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Space Suggestion: [Disconnected Deleted Comb Space]

Open hew-wolff opened this issue 6 months ago • 17 comments

Space Suggestion

Take $K = \{1, 1/2, 1/3, ...\}$. Then the space is $A = K \times [0, 1] \cup \{(0, 0), (0, 1)\}$.

The interesting property is this: $A$ has a quasi-component which is not a (connected) component. (Quasi-components in general form a coarser partition of a space than connected components.)

Proof. Look at $a_0 = (0, 0)$ and $a_1 = (0, 1)$. Any set containing $a_0$ and some point $(x, y)$ for $x > 0$ can clearly be disconnected in $A$, and similarly for $a_1$. Also the subspace topology on $\{a_0, a_1\}$ is discrete. So each $\{a_i\}$ is a component. However, $a_0$ and $a_1$ cannot be separated by a disconnection of the entire space $A$, because any neighborhoods $U_i \ni a_i$ both meet infinitely many of the $k \times [0, 1]$ components and therefore both meet one of those components. So $\{a_0, a_1\}$ is a quasi-component.

The name is not great but more or less follows Wikipedia "Comb space".

There is a more complicated version of this space described in MSE "Countability of quasicomponents/components/path-components". There is also a similar counterexample in Dugundji's Topology (Chapter V problem 3.5) which is a bit more complicated.

I could not find "quasi-component" in a search, so I guess we would also add a property like "has a nontrivial quasi-component". (There is more to add on this property: apparently a space does not have this property if it's compact Hausdorff or if it is locally connected.)

hew-wolff avatar Jun 03 '25 17:06 hew-wolff

I don't like the word "non-trivial" here, since that usually means its either empty or whole space, while you mean to say that it has a quasi-component which is not a component.

I think a better name would be "quasi-components are components" for example.

Moniker1998 avatar Jun 03 '25 21:06 Moniker1998

@hew-wolff See also example 2 in https://en.wikipedia.org/wiki/Locally_connected_space#Quasicomponents. It's the same idea as yours. Actually, your example is exactly example 6.1.24 in Engelking. He also gives a locally compact variation.

What other examples do we currently have in pi-base of spaces with some quasi-component that is not connected?

prabau avatar Jun 03 '25 21:06 prabau

I think a better name would be "quasi-components are components" for example.

Another possible name: "Has connected quasi-components" (meaning that every quasi-component is connected, i.e., every quasi-component is a connected component)

prabau avatar Jun 03 '25 21:06 prabau

I think a better name would be "quasi-components are components" for example. Another possible name: "Has connected quasi-components"

"Has all quasi-components connected"? "Connected in all quasi-components"?

hew-wolff avatar Jun 03 '25 22:06 hew-wolff

@hew-wolff See also example 2 in https://en.wikipedia.org/wiki/Locally_connected_space#Quasicomponents. It's the same idea as yours. Actually, your example is exactly example 6.1.24 in Engelking. He also gives a locally compact variation.

Nice.

What other examples do we currently have in pi-base of spaces with some quasi-component that is not connected?

I'm not sure how to look--maybe comb through "not locally connected"? I'm OK using some other space, whether already in pi-base or not, but I'd like to have some interesting quasi-components somewhere, preferably nice and simple as they are for this space.

hew-wolff avatar Jun 03 '25 22:06 hew-wolff

I think a better name would be "quasi-components are components" for example. Another possible name: "Has connected quasi-components"

"Has all quasi-components connected"? "Connected in all quasi-components"?

"Has connected quasi-components" follows the same pattern as "Has points $G_\delta$", which means exactly that all points are G_delta. It is a little shorter than the alternatives and clear in my opinion. @StevenClontz opinion?

(also, "Has closed retracts" = all retracts are closed)

prabau avatar Jun 04 '25 00:06 prabau

Example 115 in Steen & Seebach = https://topology.pi-base.org/spaces/S000112 (Nested rectangles)

prabau avatar Jun 04 '25 00:06 prabau

https://math.stackexchange.com/a/5072215/476484 I've also constructed a locally compact Hausdorff space for any possible triples of cardinals for which a space with given amount of quasi-components, components and path-components exists.

Moniker1998 avatar Jun 04 '25 10:06 Moniker1998

Example 115 in Steen & Seebach = https://topology.pi-base.org/spaces/S000112 (Nested rectangles)

Nice. This is essentially Dugundji's example too. Perhaps my space would be redundant (although more appealing in my opinion).

hew-wolff avatar Jun 04 '25 13:06 hew-wolff

https://math.stackexchange.com/a/5072215/476484 I've also constructed a locally compact Hausdorff space for any possible triples of cardinals for which a space with given amount of quasi-components, components and path-components exists.

Cool! I wonder how to express that in pi-base properties.

hew-wolff avatar Jun 04 '25 13:06 hew-wolff

Example 115 in Steen & Seebach = https://topology.pi-base.org/spaces/S000112 (Nested rectangles)

Nice. This is essentially Dugundji's example too. Perhaps my space would be redundant (although more appealing in my opinion).

All these examples are kind of similar. We have an infinite number of connected components clustering to at least two different connected components.

Your example is one of the simplest one could think of, so it seems fine to add it to pi-base if we want to. As for the name of the space, I would not use "comb" here. How about something more descriptive like one of these:

  • Sequence of intervals in the plane plus two accumulation points
  • Sequence of intervals plus two accumulation points (a little shorter, and descriptive enough?)

prabau avatar Jun 04 '25 18:06 prabau

https://math.stackexchange.com/a/5072215/476484 I've also constructed a locally compact Hausdorff space for any possible triples of cardinals for which a space with given amount of quasi-components, components and path-components exists.

@Moniker1998 Nice result.

prabau avatar Jun 04 '25 18:06 prabau

Cool! I wonder how to express that in pi-base properties.

@hew-wolff Note that not all topological facts and results need to be part of pi-base in full generality. pi-base will display interesting examples and counterexamples and theorems of the form (bunch of properties imply some other property), and people can follow up references to dig deeper.

In any case, I would add the property "Has connected quasi-components" with related theorems, and also (in a separate PR) add your example, and possibly another PR for an example with a quasi-component splitting into uncountably many components.

Would you be interested in writing a PR about that?

prabau avatar Jun 04 '25 20:06 prabau

Cool! I wonder how to express that in pi-base properties.

I don't think it can be 😛 but I still wanted to share, since I've noticed its probably your question as well

Moniker1998 avatar Jun 04 '25 21:06 Moniker1998

In any case, I would add the property "Has connected quasi-components" with related theorems, and also (in a separate PR) add your example, and possibly another PR for an example with a quasi-component splitting into uncountably many components. Would you be interested in writing a PR about that?

Yes but I don't know when I can get to it.

hew-wolff avatar Jun 05 '25 21:06 hew-wolff

I like "Has connected quasi-components"

StevenClontz avatar Jun 05 '25 21:06 StevenClontz

possibly of interest: https://math.stackexchange.com/questions/4474874/topological-spaces-whose-quasi-components-are-connected

prabau avatar Sep 29 '25 02:09 prabau