Timeline for Does existence of midpoints imply intrinsic?
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 19, 2017 at 21:19 | vote | accept | erz | ||
| Jan 19, 2017 at 16:44 | answer | added | Will Brian | timeline score: 7 | |
| Jan 13, 2017 at 22:41 | history | edited | erz | CC BY-SA 3.0 |
Deleted an incorrect example.
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| Jan 13, 2017 at 22:33 | comment | added | erz | Yes, you are right, the example in invalid. | |
| Jan 13, 2017 at 8:39 | comment | added | Benoît Kloeckner | Why is $[-1,1]\times [0,1]\setminus \{(0,0)\}$ with the $L^\infty$ norm not strictly intrinsic? The curves that are piecewise affine with slope at most one have length equal to the distance between their endpoints. Did you want to use either a rotated rectangle or the $L^1$ metric? | |
| Jan 13, 2017 at 5:42 | comment | added | erz | you have identified the metric correctly. The locally compact example is such because it is just a closed rectangle minus a point (the topology in this metric is the same as with the Euclidean); the last example is not locally compact, and there was a slight error in it (which I fixed now). | |
| Jan 13, 2017 at 5:40 | history | edited | erz | CC BY-SA 3.0 |
fixed one example
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| Jan 12, 2017 at 19:18 | comment | added | Ali Taghavi | what is the $L_{\infty}$ metric? do you mean $\parallel \; \parallel_{\infty}$ on the plane? If yes, why your example in locally compact(According to your edit) | |
| Jan 12, 2017 at 2:53 | history | edited | erz | CC BY-SA 3.0 |
added a counterexample and changed the question accordingly
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| Jan 12, 2017 at 2:48 | history | edited | erz | CC BY-SA 3.0 |
added a counterexample and changed the question accordingly
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| Jan 11, 2017 at 2:20 | history | edited | erz | CC BY-SA 3.0 |
Found a counterexample to the second question. Modified accordingly.
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| Jan 8, 2017 at 7:51 | history | asked | erz | CC BY-SA 3.0 |