A scheme $X$ is quasi separated if intersection of two quasi compact subsets is a quasi compact subset.
Question is to prove that :
A scheme $X$ is quasi separated if intersection of two affine open subsets is a finite union of affine open subsets.
Let $U,V$ be quasi compact subsets of $X$ and let $\{\text{Spec}(A_i)\}$ be an open cover for $U\cap V$.
We have $\text{Spec}(A_i)\subseteq U$ for every $i$. So, $\bigcup \text{Spec}(A_i)\subseteq U$. We complete this cover by adding some more open subsets namely $\{\text{Spec}(B_j)\}$. We thus have
$$U=\left(\bigcup \text{Spec}(A_i)\right)\bigcup \left(\bigcup \text{Spec}(B_j)\right).$$
Similarly for $V$ we will add some more open subsets, namely $\{\text{Spec}(C_k)\}$ and get an open cover for $V$. We thus have
$$V=\left(\bigcup \text{Spec}(A_i)\right)\bigcup \left(\bigcup \text{Spec}(C_k)\right).$$
As $U$ is quasi compact, this open cover has a finite subcover. Suppose we have $$U=\text{Spec}(A_1)\bigcup \text{Spec}(B_1).$$
As $V$ is quasi compact, this open cover has a finite subcover. Suppose we have $$U=\text{Spec}(A_2)\bigcup \text{Spec}(C_2).$$
We thus have,
$$U\cap V=(\text{Spec}(A_1)\cap \text{Spec}(A_2))\bigcup(\text{Spec}(A_1)\cap \text{Spec}(C_2))\bigcup(\text{Spec}(B_1)\cap \text{Spec}(A_2))\bigcup(\text{Spec}(B_2)\cap \text{Spec}(C_2)).$$
clearly, first three terms in the union is contained in $$\text{Spec}(A_1)\bigcup \text{Spec}(A_2).$$
It is the last term which we have to thik about. As it is given that intersection of two affine open subsets is finite union of affine open subsets, we have
$$\text{Spec}(B_2)\cap \text{Spec}(C_2)=\text{Spec}(D_1)\bigcup\text{Spec}(D_2).$$
Clearly, $\text{Spec}(B_2)\cap \text{Spec}(C_2)\subseteq U\cap V$ i.e.,
$\text{Spec}(D_1)\bigcup \text{Spec}(D_2)\subseteq U\cap V=\bigcup \text{Spec}(A_i)$.
As $\text{Spec}(D_1)$ is quasi compact and $\text{Spec}(D_1)\subseteq \bigcup \text{Spec}(A_i)$ we can assume that
$$\text{Spec}(D_1)=\text{Spec}(A_3)\bigcup \text{Spec}(A_4).$$
As $\text{Spec}(D_2)$ is quasi compact and $\text{Spec}(D_2)\subseteq \bigcup \text{Spec}(A_i)$ we can assume that
$$\text{Spec}(D_2)=\text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$
So,
$$\text{Spec}(D_1)\bigcup \text{Spec}(D_2)\subseteq \text{Spec}(A_3)\bigcup \text{Spec}(A_4)\bigcup \text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$
Thus, $$U\cap V\subseteq \text{Spec}(A_1)\bigcup \text{Spec}(A_2)\text{Spec}(A_3)\bigcup \text{Spec}(A_4)\bigcup \text{Spec}(A_5)\bigcup \text{Spec}(A_6).$$
So, given an open cover $\text{Spec}(A_i)$ for $U\cap V$ we have a finite sub cover $\{\text{Spec}(A_i)\}_{i=1}^6$ for $U\cap V$.
Thus, $U\cap V$ is quasi compact.
I would like to know if this justification is correct.
Best Answer
A topological space is called quasi-separated if the intersection of any two quasi-compact open (not just any) subsets is quasi-compact. Check your definition.
Let $X$ be a scheme. Suppose the intersection of any two affine open subsets is a finite union of affine open subsets. Take two quasi-compact open subsets $U, V\subset X$. We want to show that $U\cap V$ is quasi-compact.
First prove the following easy lemmas:
Lemma 1 Every affine scheme is quasi-compact.
Lemma 2: If $Y$ is a topological space which is a finite union of quasi-compact spaces, then $Y$ is quasi-compact.
Lemma 3: Let $X$ be a scheme, then the affine open subsets form a base for Zariski topology.
Using the thrid lemma both $U,V$ are unions of affine opens. Using the fact that $U,V$ are quasi-compact, we find that $U,V$ are finite union of affine opens, say $U=\bigcup_{i=1}^n \mathrm{Spec}A_i$ and $V=\bigcup_{j=1}^m \mathrm{Spec}B_j$. Then $$ U\cap V= \bigcup_{i=1}^n \bigcup_{j=1}^m \mathrm{Spec}A_i\cap \mathrm{Spec}B_j $$ since the intersection of any two affine opens is a finite union of affine opens, then $U\cap V$ is a finite union of affine opens. Since every affine schme is quasi-compact (lemma 1), and since $U\cap V$ is a finite union of affine opens/quasi-compacts, then $U\cap V$ is quasi-compact (lemma 2).