Timeline for Checkmate in $\omega$ moves?
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| S Mar 11, 2019 at 9:15 | history | suggested | Glorfindel | CC BY-SA 4.0 |
broken image fixed (click 'rendered output' or 'side-by-side' to see the difference); for more info, see https://meta.mathoverflow.net/a/4058/70594
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| Mar 11, 2019 at 8:30 | review | Suggested edits | |||
| S Mar 11, 2019 at 9:15 | |||||
| Jan 23, 2012 at 0:16 | comment | added | Joel David Hamkins | Noam, in this and your other answers, you should write $\omega3$ and $\omega4$, rather than $3\omega$ and $4\omega$, since in the usual notation, which goes back to Cantor, we have $\omega\cdot 3=\omega+\omega+\omega$, but $3\omega=3+3+\cdots=\omega$, which is not what you mean. | |
| Jan 5, 2012 at 21:59 | comment | added | Noam D. Elkies | $\omega$ means something much more precise and specific than "an arbitrarily large integer". | |
| Aug 19, 2011 at 17:57 | comment | added | user13113 | While black does have two chances to make the game arbitrarily long, it's still a mate in $\omega$, because the sum of two arbitrarily large integers is still just an integer. | |
| May 4, 2011 at 3:39 | history | edited | Noam D. Elkies | CC BY-SA 3.0 |
Added (link to) diagram of the "mate in $2\omega$
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| May 3, 2011 at 3:25 | history | answered | Noam D. Elkies | CC BY-SA 3.0 |