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I'm reading a book on ultrafilters and it tells me that showing addition on ultrafilters over the naturals (addition on the set of ultrafilters '$\beta \mathbb{N}$', defined in the standard way) is not commutative is a 'nice exercise'. I am curious to see a proof: however, this result isn't particularly important for what I need ultrafilters for, I don't really want to spend a long time trying to obtain the result.

I have seen that addition is left-continuous, so I suppose the equivalent statement is that addition is not right-continuous (as it would be if $+$ was commutative) though I suspect non-right-continuous and non-commutative immediately follow from one another; could anyone perhaps direct me to a source with a proof that addition is not commutative on $\beta \mathbb{N}$, if they know one? (Or perhaps if the proof is short and it's no more trouble, you could explain the proof yourself - I am happy with either).

Many thanks,

  • 1
    You might check Theorem 4.27 in Hindman, Strauss: Algebra in the Stone-Cech compactification, [p.81](http://books.google.com/books?id=KYXgdiegKDsC&pg=PA81).2011-11-30
  • 0
    Just out of curiosity: What *book on ultrafilters* are you reading?2012-03-19
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    BTW this is in fact part of [this question](http://math.stackexchange.com/questions/31147/sum-and-product-of-ultrafilters).2012-06-20

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