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What are some examples of topological spaces where open compact subsets form base for the topology? Where can I get discussion on such spaces?

For example, every compact zero-dimensional space has this property, but there might be some other spaces with such base.

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    $\begingroup$ Why is this closed? It's a completely standard question. I wrote an answer answering all of it and now I cannot publish the answer. The spaces in question are important, as they appear in Stone duality theorems. I am forced to publish my answer in the comments. $\endgroup$ – Andrej Bauer Mar 18 '17 at 13:17
  • $\begingroup$ Recall that a space is $0$-dimensional iff it has a base consisting of clopens (closed and open). Let us first think about Hausdorff spaces. If $X$ is Hausdorff and has a base $\mathcal{B}$ of compact opens, then the basic opens are also closed (because in a Hausdorff space a compact set is closed), therefore it is $0$-dimensional. Moreover, it is locally compact since every point has an arbitrarily small compact neighborhood (namely a basic one). $\endgroup$ – Andrej Bauer Mar 18 '17 at 13:18
  • $\begingroup$ Conversely, suppose $X$ is Hausdorff, locally compact and $0$-dimensional. We claim that the compact open sets form a basis. For this purpose, consider any open set $U$ and a point $x \in U$. It suffices to find a compact open $V$ such that $x \in V \subseteq U$. Because $X$ is locally compact, there is a compact neighborhood $K$ such that $x \in K \subseteq U$. Because $X$ is $0$-dimensional, there is a clopen neighborhood $V$ such that $x \in V \subseteq K$. But $V$ is actually compact, as it is a closed subset of a compact set $K$. We are done. We proved: $\endgroup$ – Andrej Bauer Mar 18 '17 at 13:19
  • $\begingroup$ Theorem: A Hausdorff space has a base of compact open sets if, and only if, it is $0$-dimensional and locally compact. $\endgroup$ – Andrej Bauer Mar 18 '17 at 13:19
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    $\begingroup$ @AndrejBauer Perhaps if you could edit the question to add a few words explaining why the question can be considered interesting, that might improve the chance of getting reopened. I have tried to find some support for reopening the post also elsewhere - we will see how this turns out in the end. (Needless to say, I am not that surprised that question containing two sentences with no motivation at all got closed.) $\endgroup$ – Martin Sleziak Oct 1 '17 at 9:02
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Note that compact subspaces of Hausdorff topological spaces are closed. Therefore, we need to either find non-Hausdorff spaces (harder since these are less intuitive) with this property, or find Hausdorff spaces with a base whose members are without boundary (hence clopen) as well as compact.

One possibility of the latter would be the discrete topology on any set. This has as a base singleton sets which are both open and compact.

For non-Hausdorff examples, we can take any topology on any finite set (all finite subspaces of a topological space are compact). The difficult task is finding infinite non-Hausdorff spaces with this property.

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  • $\begingroup$ Does there exist examples of infinite non hausdroff space with this property? $\endgroup$ – Jave Mar 18 '17 at 9:29
  • $\begingroup$ I have no idea. =/ $\endgroup$ – Kaj Hansen Mar 18 '17 at 9:30
  • $\begingroup$ @Jave the cofinite topology on an infinite set has your property, and is not Hausdorff $\endgroup$ – Forever Mozart Mar 18 '17 at 9:43
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This is only a temporary answer - to be deleted soon. It was posted to make copy-pasting of the text posted in the comments easier.

The spaces in question are important, as they appear in Stone duality theorems. I am forced to publish my answer in the comments. -- Andrej Bauer

Recall that a space is $0$-dimensional iff it has a base consisting of clopens (closed and open). Let us first think about Hausdorff spaces. If $X$ is Hausdorff and has a base $\mathcal{B}$ of compact opens, then the basic opens are also closed (because in a Hausdorff space a compact set is closed), therefore it is $0$-dimensional. Moreover, it is locally compact since every point has an arbitrarily small compact neighborhood (namely a basic one). -- Andrej Bauer

Conversely, suppose $X$ is Hausdorff, locally compact and $0$-dimensional. We claim that the compact open sets form a basis. For this purpose, consider any open set $U$ and a point $x \in U$. It suffices to find a compact open $V$ such that $x \in V \subseteq U$. Because $X$ is locally compact, there is a compact neighborhood $K$ such that $x \in K \subseteq U$. Because $X$ is $0$-dimensional, there is a clopen neighborhood $V$ such that $x \in V \subseteq K$. But $V$ is actually compact, as it is a closed subset of a compact set $K$. We are done. We proved: -- Andrej Bauer

Theorem: A Hausdorff space has a base of compact open sets if, and only if, it is $0$-dimensional and locally compact. -- Andrej Bauer

Such spaces abound. For example, the Cantor space, the interval topology on an ordinal $\alpha$, the one-point compactification of $\mathbb{N}$ (which is the interval topology on $\omega + 1$), etc. -- Andrej Bauer

How about non-Hausdorff spaces? (Contrary to the popular opinion arising from how topology is taught, these can be perfectly "intuitive" as well.) Here I do not see an immediate characterization, but examples are easy enough to come by. For instance, take and poset $(P, {\leq})$ and consider the topology on $P$ consisting of upper sets (a set $L \subseteq P$ is upper if $x \in L$ and $x \leq y$ implies $y \in L$). The opens of the form ${\uparrow}x = \{y \in P \mid x \leq y\}$ form a basis, and they are compact because as soon as you cover $x$ you've covered ${\uparrow}x$. -- Andrej Bauer

Good post @AndrejBauer. I would've loved to +1 that as a proper answer. -- Kaj Hansen

@AndrejBauer wow! Any reference suggestion ? -- Jave

Sure, if you look up anything regarding Stone duality and generalized Boolean algebras, you'll get plenty. -- Andrej Bauer

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