Timeline for Maximal size of an almost-disjoint linearly independent family in $K^{\mathbb{N}}$
Current License: CC BY-SA 3.0
14 events
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| Aug 19, 2013 at 15:36 | comment | added | Asaf Karagila♦ | Cross-posted: math.stackexchange.com/questions/471292/… | |
| Aug 19, 2013 at 15:29 | history | edited | user38700 |
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| Aug 19, 2013 at 11:13 | history | edited | user38700 | CC BY-SA 3.0 |
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| Aug 19, 2013 at 11:12 | history | edited | Ricardo Andrade |
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| Aug 19, 2013 at 11:09 | history | edited | user38700 | CC BY-SA 3.0 |
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| Aug 17, 2013 at 16:54 | comment | added | Joel David Hamkins | Yes, I see. Perhaps one can hope to modify the proof of Erdos-Kaplansky to answer the general case of your question. | |
| Aug 17, 2013 at 16:48 | comment | added | user38700 | Excuse me, I meant it for an infinite set $I$. So again, Erdos-Kaplansky theorem states that for any field $K$ and any infinite set $I$, the linear dimension of the vector space $K^I$ is equal to its cardinality: $dim_K(K^I)=|K^I|$ | |
| Aug 17, 2013 at 16:37 | comment | added | Joel David Hamkins | That can't be correct the way you say it, since for example $I$ could be finite. | |
| Aug 17, 2013 at 16:20 | comment | added | user38700 | The Erdos-Kaplansky theorem states that for any field $K$ and any set $I$, the linear dimension of the vector space $K^I$ is equal to its cardinality: $dim_K (K^I)=|K^I|$. | |
| Aug 17, 2013 at 12:26 | comment | added | Joel David Hamkins | Forgive me, but what is the dimension of $K^{\mathbb{N}}$ as a $K$-vector space? (Of course at least continuum, and exactly continuum when $K$ has size at most continuum; but what is it when $K$ is larger?) | |
| Aug 17, 2013 at 11:48 | history | edited | user38700 |
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| Aug 16, 2013 at 18:46 | answer | added | Joel David Hamkins | timeline score: 3 | |
| Aug 16, 2013 at 18:04 | review | First posts | |||
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| Aug 16, 2013 at 17:49 | history | asked | user38700 | CC BY-SA 3.0 |