R-omega

[Euclid]

Does R-omega satisfy the first countability axiom?
(in the box topology)

[joeboo]

Does R-omega satisfy the first countability axiom?
(in the box topology)
The short answer is: No.

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

\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}^+ \}

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

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

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

However, writing x_i as:
x^i = ( x^i_1, x^i_2, x^i_3, \dots )
and for each i, letting:
V_i = ( -x^i_i , x^i_i ) \subset \mathbb{R} ( not tensor notation )
we consider the set:
V = V^1_1 \times V^2_2 \times V^3_3 \times ... \subset \mathbb{R}^\omega
( 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 \mathbb{R}, and therefore, is open in \mathbb{R}^\omega
It should be obvious that \forall i, x^i \notin V ( they lie on the boundary of V ). Thus, V is a neighborhood of \vec{0} disjoint from \{x^i\}. Therefore:
x^i \nrightarrow \vec{0}
Thus we have a contradiction, and there can be no countable local base for \vec{0} \in \mathbb{R}^\omega, so \mathbb{R}^\omega 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 )