Proving $(-1,1)^{\mathbb{N}}$ is not open in the product topology of $\mathbb{R}^{\mathbb{N}}$

Question

Clarification: here $\mathbb{R}^{\mathbb{N}} = \mathbb{R}\times \mathbb{R} \times \cdots$, i.e, countably many copies of $\mathbb{R}$. $(-1,1)^{\mathbb{N}}$ is completely analagous.

I don't want a proof using sequences, so here's what I've done:

Suppose $(-1,1)^{\mathbb{N}}$ is open in the product topology. Then it contains a basis element of the product topology. Since any basis element of the product topology in this case is of the form $B = \displaystyle{\prod_{\alpha \in J}} U_\alpha$, where $U_\alpha = \mathbb{R}$ for all but finitely many values of $\alpha$, we have a contradiction, since obviously there are elements of $\mathbb{R}$ that aren't coordinates of $(-1,1)^{\mathbb{N}}$ (and, by definition, for a set to be open in the product topology, all of it's coordinates have to be in all of the $U_\alpha$). Therefore $(-1,1)^{\mathbb{N}}$ doesn't contain any element of the product topology and it follows that it's not open.

Now, have I done anything wrong here or made myself unclear? How could I improve what I wrote? Is it alright? I appreciate any help.

Answer 1

The idea is correct as you wrote it: suppose that $$B=\prod_{n \in \mathbb{N}} U_n \subseteq (-1,1)^{\mathbb{N}}$$ for some non-empty basic open subset $B$ so that there is a finite subset $F$ of $\mathbb{N}$ such that $U_n = \mathbb{R}$ for all $n \notin F$, and all other $U_n$ non-empty open subsets of $\mathbb{R}$.

Then the point $(p_n)$ defined by picking $p_n\in U_n$ for $n\in F$ and setting $p_n = 2$ for $n \notin F$, obeys $p \in B$ by construction, but $p \notin (-1,1)^{\mathbb{N}}$ as witnessed by any coordinate $n \notin F$ $(p \in (-1,1)^{\mathbb{N}}$ iff $\forall n : -1 < p_n < 1$, of course), contradicting the supposed inclusion. So the interior of $(-1,1)^{\mathbb{N}}$ is empty.