Is it a base for a Topology in R^N

**huberscher** · November 9th 2012, 04:57 AM

Hi there

Prove this statement or the opposite:

Consider the set $\mathbb{R}^{\mathbb{N}} := \{ (x_n)_{n \in \mathbb{N}} : x_n \in \mathbb{R} \forall n \in \mathbb{N} \}$ . Define $U_{y_1,...,y_m}(r):= \{ (x_n)_n \in \mathbb{R}^{\mathbb{N}} : |x_i - y_i| < r \text{ for every } i=1,...,m \}$ .
The collection $B:= \{ U_{y_1,...,y_m}(r) : y_1 ,...,y_m \in \mathbb{R}, m \in \mathbb{N},r>0 \}$ is a base for a topology on $\mathbb{R}^\mathbb{N}$ .

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

**HallsofIvy** · November 9th 2012, 05:41 AM

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

**Plato** · November 9th 2012, 07:32 AM

Originally Posted by huberscher

Prove this statement or the opposite:
Consider the set $\mathbb{R}^{\mathbb{N}} := \{ (x_n)_{n \in \mathbb{N}} : x_n \in \mathbb{R} \forall n \in \mathbb{N} \}$ . Define $U_{y_1,...,y_m}(r):= \{ (x_n)_n \in \mathbb{R}^{\mathbb{N}} : |x_i - y_i| < r \text{ for every } i=1,...,m \}$ .
The collection $B:= \{ U_{y_1,...,y_m}(r) : y_1 ,...,y_m \in \mathbb{R}, m \in \mathbb{N},r>0 \}$ is a base for a topology on $\mathbb{R}^\mathbb{N}$ .

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

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

**huberscher** · November 11th 2012, 03:35 AM

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)= $\emptyset$ and so it is true that $U \subset S \cap T$ .
Hmm, I am not convinced of this argument because one could always head this statement for S, T arbitrary. There was always $\emptyset \subset S \cap T$ ?

non-disjoint case:
$U_S := \{ (x_n)_n \in \mathbb{R}^\mathbb{N} : | x_i-s_i | < r_s \}$
$U_T := \{ (x_n)_n \in \mathbb{R}^\mathbb{N} : | x_i-t_i | < r_t \}$

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

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

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

So the topology is $\mathbb{R}^\mathbb{N}$
Obviously the family of balls
$\{ U_{(\frac{n}{m},\frac{n}{m},\frac{n}{m},...)}( \epsilon) \}_{m \in \mathbb{N}, n \in \mathbb{Z}}$ where $\epsilon>0$ covers $\mathbb{R}^\mathbb{N}$

What do you think?

Regards

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