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

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.

• An open set doesn't "belong to" a basis element, but (if nonempty) it contains a basis element. Jan 16, 2019 at 20:43
• Thank you! You're right, I've corrected that. Is there anything else wrong or is the proof fine now? Jan 16, 2019 at 20:46
• @MatheusAndrade Aha! Good work then. The distinction is a great but also terrible one since in some sense the box topology is the more obvious one. Such is life when all you can touch and feel is finite dimensions. Humans are awful at guessing at what should be right for infinite dimensions and this is a shining example of that. Jan 16, 2019 at 20:54
• @MatheusAndrade I think about these typologies as follows: If you consider $\mathbb{R}^\mathbb{N}$ as the set of all real sequences $a_n$. Then a sequence of sequences $(a_n)_m$ convergence to a sequence $(a_n)_0$ in the box topology if it is convergence pointwise for every $n\in\mathbb{N}$. While it is convergence in the box topology if it convergence uniformly for all $n\in\mathbb{N}$. (This has nothing to do with your question but it helps me understand these typologies better) Jan 16, 2019 at 21:03
• @Yanko No not uniform convergence, that is handled by the uniform-metric topology on that space which lies between the product and the box topology. There are uniformly convergent sequences that are not box convergent. Jan 16, 2019 at 21:50

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.