I was given the following problem and I'm struggling to work my way through it. Any help would be appreciated. On each of these I am supposed to show that this is either true or false.
Let $X = \bigcup\limits_{n\in\mathbb{N}} \mathbb{N}^n$. Then for $a\in \mathbb{N}^n$ and $F\subseteq \mathbb N ^{n+1}$ where $F$ is finite, let $U_a(F) = \{z\in X : \pi_{[1,n]}(z)=a$ and $\pi_{[1,n+1]}(z)\notin F\}$, where $\pi_{[1,n]}(z)$ is the projection of the vector $z$ on to the first $n$ coordinates; e.g. $\pi_{[1,3]}(\left<1,-2,1,4\right>) = \left<1,-2,1\right>$.
(1) Show that $B=\{ U_a(F): \exists n\in\mathbb N$ such that $a\in \mathbb N^n$ and $F\subseteq \mathbb N^{n+1}$ where $|F|\in \{0,1,2,...\}\}$ is a basis for a topology on $X$.
(2) Show that each $\mathbb N^n$ is nowhere dense in $X$.
(3) Show $cl(\mathbb N^n) = \bigcup\limits_{i\le n} \mathbb N^i$.
(4) Show if $A_n = \mathbb N^{2n}$ then $\bigcup\limits_{n\in\mathbb N} A_n$ has a void interior while $\bigcup\limits_{n\in\mathbb N} cl(A_n) = X$.
I know that for (1) I need to let $x\in X$ and then show that $B$ contains some element $u$ with $x\in u$. I also would need to show that if $u,v\in B$ and $x\in u\cap v$ then $\exists w\in B$ such that $x\in w\subseteq u\cap v$. For starters, I let $x\in X$, so then I know that $x\in \mathbb N^n$ for some $n$. However, I'm honestly not quite sure how to get started beyond this.
I think I've solved 2 and 3 (I solved 3 independently of 2 and used 3 to solve 2.) Can anyone confirm if this is correct or if I'm making a mistake?
(2) We need to show that each $\mathbb{N}^n$ is nowhere dense in $X$, so we show that the closure of $\mathbb{N}^n$ has an empty interior for each $n$. We have from below (proved independently from this problem's result, so it can be assumed) that $cl(\mathbb{N}^n) = \bigcup\limits_{i\le n} \mathbb{N}^i$. We note that the interior of a set $S$ is defined as the complement of the closure of the complement of $S$. We let $S=cl(\mathbb{N}^n)$ and want to show $S$ has an empty interior.
We have $cl(\mathbb{N}^n) = \bigcup\limits_{i\le n} \mathbb{N}^i$, so the complement $cl(\mathbb{N}^n)^c = \bigcup\limits_{i> n}\mathbb{N}^i$ because $X= \bigcup\limits_{n\in\mathbb{N}} \mathbb{N}^n$. \Then the closure of the complement will be $\bigcup\limits_{n\in\mathbb{N}} \mathbb{N}^n$ because of the problem below: the closure of $\mathbb{N}^i$ is the union of the lower dimension product spaces of the natural numbers, so the union of the closures for at infinitely increasing $i$ will cover all of $X=\bigcup\limits_{n\in N}\mathbb{N}^n$. \Then the complement of the closure of the complement is simply the empty set, because the closure of the complement was the entire space $X$. Thus the interior of $S$, which is the complement of the closure of the complement of $S$ is empty Therefore $\mathbb{N}^n$ is nowhere dense in $X$ for all $n\in\mathbb{N}$.
(3) We prove by induction.
First, we have that the limit points of $\mathbb{N}$ are all elements of $\mathbb{N}$ because any sequence $a_k$ converges to $n\in\mathbb{N}$ if and only if $\exists j\in\mathbb{N}$ such that $\forall k>j,$ we have $a_k = n$. So $cl(\mathbb{N}^1) = \bigcup\limits_{i\le 1} \mathbb{N}^i$.
Second, assume $cl(\mathbb{N}^n) = \bigcup\limits_{i\le n} \mathbb{N}^i$.\ We have that $cl(\mathbb{N}^n) \subseteq cl(\mathbb{N}^{n+1})$ because any limit point $v=\left< v_1, v_2, \cdots, v_n\right>$ of $\mathbb{N}^n$ must also be a limit point of $\mathbb{N}^{n+1}$, since the sequence $u^{(k)} = \left< v_1, v_2,\cdots, v_n, k\right>$ converges to $v$ and each $u^{(k)} \in\mathbb{N}^{n+1}$. Therefore we have $\bigcup\limits_{i\le n} \mathbb{N}^i \subseteq cl(\mathbb{N}^{n+1})$.
Further, let $v=\left< v_1, v_2, \cdots, v_{n+1}\right>\in \mathbb{N}^{n+1}$. Then the sequence $v^{(k)} = v$ for all $k$ converges trivially to $v$. Thus $v$ is a limit point of the set as well, so $v\in cl(\mathbb{N}^{n+1})$. Thus, because $v$ was arbitrary, we have that $\mathbb{N}^{n+1}\subseteq cl(\mathbb{N}^{n+1})$.
Therefore, we conclude from these things that $$cl(\mathbb{N}^{n+1}) = \mathbb{N}^{n+1} \cup \bigcup\limits_{i\le n} \mathbb{N}^i=\bigcup\limits_{i\le n+1} \mathbb{N}^i.$$
Thus we conclude by the principle of mathematical induction that $$cl(\mathbb{N}^n) = \bigcup\limits_{i\le n} \mathbb{N}^i.$$
EDIT: I have also solved part of #4, showing that the union of closures of $A_N$ is the topological space.
Let $A_n = \mathbb{N}^{2n}$.
We show that $\bigcup\limits_{n\in\mathbb{N}} cl(A_n) = X$. We have from above that $cl(\mathbb{N}^n) = \bigcup\limits_{i\le n} \mathbb{N}^i$, so we can write $\bigcup\limits_{n\in\mathbb{N}} cl(A_n) = \bigcup\limits_{n\in\mathbb{N}} cl(\mathbb{N}^{2n}) = \bigcup\limits_{n\in\mathbb{N}} \left( \bigcup\limits_{i\le 2n} \mathbb{N}^i\right)$.
We let $x\in \bigcup\limits_{n\in\mathbb{N}} cl(A_n)$. Then from above it follows that $x\in \bigcup\limits_{i\le 2n} \mathbb{N}^i$ for some $n\in\mathbb{N}$. Then from this it follows that $x\in \mathbb{N}^i$ for some $i\in\mathbb{N}$. Thus it follows that $x\in \bigcup\limits_{n\in\mathbb{N}} \mathbb{N}^n = \mathbb{X}$. Therefore $\bigcup\limits_{n\in\mathbb{N}} cl(A_n) \subseteq \mathbb{X}$.
\We now let $y\in \mathbb{X}=\bigcup\limits_{n\in\mathbb{N}}\mathbb{N}^n$, then $y\in \mathbb{N}^k$ for some $k$. Then it follows that $y \in \bigcup\limits_{i\le 2k} \mathbb{N}^i$. Thus $y\in cl(\mathbb{N}^{2k})=cl(A_{2k})$. Therefore it follows that $y\in \bigcup\limits_{n\in\mathbb{N}} cl(A_n)$. Therefore $\mathbb{X}\subseteq \bigcup\limits_{n\in\mathbb{N}} cl(A_n)$.
\noindent Thus we conclude that $\mathbb{X} = \bigcup\limits_{n\in\mathbb{N}}