The Zariski density for two given sets. Let $A$ and $B$ be two subsets of $\mathbb{C}^n$：
$ A = \mathbb{Z}^n$, and $B=\{ (z_1,z_2, \dots , z_n) \in A \text{ such that } z_1>z_2>\cdots> z_n\}$. 
My questions：
Are these two subsets Zariski dense in $\mathbb{C}^n$? Thanks very much!
 A: In the following I don't distinguish between polynomials an their associated functions ${\mathbb C}^n\to{\mathbb C}$ - that's ok since we're in characteristic $0$.

Claim: Suppose $f\in{\mathbb C}[x_1,...,x_n]$ vanishes on $A$. Then $f=0$.

Proof: Write $f = \sum_i g_i x_n^i$ with $g_i\in{\mathbb C}[x_1,...,x_{n-1}]$. Then, given any $a_1,...,a_{n-1}\in{\mathbb Z}$, $f(a_1,...,a_{n-1})\in{\mathbb C}[x_n]$ is a polynomial with infinitely many zeros, hence the zero polynomial. Therefore $g_i(a_1,...,a_{n-1})=0$ for all $a_1,...,a_{n-1}\in{\mathbb Z}$, and by induction, each $g_i$ is the zero polynomial.

Claim: Suppose $f\in{\mathbb C}[x_1,...,x_n]$ vanishes on $B$. Then $f=0$.

Proof: Consider $\varphi: {\mathbb C}^n\to{\mathbb C}^n$ defined by $$\varphi(z_1,...,z_n) := (z_1,z_1-z_2^2-1,z_1-z_2^2-z_3^2-2,...,z_1-z_2^2-...-z_n^2-n+1).$$ Then $\varphi$ is surjective and maps $A$ into $B$. Hence, by what we have just seen, $f\circ\varphi$ is a polynomial function vanishing on $A$, hence $f\circ\varphi=0$, hence $f=0$ by surjectivity of $\varphi$.
A: Yes, they are. 
Some general facts that will take care of the answer: 
A set $A$ of form
$A= A_1 \times A_2 \times \ldots \times A_n$ is Zariski dense if all of the $A_i$ are infinite. That takes care of the $\mathbb{Z}^n$. 
A nonvoid set  open in the standard topology is Zariski dense. This is because it contains an open cube $S \times S \times \ldots \times S$ and you apply the previous result. That will take care of your second set. 
$\bf{Added:}$
I misread the statement, $B$ is a subset of $\mathbb{Z}^n$. Yes, it is still Zariski dense. Consider $\mathbb{N}_{>0}^n$. This is Zariski dense. Now we should notice that the image on a Zariski dense subset under a linear bijection is again Zariski dense. Hence the set 
$$\{ (a_1, a_1 + a_2, \ldots, a_1 + a_2 + \ldots + a_n )\ | a_i \in \mathbb{N}_{>0} \}$$  is Zariski dense.  
Reverse the order of the coordinates and get a Zariski dense subset contained in $B$. 
