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

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.

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.