Box topology not first countable (constructive proof) Let $(\mathbb{R}^\omega,\tau)$ be the box topology space. I have found that everyone proofs that this topology is not first countable by finding a contradiction, but I thought of a constructive way. Would appreciate comments if I'm missing something. Goes like this:
Let $x\in\mathbb{R}^\omega$ and let $B(x)$ be a local basis for $x$. Then take $a_{i\varepsilon}=x_i-\varepsilon$ and $b_{i\varepsilon}=x_i+\varepsilon$, and let
$$
I_\varepsilon=\prod_{i\in\mathbb{N}}\,
    (a_{i\varepsilon},b_{i\varepsilon})
$$
So, in particular, we get that $\forall\varepsilon\in (0,1),\,\,$ $I_\varepsilon$ are neighbourhoods of $x$. So $\forall\varepsilon\in (0,1)$ $\exists B_\varepsilon\in B(x)$ s.t. $x\in B_\varepsilon\subseteq I_\varepsilon$. Thus there is at least a continnum of elements in $B(x)$, i.e., $|B(x)|\ge c$ and thereby $\mathbb{R}^\omega$ is not first countable.
 A: Can you make the same argument for $\mathbb{R}$? $\mathbb{R}$ is first countable, of course. 
The point is you can say essentially the same words, so there is a mistake in your argument.
Hint: 

 You implicitly claim a map is injective, but it isn't.

A: Your argument is incorrect as shown by AreaMan's answer.
Identify $\mathbb R^{\omega}$ with the set of $f:\omega \to \mathbb R.$ A local base at $f$ is $B_f=\{B(f,g)| g:\omega \to \mathbb R^+\}$ where $B(f,g)=\{h\in \mathbb R^{\omega}\;|\;\forall n \in \omega\;(\;|f(n)-h(n)|<g(n)\;)\}.$
We can show that any local base $C_f$ at $f$ is uncountable as follows: For each $c\in C_f$ let $c\supset B(f,g_c)\in B_f.$ Now if $\{c(j)\;|\;j\in\omega\}\subset C,$ then for $n\in \omega$ let $G(n)=\frac {1}{2}\min \{g_{c(j)}(n)\;|\;j\leq n\}.$
No member of $\{c(n):n\in \omega\}$ is a subset of $B(f,G)$ because for $n\in \omega$ we have $$f+\frac {3}{4}g_{c(n)}\in c(n) \backslash B(f,G)$$.....because $f+\frac {3}{4}g_{c(n)}\in B(f,g_{c(n)})\subset c(n),$ but $\frac {3}{4}g_{c(n)}(n)>G(n).$ 
So no countable subset of $C$ is a local base at $f.$
