Timeline for Abstract connectedness
Current License: CC BY-SA 3.0
32 events
| when toggle format | what | by | license | comment | |
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| Mar 8, 2015 at 7:07 | comment | added | Włodzimierz Holsztyński | Perhaps the empty set should be considered neither connected nor disconnected. | |
| Mar 8, 2015 at 7:05 | comment | added | Włodzimierz Holsztyński | Mike, nice, yes, prime numbers :-) | |
| Mar 8, 2015 at 5:32 | comment | added | Mike Shulman | (1) the same reasons that 1 is not a prime number. For instance, if $\emptyset$ is connected then the decomposition of a space into connected components is not unique since you can always add more copies of $\emptyset$. (2) What motivated you to write down these axioms? How did you think of them? | |
| Mar 7, 2015 at 6:30 | comment | added | Włodzimierz Holsztyński | Mike, about your (1) above, what would be your reason to let $\ \emptyset\ $ be disconnected? About (2), could you be a bit more specific? About (3), please, I'll try to say something later (if I can :-). | |
| Mar 7, 2015 at 5:06 | comment | added | Mike Shulman | This is very interesting, although it will take me some time to digest it. Three comments/questions for now: (1) As I said to Todd, I think the empty set should not be connected. (2) Can you say anything about where these axioms came from? (3) One possible downside of these axioms is that they aren't clearly "closure conditions", which means it's not obvious that the category of connectivity spaces will have initial and terminal structures, or even limits and colimits. Does it? | |
| Mar 6, 2015 at 22:11 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
cosmetic
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| Mar 6, 2015 at 21:02 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
another nasty big typo
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| Mar 6, 2015 at 20:13 | comment | added | Emil Jeřábek | Sorry to hear you’re unwell. Actually, I forgot to fix axiom 2 accordingly; one would need to either add the assumption $A\ne\emptyset$ there, or use Eric Wofsey’s suggestion. | |
| Mar 6, 2015 at 19:50 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
formatting typo
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| Mar 6, 2015 at 19:43 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
formatting micro-details
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| Mar 6, 2015 at 18:16 | comment | added | Włodzimierz Holsztyński | @EmilJeřábek -- you must be right, thank you; I am in poor shape, and it's hard for me to take care of pests like the empty set. If axiom 1 is not in error then hopefully I'd like to come back to it a bit later too make thing smoother. | |
| Mar 6, 2015 at 18:02 | comment | added | Emil Jeřábek | A simple way how to handle the empty set correctly is to drop the assumption $A\ne\emptyset\ne B$ from axiom 1. | |
| Mar 6, 2015 at 17:38 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
cosmetic
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| Mar 6, 2015 at 17:32 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
point-free (1-word explanation)
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| Mar 6, 2015 at 11:50 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
Example -- extra
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| Mar 6, 2015 at 11:44 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
Example -- explicit notation from the beginning
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| Mar 6, 2015 at 11:28 | comment | added | Włodzimierz Holsztyński | @EricWofsey -- I considered both (basically equivalent) versions. I felt that the 2-point version was somewhat simpler. I remembered a similar situation for a (generalized) convex structure--analogous two conditions occur there too (but the 2-point condition is stronger in the convex case). | |
| Mar 6, 2015 at 11:24 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
Remark aboit the empty set
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| Mar 6, 2015 at 11:11 | comment | added | Włodzimierz Holsztyński | @sure -- a non-continuous map from a circle to $\ \mathbb R\ $ or (even to the circle) may still have a connected graph. Connected graph is not a local property. But, as we saw, even a local connectedness of a graph doesn't force continuity. | |
| Mar 6, 2015 at 10:43 | comment | added | Eric Wofsey | An equivalent restatement of (2) (in the presence of (1)) that is perhaps more natural and also correctly handles the empty set is that a directed union of connected sets is connected. | |
| Mar 6, 2015 at 10:00 | comment | added | ACL | @sure. I had forgotten the precision “closed and connected“; I edited the comment. | |
| Mar 6, 2015 at 9:59 | comment | added | sure | @WłodzimierzHolsztyński: I guess you could take the topological sum to make sense of this, that is, connectivity "by part"? | |
| Mar 6, 2015 at 9:58 | comment | added | sure | @ACL: no there are no such obvious links. For example, the topologist sin curve is not at all continuous at 0, yet its graph is connected! Oh sorry, you said closed too | |
| Mar 6, 2015 at 9:56 | comment | added | ACL | @sure. This is a nice try, but it does not work. It is true that a function from $\mathbf R$ to itself is continuous if and only if its graph is closed and connected. However, I remember that there are counterexamples for functions from $\mathbf R^2$ to $\mathbf R$. | |
| Mar 6, 2015 at 9:54 | comment | added | Włodzimierz Holsztyński | @sure -- What about a disconnected $\ A\ $ ? | |
| Mar 6, 2015 at 9:39 | comment | added | Włodzimierz Holsztyński | @EricWofsey -- thank you again; I fixed the issue of the empty set. (Perhaps there are more things to fix), | |
| Mar 6, 2015 at 9:37 | history | edited | Włodzimierz Holsztyński | CC BY-SA 3.0 |
a math triviality fixed
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| Mar 6, 2015 at 8:57 | comment | added | sure | I always wondered what would happen if we took as definition of continuity, something like that: A function $f : A \rightarrow B$ is continuous iff its graph (or even its closure with product topology) $\Gamma(f)$ is connected in $A\times B$. This, in my opinion, captures the notion of continuity of "walking" of any ball of radius $\epsilon$, you stay inside your space (in the sense of metric space). | |
| Mar 6, 2015 at 8:51 | comment | added | Eric Wofsey | Note that these axioms do not imply singletons are connected, nor do they rule out having only singletons and the whole space being connected (which is impossible for topological spaces with more than two points by this answer). | |
| Mar 6, 2015 at 8:47 | comment | added | Włodzimierz Holsztyński | I was just about to fix this nasty typo. Thank you Eric! | |
| Mar 6, 2015 at 8:46 | history | edited | Eric Wofsey | CC BY-SA 3.0 |
fixed a couple obvious errors
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| Mar 6, 2015 at 8:36 | history | answered | Włodzimierz Holsztyński | CC BY-SA 3.0 |