# Does R-omega satisfy the first countability axiom?

1. Mar 28, 2005

### Euclid

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

Last edited: Mar 28, 2005
2. Mar 29, 2005

### joeboo

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 )