Is it a base for a Topology in R^N

Hi there

Prove this statement or the opposite:

Consider the set $\displaystyle \mathbb{R}^{\mathbb{N}} := \{ (x_n)_{n \in \mathbb{N}} : x_n \in \mathbb{R} \forall n \in \mathbb{N} \} $ . Define $\displaystyle U_{y_1,...,y_m}(r):= \{ (x_n)_n \in \mathbb{R}^{\mathbb{N}} : |x_i - y_i| < r \text{ for every } i=1,...,m \} $ .

The collection $\displaystyle B:= \{ U_{y_1,...,y_m}(r) : y_1 ,...,y_m \in \mathbb{R}, m \in \mathbb{N},r>0 \}$ is a base for a topology on $\displaystyle \mathbb{R}^\mathbb{N} $ .

Is this true or not? Why? I'm sorry but I don't understand a lot after reading this multiple times?? How do I show that it is a base of a topology and which one?

Could maybe please someone give me a hint?

Regards

Re: Is it a base for a Topology in R^N

What definition of "base" for a topology are you using?

Re: Is it a base for a Topology in R^N

Quote:

Originally Posted by

**huberscher** Prove this statement or the opposite:

Consider the set $\displaystyle \mathbb{R}^{\mathbb{N}} := \{ (x_n)_{n \in \mathbb{N}} : x_n \in \mathbb{R} \forall n \in \mathbb{N} \} $ . Define $\displaystyle U_{y_1,...,y_m}(r):= \{ (x_n)_n \in \mathbb{R}^{\mathbb{N}} : |x_i - y_i| < r \text{ for every } i=1,...,m \} $ .

The collection $\displaystyle B:= \{ U_{y_1,...,y_m}(r) : y_1 ,...,y_m \in \mathbb{R}, m \in \mathbb{N},r>0 \}$ is a base for a topology on $\displaystyle \mathbb{R}^\mathbb{N} $ .

It is fairly clear that $\displaystyle \bigcup B = \mathbb{R}^\mathbb{N} $.

Can you show that $\displaystyle S \in B\;\& \,T \in B \Rightarrow \left( {\exists U \in B} \right)\left[ {U \subseteq S \cap T} \right]~?$

Re: Is it a base for a Topology in R^N

Thank you.

So I have to differ between the disjoint and the non-disjoint case, don't I ?

disjoint case:

For r=0 U(0)=$\displaystyle \emptyset $ and so it is true that $\displaystyle U \subset S \cap T$ .

Hmm, I am not convinced of this argument because one could always head this statement for S, T arbitrary. There was always $\displaystyle \emptyset \subset S \cap T$ ?

non-disjoint case:

$\displaystyle U_S := \{ (x_n)_n \in \mathbb{R}^\mathbb{N} : | x_i-s_i | < r_s \}$

$\displaystyle U_T := \{ (x_n)_n \in \mathbb{R}^\mathbb{N} : | x_i-t_i | < r_t \}$

Then

$\displaystyle U_U := \{ (x_n)_n \in \mathbb{R}^\mathbb{N} : | x_i-u_i | < \min\{ |u_i-s_i|,|u_i-t_i|.| r_s-|u_i-s_i| |,|r_i-|u_i-t_i|| \} \}$ fullfills the condition above.

So $\displaystyle |x_i-u_i|<|u_i-t_i|<r_t$ since $\displaystyle (u_n)_n \in U_T$ ??

So $\displaystyle |x_i-u_i|<|u_i-s_i|<r_s$ since $\displaystyle (u_n)_n \in U_s$ ??

What about the other parameters used by "min". They make sure that the whole Ball is contained in $\displaystyle S \cap T $. How can I show formally that they are needed and make sure that the statement is true?

So the topology is $\displaystyle \mathbb{R}^\mathbb{N}$

Obviously the family of balls

$\displaystyle \{ U_{(\frac{n}{m},\frac{n}{m},\frac{n}{m},...)}( \epsilon) \}_{m \in \mathbb{N}, n \in \mathbb{Z}} $ where $\displaystyle \epsilon>0 $ covers $\displaystyle \mathbb{R}^\mathbb{N}$

What do you think?

Regards