# Hartshorne Chapter 1 Exercise 1.10 (a)

In Chapter 1 of Robin Hartshorne's "Algebraic Geometry," exercise 1.10 is:

If $$Y$$ is any subset of a topological space $$X$$, then $$\dim Y \le \dim X$$.

Here, $$\dim X$$ is the supremum of integers $$n$$ such that there exists a chain of distinct irreducible closed subsets $$X_0\subsetneq X_1\subsetneq \cdots \subsetneq X_n$$ of $$X$$.

I figured that if I can create a chain of irreducible closed subsets in $$X$$ from a chain in $$Y$$, $$Y\cap X_0\subsetneq Y\cap X_1\subsetneq \cdots \subsetneq Y\cap X_n$$, where the $$X_i$$ are closed subsets of $$X$$, since then, $$n\le \dim X$$.

However, the only chain in $$X$$ I can create from this chain in $$Y$$ is $$X_0\subsetneq X_0\cup X_1\subsetneq \cdots \subsetneq \bigcup_{i=0}^n X_i,$$ which I cannot prove consists of irreducible sets.

I feel like this exercise should be quite trivial, but I can't figure it out, so I would appreciate any help you can give me.

• If you start with a chain in $Y$, say $Y_i := Y \cap X_i$, then as stated $X_i$ is not well-defined. There is, however a canonical choice of $X_i$ starting from $Y_i$, and you don't need to further manipulation (i.e. you get $X_0 \subset X_1 \subset \cdots \subset X_n$ for free, although you will need to justify why they are distinct). Can you see what $X_i$ to choose? – Pig Jul 2 '19 at 3:56
• $X_i=\bar{Y_i}$? – Kenta S Jul 2 '19 at 3:59
• :) The question is then 1. the sets are distinct and 2. why they are irreducible – Pig Jul 2 '19 at 5:17
• Thanks, it was very helpful. – Kenta S Jul 2 '19 at 5:40

Let $$Y_0\subsetneq Y_1 \subsetneq\cdots\subsetneq Y_n$$ be an arbitrary chain of irreducible, closed subsets of $$Y$$. Let $$\overline{Y_i}$$ denote the closure of $$Y_i$$ in $$X$$. $$\overline{ Y_{i-1}}\subsetneq \overline{ Y_i}$$, since, otherwise, $$Y_{i-1}=Y\cap \overline{Y_{i-1}}=Y\cap \overline{Y_i}=Y_i,$$ contrary to the hypothesis.
If $$\overline{Y_i}$$ is irreducible in $$X$$ by Proposition 1.6 of Robin Hartshorne's "Algebraic Geometry":
Any nonempty open subset of an irreducible topological space is dense and irreducible, If $$Y$$ is a subset of a topological space $$X$$, which is irreducible in its induced topology, then the closure $$\overline Y$$ is also irreducible.
Hence, we have created a chain $$\overline{Y_0}\subsetneq \overline{Y_1} \subsetneq\cdots\subsetneq \overline{Y_n}$$ in $$X$$. By the definition of $$X$$'s dimension, $$n\le \dim X$$. Since $$Y_0\subsetneq Y_1 \subsetneq\cdots\subsetneq Y_n$$ was arbitrary, $$\dim Y\le \dim X.$$