This is a homework problem and I am looking for clarification of some of my doubts. (not solutions)
Let $X\subset A^n$ or $P^n$, where $X$ is a non-empty algebraic set. Open sets of $X$ are given as $\mathcal{O}\cap X$, where $\mathcal{O}$ is an open set of the $n$-space. This shows $X$ has an induced Zariski topology. Show equivalence of the 3 statements:
a) $X$ is irreducible
b) If $\mathcal{O_i}$ are non-empty open subsets of $X$, then $\mathcal{O}_1\cap \mathcal{O}_2 \neq \emptyset$
c) Any non-empty open subset $\mathcal{O}$ of $X$ is dense in $X$. Also described as: "closure of $\mathcal{O}$ in $X$ equals $X$".
Question 1: $X$ is closed in Zariski topology in $A^n$. But since we are looking at the induced topology by $X$, does it mean the whole space is $X$ and hence $X$ is now both closed and open set?
Part of my proof for $a\implies b$:
Open sets in $A^n$ are given as $\mathcal{O_i}=A^n\backslash V(I_i)$ for some ideal $I_i$.
$X$ is an algebraic set $\implies X=V(I_x)$ and $X$ is closed.
In $X$, $\mathcal{O_i}=(A^n\backslash V(I_i)) \cap X=(A^n\backslash V(I_i))\cap V(I_x)=V(I_x)\backslash V(I_i)$
Question(s) 2: Does the set-theoretic operations make sense for the last statement?
I noticed that $A^n\backslash V(I_i)$ is an open set while $V(I_x)$ is closed. Why is $V(I_x)\backslash V(I_i)$ not a closed set? (The question says $\mathcal{O}_i$ is an open set)
Question 3: (For $c$) What exactly is meant by a "dense" subset? My loose interpretation is that for a subset $S$ in $X$ to be dense, all points of $X$ are either "touching" $S$ or in $S$.