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 $S$.


Q1: In every topological space, the space itself is both open and closed. Note however that the notions "open/closed subset of $X$" and "open/closed subset of $A^n$" differ.

Q2: $A^n\setminus V(I_i)$ is an open set in $A^n$, $V(I_x)$ is closed in $A^n$. The intersection of a closed and an open set need not be either - as a ubset of $A^n$. But by definition of relative topology, this intersection is an open subset of the topological space $V(I_x)$.

Q3: Your loose interpretation is loosely correct. $S$ is dense in $X$ if every open subset of $X$ intersects $S$. Equivalently, the closure of $S$ (i.e. the smallest closed subset of $X$ containing $S$) is all of $X$.

  • $\begingroup$ +1 for speed and "Your loose interpretation is loosely correctly". =) $\endgroup$ – Yong Hao Ng Sep 1 '12 at 15:06

(1) Your interpretation is correct: we are viewing $X$ as a topological space with the subspace topology induced by the inclusion $X \subseteq \mathbb A^n$ (or $X \subseteq \mathbb P^n$).

(2) Your set manipulation makes sense, but I’m not sure that it helps you as it stands. In fact the question is purely topological, so you can avoid saying the word “ideal” entirely. To prove that (a) implies (b), for example, assume that $X$ is irreducible and let $\mathcal O_1, \mathcal O_2$ be non-empty open subsets of $X$. The $X \setminus \mathcal O_i$ are closed and proper subsets of $X$; their union cannot be all of $X$, since $X$ is irreducible. Write down what this means and use elementary set theory to conclude that $\mathcal O_1 \cap \mathcal O_2 \neq \emptyset$.

To the question about $V(I_X) \setminus V(I_i)$, there is very little reason to expect this to be closed. [It is certainly an open subset of $X$, and you can check that irreducible sets are connected.] Here is an example from the subject at hand. Suppose $X = \mathbb A^1$ (so $I_X = 0$) and we subtract the closed set $\{0\} = V(x)$. If our underlying field is infinite then this set is not closed: if a polynomial vanishes on the infinite set $\mathbb A^1 \setminus \{0\}$, then it vanishes everywhere.

(3) The problem states that “the closure of $\mathcal O$ in $X$ is $X$”, which seems pretty exact to me! There's more information on Wikipedia. It sounds like you might favor “every point of $X$ belongs to $\mathcal O$ or is a limit point of $\mathcal O$”.

  • $\begingroup$ +1 for a more detailed answer. Am I suppose to wait for a period of time and choose between answers? $\endgroup$ – Yong Hao Ng Sep 1 '12 at 15:27
  • $\begingroup$ @Yong I'm quite happy with the ways things are in this case. There are some relevant threads on meta that I will try to dig up, but it's not something I would worry much about. Cheers, $\endgroup$ – Dylan Moreland Sep 1 '12 at 15:31

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