Is the sequence of real number that are $0$ from some point Sense in the box or product topologies

Question

Let $A=\{(x_n)_{n\in\mathbb{N}}\in\mathbb{R}^\mathbb{N}|\exists M\in\mathbb{N} ,\forall n>M, x_n=0 \}\subset\mathbb{R}^\mathbb{N}$, series of real numbers that are zero from some point forward.

Let $X$ be $\mathbb{R}^\mathbb{N}$ with the Product Topology and $Y$ be $\mathbb{R}^\mathbb{N}$ with the Box Topology.

Is A Dense in X? In Y?

I'm trying to get some "feel" to the product and box topologies, So I will value any intuition you can give me to solve question such as this.

I don't know how to prove it but I think it is dense in the Product topology, since it is of the same "nature" as of the open sets of the product topology: From some $n$ the open subsets are not restricted thus they can be $0$ as $A$ series demands.

On the other hand in the Box topology I can limit the value of each coordinate to the interval $(1,2)$ for example, thus it can't ever be $0$.

Is my intuition correct? Will you please assist me with a rigorous proof?

Answer 1

Yes, your intuition is correct.

Box topology: The set $(1,2)\times(1,2)\times(1,2)\times\cdots$ is an open set which has no element of $A$. Therefore, $A$ is not dense.

Product topology: If $A_1,A_2,\ldots,A_n$ are non-empty open subsets of $\mathbb R$, then$$A_1\times A_2\times\cdots\times A_n\times\mathbb{R}\times\mathbb{R}\times\cdots$$clearly contains elements of $A$. And since every non-empty open subset of $\mathbb{R}^{\mathbb N}$ can be obtained as an union of such sets, every non-empty open subset of $\mathbb{R}^{\mathbb N}$ contains elements of $A$. Therefore, $A$ is dense.

Answer 2

Yes, your intuition is correct: $A$ is product-dense and box-nowhere-dense:

The set $A$ is indeed dense in the product topology on $\mathbb{R}^\mathbb{N}$: let $O$ be a non-empty basic open subset of the product so

$O = \prod_n O_n$ where we have a finite subset $F \subseteq \mathbb{N}$ such that $O_n = \mathbb{R}$ for $n \notin F$ and $O_n$ is some non-empty open subset of the reals for all $n \in F$. Pick some $a_n \in O_n$ for those finitely many $n$ and set $a_n= 0$ for $n \notin F$, then $(a_n)_n \in A \cap O$ and so $A$ is dense.

In the box topology, the set $A$ is even closed (And as it's not equal to the whole space, it's certainly not dense, it even has empty interior:)

Let $(x_n) \notin A$. This means that there is an infinite sequence of indices $N \subseteq \mathbb{N}$ such that $x_n \neq 0$. For each of those $n$ define $O_n = B(x_n, |x_n|)$ which is an open subset of $\mathbb{R}$ that does not contain $0$ and set $O_n = \mathbb{R}$ for all other $n$. Then for any $(y_n)_n \in O:= \prod_n O_n$, which is box-open, we have that $y_n \neq 0$ for all $n \in N$ too, and so $y \notin A$. So $O$ shows that $(x_n)$ is an interior point of $X\setminus A$ and so $A$ is closed.

If $(a_n)_n \in A$ and if there would be a box open $(a_n)_n \in O=\prod_n O_n$, then for all $n \in \mathbb{N}$ we can pick $a'_n \in O_n$ such that $a'_n \neq 0$ (because $a_n \in O_n$ and $O_n$ is open, if already $a_n \neq 0$ we just use $a'_n = a_n$ otherwise there is some interval $(-r,r)\subseteq O_n$ and we have plenty of points to choose). Then $(a'_n)_n \in O$ and $a'_n \notin A$. So $A$ contains no non-empty open subset at all, it's so-called nowhere dense ( $\operatorname{int}(\overline{A}) = \emptyset$ ).