Proving the topology of such sets: $V_{f,x,\epsilon}=\{g \in E| \forall i \in \{1, \dots, N \}, |f(x_i)-g(x_i)| < \epsilon \}$

Question

Let $E$ be the vector space of applications from $[0,1]$ to $\mathbb{R}$. If $f \in E, N \in \mathbb{N}^*, x = (x_1, \dots x_N)\in [0,1]^N \text{ and } \epsilon = (\epsilon _1, \dots, \epsilon _N) \in (\mathbb{R}^*_+)^N$, we define:$$V_{f,x,\epsilon}=\{g \in E| \forall i \in \{1, \dots, N \}, |f(x_i)-g(x_i)| < \epsilon_i \}$$ We define $\Theta $ as the set of unions of the sets defined previously. Show that $\Theta $ is a topology on $E$.

So in order to show that $\Theta$ is a topology on $E$, I gotta show that it's non empty, that any union of its elements is an element of $\Theta$ and that any finite intersection of $\Theta$ elements is an element of $\Theta$.

I am interested in showing the last $2$ points. In order to show that any union is an element of $\Theta$ I have to construct a set that would equal to the union, but I struggle to do it in a simple situation when we work with only two sets.

Suppose that we have $V_{f,x,e}$ and $V_{f',x',e',}$, I need to construct a set $V_{g,y,\delta}$ such that $V_{f,x,e} \cup V_{f',x',e'} = V_{g,y,\delta}$ or such that $V_{f,x,e} \cup V_{f',x',e'} \subset V_{h,y,\delta}$. I would first define $\delta$ as $\delta _i = \text{max}\{e_i, e_i'\}$, and then $y$ which is where I start to struggle. And then I would need to construct the $g$ functions.

And I don't have more luck with the intersection point. Can anyone show me how to proceed?

Answer 1

You want $\Theta$ to be a topology. This means that you must show that the set $\mathcal{B}$ of all $V_{f,x,\epsilon}$ must be a base for a topology.

This means that you need to show that $\cup \mathcal{B} =E$ (which is clear as $f \in V_{f,x,\epsilon}$) and most importantly:

If $f \in V_{f,x_1,\epsilon_1} \cap V_{f_2,x_2,\epsilon_2}$ there exists $V_{f_3,x_3,\epsilon_3}$ such that $$f \in V_{f_3,x_3,\epsilon_3} \subseteq V_{f,x_1,\epsilon_1} \cap V_{f_2,x_2,\epsilon_2}$$

This is not too hard, try it. Hint: we can take $f_3 =f$, $x_3$ the "union" of $x_1$ and $x_2$ and be careful on common points of $x_1$ and $x_2$, taking some minima to define $\epsilon_3$, where appropriate.