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 $\mathbb{R}^\omega$ 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:
[tex]U_i = \bigcap^n_{k=1}B_k[/itex]<br /> Then construct a sequence as follows:<br /> For each <i>i</i>, choose $x^i \in U_i \cap \mathbb{R}^\omega_+$ ( i is a superscripted index, not an exponent )<br /> 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.<br /> 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:<br /> $$x^i \rightarrow \vec{0}$$<br /> <br /> However, writing $x_i$ as:<br /> $$x^i = ( x^i_1, x^i_2, x^i_3, \dots )$$<br /> and for each <i>i</i>, letting:<br /> $$V_i = ( -x^i_i , x^i_i ) \subset \mathbb{R}$$ ( not tensor notation )<br /> we consider the set:<br /> $$V = V^1_1 \times V^2_2 \times V^3_3 \times ... \subset \mathbb{R}^\omega$$<br /> ( note the similarity to Cantor's Diagonalization Argument for the uncountablilty of the reals )<br /> Now, we use box topology: V is the countable product of sets in the basis of $\mathbb{R}$, and therefore, is open in $\mathbb{R}^\omega$<br /> 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:<br /> $$x^i \nrightarrow \vec{0}$$<br /> 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.<br /> <br /> -joeboo<br /> <br /> ( 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 )[/tex]

 

[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.