Chapter 1, Exercise 1.10: Topological properties of dimension

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

(b) If $X$ is a topological space which is covered by a family of open sets $\{U_i\}$, then $\dim X=\sup\dim U_i$.

(c) Give an example of a topological space $X$ and a dense open subset $U$ with $\dim U<\dim X$.

(d) If $Y$ is a closed subset of an irreducible finite-dimensional topological space $X$, and if $\dim Y=\dim X$, then $Y=X$.

(e) Give an example of a noetherian topological space of infinite dimension.

(a) This is quite trivial: any closed irreducible chain in $Y$ is also such a chain in $X$, and as such cannot be longer than $\dim X$.

(b) By (a), each $\dim U_i$ is $\leq\dim X$ and thus $\sup\dim U_i\leq X$. Consider some chain $X_0\subset\dots\subset X_n$ in $X$, where $n=\dim X$. This has maximal length, thus $X_0$ must be a minimal closed irreducible subset of $X$. Thus $X_0$ is a point (more precisely, a singleton). We have $X_0\in U_i$ for some $i$. Then $X_j\cap U_i\neq\emptyset$ for each $j\leq n$, since each $X_j$ contains $X_0$, as does $U_i$. Further, each $U_i\cap X_j\subseteq X_j$ is irreducible and dense in $X_j$ by Exercise 1.6, and we see that $X_j\neq X_{j-1}$. Thus we can construct a chain $U_i\cap X_0\subset\dots\subset U_i\cap X_n$ of closed distinct irreducible sets in $U_i$, but this chain must have length $\leq\sup\dim U_i$.

(c) When looking for strange counterexamples involving "small" dense sets, the usual place to look is the Sierpinski space $X=\{0,1\}$ with open sets $\emptyset,\{1\},\{0,1\}$. Note that the longest applicable chain in $X$ is $\{1\}\subset\{0,1\}$, so $\dim X=2$, but $\dim\{1\}=1$ in $X$. Now, the only closed set containing $\{1\}$ is $X$ itself, so $\{1\}$ is dense but $\dim\{1\}<\dim\{0,1\}$.

(d) Suppose $Y\neq X$, and $Y_0\subset\dots\subset Y_n$ is a maximal chain in $Y$ (so $n=\dim Y$). Then $Y_0\subset\dots\subset Y_n\subset X$ is a longer chain in $X$ satisfying the hypotheses since $X$ is irreducible.

(e) Take any well-ordered set $S$ and give it the topology with closed sets being all initial segments. For example, choose $S=\mathbb N=\{0,1,2,\dots\}$. This satisfies a descending chain condition, since every chain will eventually reach $\{0\}$. However, it is infinite-dimensional since $\{0\}\subset\{0,1\}\subset\dots$ is a valid infinite ascending chain.