In Chapter 1 of Robin Hartshorne's $\textit{Algebraic Geometry}$, 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 $X$, 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.
Best Answer
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":
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, $\dim Y\le \dim X.$