Timeline for Classification of 1-dimensional manifolds (not second-countable)
Current License: CC BY-SA 3.0
13 events
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| Nov 7, 2013 at 12:00 | comment | added | Todd Trimble | @ziggurism In any case, the point is that Gerald Edgar's point in his comment is applicable to this case: such a manifold has to be second-countable, and so we can just invoke the usual classification of second-countable Hausdorff connected topological 1-manifolds (as in Gale's article). | |
| Nov 7, 2013 at 11:52 | comment | added | Todd Trimble | @ziggurism This is a sketch; I think Gale's article would have more details. There is a neighborhood of the point $y$ to be removed that is homeomorphic to a closed interval $I$. Let $x, z$ be the two endpoints of the interval. Now remove $y$; the result is connected, but manifolds are locally path connected, so the result is path connected. Hence there is a path $J$ connecting $x$ and $z$ in the complement of the point (and containing no interior point of $I$). I claim $I \cup J$ is the whole original manifold. But this union of two intervals is homeomorphic to $S^1$. | |
| Nov 7, 2013 at 0:47 | comment | added | ziggurism | @ToddTrimble: but how do you know that $S^1$ is the only 1-manifold which satisfies this property, that removing a point leaves one connected component? That seems like the most important part of the classification to me. | |
| Nov 6, 2013 at 3:07 | comment | added | Todd Trimble | @ziggurism Certainly you're right; thanks. Fixed. | |
| Nov 6, 2013 at 3:06 | history | edited | Todd Trimble | CC BY-SA 3.0 |
fixed a little glitch mentioned by Joe Hannon
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| Nov 5, 2013 at 22:51 | comment | added | ziggurism | @ToddTrimble: you start out with "let $Y$ be a connected nonempty topological 1-manifold. Then $Y-\{y\}$ has two components." I don't understand. $S^1$ is a connected nonempty topological 1-manifold, but $S^1-\{\text{pt}\}$ has one component...? | |
| Sep 8, 2012 at 14:30 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added 33 characters in body
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| Sep 8, 2012 at 14:04 | comment | added | Todd Trimble | @Andreas: oops, my bad (and thanks). I hope it's fixed now. | |
| Sep 8, 2012 at 14:01 | history | edited | Todd Trimble | CC BY-SA 3.0 |
deleted 341 characters in body
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| Sep 8, 2012 at 13:36 | history | edited | Todd Trimble | CC BY-SA 3.0 |
added 18 characters in body; added 14 characters in body
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| Sep 8, 2012 at 11:54 | comment | added | Andreas Blass | I don't understand the argument showing the every discrete subset $D$ of your space is well-ordered. Even when $X$ is $[0,1]$, there is a decreasing, discrete $\omega$-sequence. Perhaps you intended $D$ to also be closed? Also, this argument (applied to the special case $S=D$) seems to say that each point $s$ in such a $D$ has only finitely many predecessors, which would mean that $D$ has order-type at most $\omega$. | |
| Sep 8, 2012 at 10:12 | vote | accept | Martin Brandenburg | ||
| Sep 8, 2012 at 7:11 | history | answered | Todd Trimble | CC BY-SA 3.0 |