(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.
(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.
1.10a is incorrect. Dimension is defined on a topological space, so the dimension of $Y\subset X$ is its dimension under its induced topology. A chain $Z_0\subset\cdots Z_n\subseteq Y$ can be irreducible and closed in $Y$, and not in X. In fact, $\overline{Z}_i$ - closure taken in X - are closed and irreducible in $X$. That fixes the proof, but proving they're irreducible takes a little effort:
ReplyDeleteChoose any pair of closed sets $A$ and $B$ with $A\cup B=\overline{Z}$ (note $A\subseteq \overline{Z}$. From elementary topology, $\overline{Z}\cap Y=Z$ (taking the closure of $Z$ closed in $Y$ can't add any points from $Y$), so we also have $(A\cap Y)\cup(B\cap Y)=Z$. These sets are closed in $Y$, so by $Z$ irreducible in $Y$, choose WLOG $A\cap Y=Z$. So $A$ is a closed set containing $Z$: it contains $Z$ and all limit points of $Z$. So $\overline{Z}\subseteq A$. ie, $A$ cannot be a proper subset.
$\overline{Z}\cap Y=Z$ also implies that the sequence $\overline{Z_0}\subset\cdots \overline{Z_n}\subset Y$ STAYS proper.
-David
I think in the proof of (c) we have dim X = 1 and dim {1} as 0, right?
ReplyDeleteI think in (b) you have not established that \(U_j\cap X_i \neq U_j\cap X_{i+1}\). What if $X_{i+1} - X_i$ lives entirely outside $U_j$? I believe the following additional logic is necessary: if $U_j$ does not meet $X_{i+1}-X_i$, then $X_i \cup (U_j^c \cap X_{i+1})=X_{i+1}$ gives a decomposition of $X_{i+1}$ into two closed subsets, contradicting that $X_{i+1}$ is assumed irreducible.
ReplyDeleteIn part (b), how do you know that points are closed? [I think your proof still goes through, though]
ReplyDeleteI would say in (c) that dim X=1 and dim{1} is undefined since {1} does not contain any irreducible closed subsets (maybe some convention of -infty, etc.)
ReplyDelete