Let $X$ be topological space and $Y\subset X$. The goal is to show that $\dim Y\leq \dim X$.
Here we use this definition of dimension.
Let $Y_0\subsetneq Y_1\subsetneq \cdots \subsetneq Y_{n}$ be a chain of strictly increasing closed irreducible subsets of $X$ then $\dim X$ is the supremum over all such chains.
This is what I've got so far. I'm just stuck on one last bit. A hint would be nice.
Let $Y_0\subsetneq Y_1\subsetneq \cdots \subsetneq Y_n$ be a chain of irreducible closed subsets of $Y$. Since each $Y_i$ is irreducible, we just need to close them in $X$. There exists $F_i$ closed in $X$ such that $Y_i=F_i\cap Y$. Then if we closed each set in $X$, $\overline{F_0\cap Y}\subset\overline{F_1\cap Y}\subset \cdots \subset \overline{F_n\cap Y}$ is a chain of irreducible closed subsets of $X$. I just need to show that they are distinct. I really don't know where to proceed. A hint would be appreciated. Thanks.
Best Answer
I'll try to use as much of what you've done as I can.
You want to know that $\overline{Y_i} \subsetneq \overline{Y_{i+1}}$. Can I find points in the latter but not the former? Yes; I claim the points of $Y_{i+1} \setminus Y_i$ will work. Find me a closed set in $X$ that contains $Y_i$ but misses all of the points in this difference. You've already written one down.
Something you can prove in the same way is that if $S$ is a subset of $Y$ then $\operatorname{cl}_Y(S) = \operatorname{cl}_X(S) \cap Y$. That's comforting.