$V_{f,x_1,\cdots,x_n,\epsilon} = \{g\in X:|f(x_i)-g(x_i)|<\epsilon, i=1,\cdots,n\}$ is a topology

Question

Let $X=C([0,1],\mathbb{R})$. Given $f\in X$, $\epsilon >0$ and $x_1,\cdots,x_n\in [0,1]$ consider

$$V_{f,x_1,\cdots,x_n,\epsilon} = \{g\in X:|f(x_i)-g(x_i)|<\epsilon, i=1,\cdots,n\}$$

a) Verify that $\tau = \{U\subseteq X: \forall f\in U, \exists\epsilon>0, \exists x_1,\cdots, x_n\in[0,1]\mbox{ such that } V_{f,x_1,\cdots,x_n,\epsilon}\subseteq U\}$ define a topology in $X$. This topology is called the pontual convergence topology.

b) Show that any sequence of elements of $X$ converge in relation to $\tau$ $\iff$ it converges pontually

c) Show that $\tau$ is not 'metrizable' (cannot be measured with a metric, I guess)

I've found Proving the topology of such sets: $V_{f,x,\epsilon}=\{g \in E| \forall i \in \{1, \dots, N \}, |f(x_i)-g(x_i)| < \epsilon \}$ but I need to prove using the conditions below:

Let X be a set and let τ be a family of subsets of X. Then τ is called a topology on X if:

1. Both the empty set and X are elements of τ.
2. Any union of elements of τ is an element of τ.
3. Any intersection of finitely many elements of τ is an element of τ.

Let's check condition 1.

The empty set is an open $U$ of $X$ because we cannot show that there is a function $g$ in the empty set such that the condition $|f(x_i)-g(x_i)|<\epsilon$.

Is $X$ a set with the property tha for all functions on $X$, there exists epsilon and $x_1,\cdots,x_n$ such that $V_{f,x_1,\cdots,x_n,\epsilon}\subseteq X$? Well, $X$ is the set of all continuous functions from $[0,1]$ to $\mathbb{R}$. Since there's nothing about norms here, I'd use some topological definition of continuity, but which one?

Now, how do I show any union of elements of $\tau$ is in $\tau$?

Let's see, suppose $U,V\in \tau$. What does it mean to take the union of this?

I'm lost.

Answer 1

Why lost? Simply take the union, it doesn't "mean" that much, but note that we need to check that any union is contained in $\tau$. Well, let $(U_i)_{i\in I}\subset\tau$ be a collection of open sets. Set $U=\bigcup_{i}U_i$. We check our definition; Let $f\in U$. We would like to find $\varepsilon>0$ and $x_1,\dots x_n\in[0,1]$ so that $V_{f,\varepsilon, x_1,\dots, x_n}\subset U$. Since $f\in U$, there exists some $i\in I$ such that $f\in U_i$. But $U_i\in\tau$, hence there exist $\varepsilon>0, x_1,\dots, x_n\in[0,1]$ so that $V_{f,\varepsilon, x_1,\dots, x_n}\subset U_i$; since $U_i\subset U$, we have that $V_{f,\varepsilon, x_1,\dots, x_n}\subset U$, as desired.

Now for finite intersections, it suffices to check that if $U_1,U_2\in\tau$ then $U_1\cap U_2\in\tau$. Set $A=U_1\cap U_2$ and let $f\in A$. Therefore $f\in U_1, f\in U_2$ and we can find $\varepsilon_1>0, \varepsilon_2>0$ and $x_1,\dots,x_n, y_1,\dots, y_m\in [0,1]$ such that $V_{f,\varepsilon_1,x_1,\dots x_n}\subset U_1$ and $V_{f,\varepsilon_2, y_1,\dots, y_m}\subset U_2$. Now let's take $\varepsilon=\min\{\varepsilon_1,\varepsilon_2\}$. Then we can easily check that $V_{f,\varepsilon, x_1,\dots, x_n, y_1,\dots y_m}\subset V_{f,\varepsilon_1, x_1,\dots, x_n}$ and that $V_{f,\varepsilon, x_1,\dots, x_n, y_1,\dots y_m}\subset V_{f,\varepsilon_2, y_1,\dots,y_m}$, hence $V_{f,\varepsilon, x_1,\dots, x_n, y_1,\dots y_m}\subset U_1\cap U_2$ and we are done.