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Math Help - Is it a base for a Topology in R^N

  1. #1
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    Is it a base for a Topology in R^N

    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
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  2. #2
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    Re: Is it a base for a Topology in R^N

    What definition of "base" for a topology are you using?
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  3. #3
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    Re: Is it a base for a Topology in R^N

    Quote Originally Posted by huberscher View Post
    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]~?
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  4. #4
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    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)= \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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