Timeline for Example of a weak Hausdorff space that is not Hausdorff?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 15, 2012 at 9:35 | vote | accept | Bob Solovay | ||
| Feb 15, 2012 at 9:29 | history | edited | Stephen S | CC BY-SA 3.0 |
add another example
|
| Feb 15, 2012 at 9:05 | comment | added | Stephen S | @Bob Solovay: Yemon Choi's definition of the one-point compactification is correct. @Yemon Choi: But your second comment is wrong, since $\mathbb{Q}$ has infinite compact subsets (e.g., any convergent sequence together with its limit forms a compact set). But every compact subset of $\mathbb{Q}$ has empty interior, so it's true that every neighbourhood of the point at infinity meets every non-empty open set. | |
| Feb 15, 2012 at 3:48 | comment | added | Yemon Choi | Thus, if I have understood things correctly: when we give Q its subspace topology from R (thus not locally compact) we find that the open neighbourhoods of the point at infinity correspond to cofinite subsets of Q, and hence they meet every non-empty open subset of Q. (Open subsets of Q in this topology are either empty or infinite.) | |
| Feb 15, 2012 at 3:46 | comment | added | Yemon Choi | @BobSolovay: I was not sure of this myself, but consulting Kelley's book tells me that for any top. space X the 1-point compactification X* has as its open sets all the open subsets of X, together with all subsets of X* whose complements are closed compact subsets of X. (TBC) | |
| Feb 15, 2012 at 2:21 | comment | added | Bob Solovay | I don't know what you mean by "the one point compactification of Q". Of course, I am familiar with the one point compactification of a locally compact space. What is the topology of this space. What are two points that don't have disjoint neighborhoods? | |
| Feb 14, 2012 at 11:38 | history | answered | Stephen S | CC BY-SA 3.0 |