1
$\begingroup$

Suppose $X$ is a Hausdorff (I'm happy to also assume "non compact") topological space that can be written as the topological direct limit

$$\varinjlim_{a\in J} K_a$$ for $J$ a directed set and $K_a$ compact Hausdorff subspaces of $X$, where all bonding maps are continuous non-surjective immersions (I'm happy to assume, additionally, that they are not surjective).

Suppose that there is another system of compact Hausdorff subspaces of $X$ with bonding maps as in the previous system,

$$\{H_a, a\in J'\}$$ for a directed set $J'$ with $J\subseteq J'$. Suppose that for every $a\in J$ we have

$$K_a\subseteq H_a.$$

Suppose that we still have

$$X = \varinjlim_{a\in J'}H_a$$ with the direct limit topology on the right side and $X$'s own topology on the left side.

Q1. Can we conclude that for every $a\in J'$ there exists a $b\in J$ such that $H_a\subseteq K_b$?

Q2. Can we conclude as in Q1 if in addition all the $K_a$'s and $H_a$'s are compact differentiable manifolds with differentiable immersions as bonding maps?

The answer by KP Hart answers Q1 in the negative.

Improve this question
$\endgroup$

1 Answer 1

Reset to default
2
$\begingroup$

Q1. Consider the space of rationals $\mathbb{Q}$. It is the direct limit of the family of finite unions of convergent sequences (including their limits), ordered by inclusion, as a set is closed iff its intersection with every convergent sequence is closed.

Now choose for every convergent sequence $\mathbf{q}=\langle q_n:n\in\mathbb{N}\rangle$ a set $S_\mathbf{q}$ of convergent sequences $\{\mathbf{r}_n:n\in\mathbb{N}\}$, where $\mathbf{r}_n$ converges to $q_n$, and has diameter not larger than $1/n$. If $K$ is a finite family of convergent sequences then let $H_K=\bigcup_{\mathbf{q}\in K}S_\mathbf{q}$. Then $K\subseteq H_K$, and $H_K$ is compact; and $\mathbb{Q}$ is also the direct limit of the system of $H_K$s. But there are no $L$ and $K$ such that $H_K\subseteq L$ ($L$ has finitely many accumulation points, $H_K$ has infinitely many).

Addendum: this construction works in every metric space without isolated points. There a set is closed iff its intersection with every convergent sequence is closed, and every point is the limit of a non-trivial convergent sequence of arbitrarily small diameter.

Improve this answer
$\endgroup$
4
  • $\begingroup$ This is very nice! Thank you! I'm wondering if my question has "yes" as an answer if we assume in addition that every $K_a$ and $H_a$ are compact differentiable manifolds and not just compact Hausdorff topological spaces. Do you see any obstruction? This example wouldn't work, of course. I'll accept this answer if after a bit of time this updated question will remain unanswered. Thanks again. $\endgroup$
    – user335418
    Commented Feb 10, 2022 at 9:16
  • $\begingroup$ This works in every space with sufficiently many convergent sequences; I will add a remark to that effect to the answer. $\endgroup$
    – KP Hart
    Commented Feb 10, 2022 at 10:42
  • $\begingroup$ As to Q2: I think that the answer is positive but I don't have time to work out the details. The point is that the interiors of the $K_a$ will cover the manifold. Given a point $x$ if all $K_a$ that contain $x$ have dimension less that that of the manifold then one can construct a sequence that converges to $x$ that hits every $K_a$ in a finite set (make it converge to $x$ perpendicularly to the $K_a$ that contain $x$. This would contradict the direct limit assumption. If the dimension of $K_a$ is the same as that of the manifold its interior is nonempty. 1/2 $\endgroup$
    – KP Hart
    Commented Feb 13, 2022 at 21:04
  • $\begingroup$ Continued: the tricky bit is to find then a $K_a$ that has $x$ in its interior, but maybe someone better versed in manifolds will see how to do this. 2/2 $\endgroup$
    – KP Hart
    Commented Feb 13, 2022 at 21:06

You must log in to answer this question.