Constructible set in a Zariski space that contains all closed points

Question

Let $X$ be a Zariski space such that the closed points are dense, and let $Y$ be constructible set which contains all closed points.

Does it follow that $Y=X$?

If the set of closed points is not dense, this is false. For example take as $X$ the spectrum of a DVR, which contains one closed $p$ and one open point $q$. Then $\{p\}$ is closed, hence constructible, but not all of $X$.

On the other hand, if the closed points are dense we know that $Y$ itself has to be dense, hence contains at least the generic points of the components of $X$ (this is an exercise in Hartshorne). But how do I conclude further?

Answer 1

I think I missed one assumption about $X$, that is:

In every closed subspace of $X$, the closed points are dense as well.

I'm not sure if this is trivially true, but if $X$ is a scheme of finite type over a field (which was my motivation), this holds for certain, as closed subspaces are again schemes of finite type.

So to the proof I found:

We know that $Y$ contains a dense set $U \subset Y \subset X$, which is open in $X$. Defining $X_1 := X \setminus U$ we know

1. $X_1 \subset X$ is a Zariski space which is closed in $X$,
2. $Y_1 := Y \cap X_1$ is constructible in $X_1$,
3. $Y_1$ contains all closed points of $X_1$, because those are exactly the closed points of $X$ contained in $X_1$,
4. the closed points of $X_1$ are dense.

Thus we obtain a descending chain of closed subsets $X = X_0 \supset X_1 \supset X_2 \cdots$ which has to stabilize at one point, because $X$ is a noetherian space. But this means that at some point in the construction, we get $U_i = Y_i$. Thus $Y_{i-1} = U_{i-1} \cup (X_{i-1} \setminus U_{i-1}) = X_{i-1}$, hence we may take $U_{i-1} = X_{i-1} = Y_{i-1}$. Proceeding like this we finally may take $U = Y = X$, finishing the proof.