I have to show that  if $g \in U(f_1,F_1,\delta_1) \cap U(f_2,F_2,\delta_2)$, then there is some basic $U(f,F,\delta)$ such that $g \in U(f,F,\delta) \subset U(f_1,F_1,\delta_1) \cap U(f_2,F_2,\delta_2)$. I have trouble proving this, since I am unable to find the function $f$. I know that $g$ is very near to $f_1$ and to $f_2$ in some finite sets $F_1,F_2$, but who can be $f$? I think it may be the same $g$, but it seems a little suspicious.
Now, for (b) my heart says "it's $\{f\}$, it's $\{f\}$..." but I really can't picture it in my mind. The clousure is $\{f\} \cup \{f\}'$, and $\{f\}'$ is the set of all the fuctions such that every neighborhood has non-empty intersection with $\{f\}$. But what is that? That definition is hard to see in a space of functions. It could also be a bunch of mysterious functions.
 A: (a) Use $g$ as $f$. That is, $g$ is certainly an element of $U(g,F,\delta)$ for any finite $F$ and any $\delta >0$, so we just need to find suitable $F$ and $\delta$ so that $U(g,F,\delta)\subset U(f_1,F_1,\delta_1) \cap U(f_2,F_2,\delta_2)$. 
Since $g\in U(f_1,F_1,\delta_1)$ we have that $|f_1(x)-g(x)|<\delta_1$, or in other words, $\mu_1(x)=\delta_1-|f_1(x)-g(x)|>0$ for each $x\in F_1$. So if we define $\mu_1=\min\{\mu_1(x):x\in F_1\}$, then $\mu_1>0$. 
Similarly if $\mu_2(x)=\delta_2-|f_2(x)-g(x)|$ then $\mu_2:=\min\{\mu_2(x):x\in F_2\}>0$. 
Let $F=F_1\cup F_2$ and $\delta = \min\{\mu_1,\mu_2\}$. 
Pick any $h\in U(g,F,\delta)$ we will show that $h\in U(f_1,F_1,\delta_1)$. Indeed, if $x\in F_1$ then $|f_1(x)-h(x)|\le|f_1(x)-g(x)| + |g(x)-h(x)| < |f_1(x)-g(x)| + \mu_1(x) = \delta_1$, hence we have that $h\in U(f_1,F_1,\delta_1)$. Similarly $h\in U(f_2,F_2,\delta_2)$, that is 
$h\in U(f_1,F_1,\delta_1)\cap U(f_1,F_1,\delta_1)$. This proves that $U(g,F,\delta)\subseteq U(f_1,F_1,\delta_1)\cap U(f_2,F_2,\delta_2)$. 
(b) To show that $\overline{\{f\}} = \{f\}$ it is enough to show that the complement of $\{f\}$ is open. Indeed, take any $h\in\Bbb R^I\setminus\{f\}$. Then there is some $x_0$ such that $f(x_0)\not=h(x_0)$. Let $F=\{x_0\}$ and $\delta=\dfrac{|f(x_0)-h(x_0)|}2$. Then $U(h,F,\delta)$ is a neighborhood of $h$ contained in the complement of $f$, which shows that the complement of $f$ is open, and hence $\{f\}$ is closed. 
A: For any finite set $F$, put $\rho_F(f) = \max_{F} |f|.$  This is a seminorm on your space.  This family $\{\rho_{F}: F \text{finite}\}$ induces a locally convex topology on the space.  
