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Denote by $\mathcal C(X)$ be the space of compact subsets of a topological space X. Let $(X_\alpha)_\alpha$ be an inverse system of topological spaces, then $(\mathcal C(X_\alpha))_\alpha$ is also an inverse system with the induced bonding maps (a compact subset goes to its image).

I'm trying to understand why $\mathcal C(\lim_\leftarrow X_\alpha)=\lim_\leftarrow\mathcal C(X_\alpha)$.

There is a map $T:C(\lim_\leftarrow X_\alpha)\to\lim_\leftarrow\mathcal C(X_\alpha)$ given by $T(H)=(p_\alpha(H))_\alpha$ where $p_\alpha:\lim_\leftarrow X_\beta\to X_\alpha$ is the projection. What I don't understand is why is this map onto?

Given an element $(H_\alpha)_\alpha\in \lim_\leftarrow\mathcal C(X_\alpha)$, we have to find a compact subset of $\lim_\leftarrow X_\alpha$ s.t. its projection of the $\alpha$th coordinate is $H_\alpha$ for all $\alpha$. I tried to take $H=\lim_\leftarrow H_\alpha=\Pi H_\alpha \cap \lim_\leftarrow X_\alpha$, but I can't prove that indeed $p_\alpha (H)=H_\alpha$.

Can someone explain this to me, or point me in the right direction? thanks!

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  • $\begingroup$ What's the topology on $\mathcal{C}(X)$? $\endgroup$
    – Dan Rust
    Commented May 25, 2014 at 12:34
  • $\begingroup$ The Vietoris topology. Sorry for the ommision, I thought it's not relevant to the specific question about surjectivity. $\endgroup$
    – ChanaG
    Commented May 25, 2014 at 12:45
  • $\begingroup$ I'm actually interested in the case of profinite groups, so we can assume for simplicity that the topology on each $X_\alpha$ is discrete (then $\mathcal C(X_\alpha))$ is also discrete). $\endgroup$
    – ChanaG
    Commented May 25, 2014 at 12:54

1 Answer 1

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You're having trouble proving this because it is not quite true, and the reason has more to do with inverse limits being complicated than with topology. I'll explain some examples where it fails and then give some hypothesis were this will hold.

Here's a nice example of an inverse system of surjective maps with empty inverse limit: consider the directed set consisting of the finite subsets of $\mathbb{R}$ ordered by inclusion; for each such finite set $\alpha$, let $X_\alpha$ be the set of injective functions $\alpha \to \mathbb{N}$. If $\beta \supseteq \alpha$, there is a surjective restriction map $X_\beta \to X_\alpha$. However, the inverse limit of the $X_\alpha$ is empty because an element of the inverse limit would glue together to give an injective function $\mathbb{R} \to \mathbb{N}$!

We can use that example to show that the claim $\mathcal{C}(\lim X_\alpha) = \lim \mathcal{C}(X_\alpha)$ can fail, even for inverse systems of surjective maps: just give each $X_\alpha$ from the above example the indiscrete topology. Then $\mathcal{C}(\lim X_\alpha) = \mathcal{C}(\emptyset)$ has only one point (corresponding to the empty space), but $\lim \mathcal{C}(X_\alpha)$ has at least two points: $(\emptyset)_\alpha, (X_\alpha)_\alpha$. (Notice that with the indiscrete topology $X_\alpha$ is compact.)

OK, so can we give additional hypothesis to make your claim true? I claim either of the following conditions make it true:

  1. The inverse system is indexed simply by the natural numbers.

  2. The indexing set is arbitrary, but all the spaces $X_\alpha$ are Hausdorff.

I'll explain in each case why, in your notation, $p_\alpha(H) = H_\alpha$, since that is the only thing you still had left to show. Of course $p_\alpha(H) \subseteq H_\alpha$, so we can focus on showing $H_\alpha \subseteq p_\alpha(H)$. A couple of points before we do that:

  • By definition of the inverse system $\mathcal{C}(X_\alpha)$, the maps in the inverse system $X_\alpha$ when restricted to the $H_\alpha$ become surjections. This simplifies things.
  • If some $H_\alpha = \emptyset$, then they are all empty, since given any $\beta$ there is some $\gamma$ such that both $H_\alpha$ and $H_\beta$ are images of $H_\gamma$. So we will assume $H_\alpha$ are non-empty.

Here are the arguments for each case:

  1. Because the maps between $H_\alpha$ are surjections, given a point $x_{\alpha_0}$ in $H_{\alpha_0}$, we can pick a preimage in $H_{\alpha_0+1}$ and then a preimage of that and so on, to construct a point in $H$ mapping to $x$ under $p_{\alpha_0}$.

  2. In this case, the $H_\alpha$ are all compact Hausdorff spaces, and as before, the maps in the inverse system $H_\alpha$ are surejections. Let $x_{\alpha_0} \in H_{\alpha_0}$, we'll show there is an $x \in H$ with $p_\alpha(x) = x_\alpha$. For $\alpha > \alpha_0$ let $$T_\alpha := \{ (t_\beta)_\beta \in \prod H_\beta : t_{\alpha_0} = x_{\alpha_0} \text{ and for all } \beta < \alpha, f_{\alpha\beta} (t_\alpha) = t_\beta \}.$$ Notice that $\bigcap_{\alpha > \alpha_0} T_\alpha \subseteq H$ and every point in this intersection has $\alpha_0$-coordinate $x_{\alpha_0}$, so all we need to do is prove this intersection is non-empty.

    Because the $H_\alpha$ are Hausdorff, the $T_\alpha$ are closed and therefore it is enough to show that every finite number of the $T_\alpha$ have non-empty intersection. OK, take indices $\alpha_1, \alpha_2, \ldots, \alpha_n$ bigger than $\alpha_0$. Since the system is directed, there is some $\beta$ bigger than all of $\alpha_1, \alpha_2, \ldots, \alpha_n$. Clearly $T_\beta \subseteq T_{\alpha_1} \cap T_{\alpha_2} \cap \cdots \cap T_{\alpha_n}$, and since all the maps in the inverse system of the $H_\alpha$ are surjective, $T_\beta$ is non-empty.

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    $\begingroup$ Great, the proof for the Hausdorff case seems exactly what I need. Thanks! $\endgroup$
    – ChanaG
    Commented May 26, 2014 at 11:45
  • $\begingroup$ do you have references for this? $\endgroup$
    – Stu Kraji
    Commented Nov 2, 2016 at 17:14
  • $\begingroup$ Sorry, @StuKraji, I don't have references. Is there any particular point in the arguments you'd like me to expand on? $\endgroup$ Commented Nov 2, 2016 at 18:18
  • $\begingroup$ Probably am I misreading but, at first sight, I don't think that $\mathcal C(\lim_\leftarrow X_\alpha)=\lim_\leftarrow\mathcal C(X_\alpha)$ holds in $\mathcal{Top}$ even if the inverse system is indexed by the natural numbers (although, the surjective part seems correct). $\endgroup$
    – Stu Kraji
    Commented Nov 3, 2016 at 12:21
  • $\begingroup$ Oh, I see. Let's try to get to the bottom of this. It would help if you gave a counterexample or pointed out a mistake in my proof, @StuKraji. $\endgroup$ Commented Nov 3, 2016 at 15:55

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