The dimension of a topological space $X$ is defined as the supremum of all integers $n\ge0$ for which there is a strictly increasing chain of $n+1$ irreducible closed subsets of $X$. Thus is the dimension of the empty topological space $-\infty$? Is the dimension $>-\infty$ for all nonempty topological spaces?
Asked
Active
Viewed 481 times
0
-
Why would the dimension of the empty set not be $-1$ by your definition? – Matt Dec 08 '18 at 08:39
-
@Matt: Because I consider it to be the supremum of a set of extended real numbers, and so the supremum of the empty set is $-\infty$. – Dec 08 '18 at 08:58
1 Answers
0
Yes and yes. For your first question, since there's no such $n \ge 1$, we have $\sup \emptyset = -\infty$. So you're right. For your second question, it suffices to show that any nonempty space $X$ has at least one irreducible closed subset, so that $\dim X = \sup \{0, \dots\} \ge 0$
Since $X$ is nonempty, we can pick some $x \in X$. By lemma 8.3 (1), the closure of any irreducible subset is irreducible. Since $\{x\}$ is irreducible, $\overline{\{x\}}$ is an irreducible closed subset of $X$. So we're done.
Note: I refer to definition 8.1 (1) for the definition of irreducible subset.
Alex Vong
- 1,654
-
-
-
Is this argument plausible? There is a minimal nonempty closed subset of $X$ by Zorn's lemma, and then this must necessarily be irreducible. – Dec 11 '18 at 02:26
-
@sai I suppose you mean to partially order all nonempty closed subsets of $X$ by reversed inclusion $\supseteq$ and then apply Zorn's lemma. I think you would run into problem in showing any nonempty chain has an upper bound since the intersection of a chain of nonempty closed sets can be empty. – Alex Vong Dec 11 '18 at 06:08
-