Timeline for Connectedness in the language of path-connectedness
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
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| May 13, 2016 at 15:15 | history | edited | Helene Sigloch | CC BY-SA 3.0 |
Added reference to research related to the question.
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| Jun 23, 2015 at 7:55 | vote | accept | Dominic van der Zypen | ||
| Jun 23, 2015 at 7:55 | |||||
| Mar 25, 2015 at 6:50 | vote | accept | Dominic van der Zypen | ||
| Jun 23, 2015 at 7:54 | |||||
| Mar 24, 2015 at 14:28 | vote | accept | Dominic van der Zypen | ||
| Mar 25, 2015 at 6:49 | |||||
| Mar 24, 2015 at 12:57 | comment | added | Helene Sigloch | $f$ is required to be continuous and to hit both $x$ and $y$. That means it has to hit the whole line segment between $x$ and $y$. | |
| Mar 24, 2015 at 12:54 | history | edited | Helene Sigloch | CC BY-SA 3.0 |
adjusted to the comments
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| Mar 24, 2015 at 12:52 | comment | added | LSpice | Why does the existence of many path components of $X$ imply as many of $C$ when $f$ is not required to be surjective? | |
| Mar 24, 2015 at 12:48 | comment | added | Helene Sigloch | @Goldstern Right. For every topological space there exists a connected topological space that it doesn't surject to. This is a nice simple argument. You should make it an answer. | |
| Mar 24, 2015 at 12:26 | comment | added | Goldstern | Very long lines exist. Let $\kappa$ be any cardinal (viewed as an ordinal). For each $\beta$ in $\kappa$ add a copy of the unit interval between $\beta$ and $\beta+1$, plus a point $\infty$. The resulting linear order is dense and complete. But if $C$ has cardinality smaller than $\kappa$, then any continuous image of $C$ that contains $0$ and $\infty$ will not be onto, hence not connected. | |
| Mar 24, 2015 at 10:48 | history | edited | Helene Sigloch | CC BY-SA 3.0 |
I think I repaired the example now.
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| Mar 24, 2015 at 10:32 | history | edited | Helene Sigloch | CC BY-SA 3.0 |
admitted that it doesn't work
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| Mar 24, 2015 at 10:07 | comment | added | Helene Sigloch | We are looking for a universal space $C$. If we can choose $X$ to have an arbitrarily large number of path components, $C$ also needs to have an aribtrarily large number of path components. Thus, in particular an arbitrarily large number of points. But we are looking for a single universal space $C$ that does it for all $X$. This $C$ would have to have a bigger number of points than any cardinal. Thus, it's not a set. | |
| Mar 24, 2015 at 10:03 | comment | added | Helene Sigloch | I tried to construct an example of a space $X$ where each open set containing $x$ and $y$ has to contain the "open interval" between $x$ and $y$ to have something precise. If now the "open interval" has a certain number of path components, its preimage needs to have the same number of path components by the very first argument in my answer. | |
| Mar 24, 2015 at 9:59 | comment | added | Helene Sigloch | Suppose you have such a space $(C,\tau ,c_0,c_1 )$ and map $f : C \rightarrow X$. | |
| Mar 24, 2015 at 9:47 | comment | added | Dominic van der Zypen | I'm not sure I follow your argument. I think it has to be shown that for any space $(C,\tau_C)$ and points $c_0,c_1\in C$ there is a space $(X,τ)$ that is connected, but not "$(C,c_0,c_1)$-connected", or vice versa. | |
| Mar 24, 2015 at 8:43 | history | answered | Helene Sigloch | CC BY-SA 3.0 |