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I was reading this paper on an algorithm for finding a dominating set in a tournament graph. The paper claims that an $n$-element set has $n^{O(\log n)}$ subsets of size at most $\log n$. The paper doesn't offer any proof of this, and I don't know how to prove that this is true.

Is there a straightforward way to show this?

Thanks!

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    $\begingroup$ I think it actually claims that an $n$-element set has $n^{O(\log n)}$ subsets of size at most $\log n$ (which is obviously true). The | sign appears to correspond to ^ in LaTeX. $\endgroup$ – Christopher May 7 '13 at 15:28
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    $\begingroup$ Oh wow, I miswrote the original question. That's definitely supposed to be $n^{O(\log n)}$. Sorry about that! $\endgroup$ – templatetypedef May 7 '13 at 15:31
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If we pull with replacement, there are $n^{\log n}$ possibilities for a draw of size $\log n$. Pulling without replacement to get a subset will decrease the possibilities. Then if we sum for the number of elements from $1$ to $\log n$ we get $\frac {n^{1+log n}-1}{n-1}\in O(n^{\log n})$

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  • $\begingroup$ My apologies... I wrote the wrong value in the question. But thanks for this answer! $\endgroup$ – templatetypedef May 7 '13 at 15:32
  • $\begingroup$ @templatetypedef: new question answered $\endgroup$ – Ross Millikan May 7 '13 at 15:42
  • $\begingroup$ Beautiful... that was very straightforward. Thanks! $\endgroup$ – templatetypedef May 7 '13 at 15:44
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You can write this number $\sum_{k=0}^{\log n} {n \choose k} \leq (\log n + 1) {n \choose \log n}$, and try to approximate this last term.

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  • $\begingroup$ How would you approximate that term? I was able to get to the point you're at now but am totally stuck on how to proceed. $\endgroup$ – templatetypedef May 7 '13 at 15:36
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    $\begingroup$ There must be something more elegant, but using Stirling's approximation should probably get you somewhere. $\endgroup$ – Clement C. May 7 '13 at 15:36

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