Timeline for The dual group of $\mathbb Q$
Current License: CC BY-SA 2.5
14 events
| when toggle format | what | by | license | comment | |
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| Aug 10, 2010 at 12:45 | comment | added | Gerald Edgar | @Hany: In your topology, $\mathbb{N}\cup\{\infty\}$ is compact, right? How about the base of neighborhoods of $\infty$ consists of the co-countable subsets? | |
| Aug 6, 2010 at 9:02 | comment | added | Hany | @Victor: Consider $X={\mathbb R}\cup\{\infty\}$ with the topology consisting of the discrete topology on $\mathbb R$ and with a base of neighborhoods at $\infty$ consisting of intervals $]k,\infty]$ with $k\in{\mathbb N}$. Compact sets in this space are finite, but the space is not discrete. | |
| Aug 4, 2010 at 13:42 | comment | added | Gerald Edgar | @Victor: A Hausdorff topological space is discrete if and only if the only compact subsets are finite. I don't believe it. First-countable, OK, but in general for Hausdorff? | |
| Aug 4, 2010 at 9:00 | comment | added | BS. |
It is easy to see that if $x$ is real and $|\exp(ix/n)-1|<\epsilon$ for all integers $n\geq1$, then for small enough $\epsilon$ ($\epsilon<1/\sqrt{2}$ should work), $x$ must be small : $|x|<2\epsilon/\pi$ (hint: consider $k\in\mathbb{Z}$ such that $|x-k\pi|\leq \pi/2$, and take $n=|k|$ if $k\neq 0$). This implies that uniform convergence on compact subsets of $\mathbb{Q}$ (in fact one compact subset) induces the standard topology on $\mathbb{R}\simeq\{t\mapsto\exp(ixt)\}_{x\in\mathbb{R}}$.
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| Aug 4, 2010 at 0:27 | comment | added | Victor Protsak | This is an artifact of the difference between the discrete topology and the induced topology of $\mathbb{Q}.$ A Hausdorff topological space is discrete if and only if the only compact subsets are finite. | |
| Aug 3, 2010 at 17:13 | comment | added | Gerald Edgar | @Johannes: Is that already an answer in the "common misconceptions" question? If not it should be. | |
| Aug 3, 2010 at 15:24 | history | edited | Torsten Ekedahl | CC BY-SA 2.5 |
deleted 474 characters in body
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| Aug 3, 2010 at 14:24 | comment | added | BS. | @Johannes : $\mathbb{Q}$ has infinite compact subsets, like a sequence converging to a rational and its limit. | |
| Aug 3, 2010 at 12:57 | comment | added | Johannes Hahn | Compact subsets of $\mathbb{Q}$ are finite. So uniform convergence on compact subsets of $\mathbb{Q}$ ist the same as convergence in all rational points. | |
| Aug 3, 2010 at 10:50 | history | edited | Torsten Ekedahl | CC BY-SA 2.5 |
Added note on topology.
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| Aug 2, 2010 at 18:43 | vote | accept | Hany | ||
| Aug 7, 2010 at 11:29 | |||||
| Aug 2, 2010 at 12:58 | comment | added | Torsten Ekedahl | Which is the argument of my reference (chosen because it is where uniformities are introduced)... | |
| Aug 2, 2010 at 12:32 | comment | added | Gerald Edgar | A continuous group homomorphism is of course uniformly continuous, thus extends to the completions. | |
| Aug 2, 2010 at 12:25 | history | answered | Torsten Ekedahl | CC BY-SA 2.5 |