Timeline for How to define compatible topology for first-order structures?
Current License: CC BY-SA 3.0
6 events
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| Apr 13, 2017 at 12:57 | history | edited | CommunityBot |
replaced http://mathoverflow.net/ with https://mathoverflow.net/
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| Aug 31, 2013 at 9:27 | vote | accept | Thomas Klimpel | ||
| Aug 30, 2013 at 19:42 | history | edited | Joseph Van Name | CC BY-SA 3.0 |
added 8460 characters in body
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| Aug 30, 2013 at 2:26 | comment | added | Joseph Van Name | And the connection between the product topology and Hausdorff spaces without order is also simple since a space is Hausdorff if and only if the diagonal is closed in $X\times X$, so this result on ordered topological spaces is a generalization of the characterization on Hausdorff spaces. | |
| Aug 29, 2013 at 16:26 | comment | added | Thomas Klimpel | Did you have some sleep? Independent of whether you will finish editing the answer or just remove that line, I have a question. I thought about the second paragraph now and agree, except for the sentence "This is equivalent to saying $\leq$ is closed in $X^2$." I don't say that it is false, just that I don't understand it yet. To simplify, assume we say that $=$ is closed in $X^2$. Since this is a special case of your statement, this should be equivalent to $X$ being Hausdorff. I remember that there was some connection between the product topology and Hausdorff spaces, but was it so simple? | |
| Aug 28, 2013 at 4:45 | history | answered | Joseph Van Name | CC BY-SA 3.0 |