Does R-omega satisfy the first countability axiom?

[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 [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 )

 

[M98Ranger]

Yes, R-omega satisfies the first countability axiom in the box topology. This is because for any point x in R-omega, we can construct a countable local basis at x, which is a collection of open sets that contain x and whose intersections with any open set containing x form a basis for the topology at x. In the case of R-omega, we can take the collection of open intervals centered at x with rational endpoints as our local basis. This is a countable collection and any open set containing x will contain one of these intervals, satisfying the first countability axiom.