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May 26, 2021 at 15:31 vote accept usul
Nov 7, 2014 at 11:31 comment added Joel David Hamkins Yes, I agree. But there is an analogous result, however, for at-most $d$-branching trees $T\subset d^{<\omega}$, using $d^n$ in place of $2^n$.
Nov 7, 2014 at 7:52 comment added Douglas Zare There is an assumption in your first theorem that is not present in the question. Your assumption that there are at most $2$ children of each node makes the question more interesting. Without this, it might be that only the first node has multiple children, in which case the number of branches is countable.
Nov 7, 2014 at 2:41 history edited Joel David Hamkins CC BY-SA 3.0
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Nov 7, 2014 at 2:33 history edited Joel David Hamkins CC BY-SA 3.0
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Nov 7, 2014 at 2:15 comment added Joel David Hamkins There is an unsettled middle ground for growth rates between $o(2^n)$ and $\Omega(2^n)$, and it isn't clear to me what is going on there.
Nov 7, 2014 at 2:05 history edited Joel David Hamkins CC BY-SA 3.0
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Nov 7, 2014 at 1:55 history edited Joel David Hamkins CC BY-SA 3.0
More precise positive and negative results
Nov 7, 2014 at 0:55 history edited Joel David Hamkins CC BY-SA 3.0
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Nov 7, 2014 at 0:50 history edited Joel David Hamkins CC BY-SA 3.0
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Nov 6, 2014 at 22:51 history answered Joel David Hamkins CC BY-SA 3.0