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Does R-omega satisfy the first countability axiom?

  1. Mar 28, 2005 #1
    Does R-omega satisfy the first countability axiom?
    (in the box topology)
     
    Last edited: Mar 28, 2005
  2. jcsd
  3. Mar 29, 2005 #2
    The short answer is: No.

    The long answer:
    Suppose [itex]\mathbb{R}^\omega[/itex] is first countable.
    Consider the set :

    [tex]\mathbb{R}^\omega_+ = \{ x \in \mathbb{R}^\omega \hspace{2 mm} \vert \hspace{2 mm} \pi_i (x) > 0 \hspace{2 mm} \forall \hspace{2 mm} i \in \mathbb{Z}^+ \}[/tex]

    Where [itex]\pi_i : \mathbb{R}^\omega \longrightarrow \mathbb{R}[/itex] is the i-th projection function.

    Clearly, the point [itex]\vec{0} = (0,0,0,\dots) \hspace{2 mm} \in \mathbb{\bar{R}}^\omega_+[/itex] ( closure )

    Now, by assumption, [itex]\mathbb{R}^\omega[/itex] is first-countable, so [itex]\vec{0}[/itex] has a countable local base, [itex] \{ B_i \}, i \in\mathbb{Z}_+[/itex]
    Define a new collection as follows:
    [tex]U_i = \bigcap^n_{k=1}B_k[/itex]
    Then construct a sequence as follows:
    For each i, choose [itex]x^i \in U_i \cap \mathbb{R}^\omega_+[/itex] ( i is a superscripted index, not an exponent )
    Because [itex]\vec{0} \in \mathbb{\bar{R}}^\omega_+[/itex], for any neighborhood [itex]U[/itex] of [itex]\vec{0}[/itex], [itex]U \cap \mathbb{R}^\omega_+ \neq \varnothing [/itex], so this process is well defined.
    Clearly, then, [itex]x^i \longrightarrow \vec{0}[/itex], as any neighborhood [itex]W[/itex] containing [itex]\vec{0}[/itex] must contain [itex]B_N[/itex] for some N ( definition of a local base ). But [itex]B_N \supset U_{N-1}\cap B_N = U_N [/itex] so that [itex]\forall i > N, x^i \subset U_N \subset W[/itex]. Therefore:
    [tex]x^i \rightarrow \vec{0}[/tex]

    However, writing [itex]x_i[/itex] as:
    [tex]x^i = ( x^i_1, x^i_2, x^i_3, \dots )[/tex]
    and for each i, letting:
    [tex]V_i = ( -x^i_i , x^i_i ) \subset \mathbb{R}[/tex] ( not tensor notation )
    we consider the set:
    [tex]V = V^1_1 \times V^2_2 \times V^3_3 \times ... \subset \mathbb{R}^\omega[/tex]
    ( note the similarity to Cantor's Diagonalization Argument for the uncountablilty of the reals )
    Now, we use box topology: V is the countable product of sets in the basis of [itex]\mathbb{R}[/itex], and therefore, is open in [itex]\mathbb{R}^\omega[/itex]
    It should be obvious that [itex]\forall i, x^i \notin V[/itex] ( they lie on the boundary of V ). Thus, V is a neighborhood of [itex]\vec{0}[/itex] disjoint from [itex]\{x^i\}[/itex]. Therefore:
    [tex]x^i \nrightarrow \vec{0}[/tex]
    Thus we have a contradiction, and there can be no countable local base for [itex]\vec{0} \in \mathbb{R}^\omega[/itex], so [itex]\mathbb{R}^\omega[/itex] cannot be first countable.

    -joeboo

    ( sorry this took so long, it was my first time ( ever! ) using LaTeX, so it took me a long while to write it out. I hope it's clear enough )
     
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