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can anyone tell me, please, two ultrafilters such that $\mathcal{U}\otimes\mathcal{V}\neq\mathcal{V}\otimes\mathcal{U}$ and others two such that $\mathcal{U}\oplus\mathcal{V}\neq\mathcal{V}\oplus\mathcal{U}$?. Thanks

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Could you explain, what $\otimes$ and $\oplus$ mean? (Or provide some link to a text, where they are defined.) –  Martin Sleziak Apr 5 '11 at 19:22
    
$A\in\mathcal{U}\otimes\mathcal{V}$ if and only if $\{i:\{j:(i,j)\in A\}\in\mathcal{V}\}\in\mathcal{U}$ –  Jacob Fox Apr 5 '11 at 19:26
    
$A\in\mathcal{U}\oplus\mathcal{V}$ if and only if $\{n:A-n\in\mathcal{V}\}\in\mathcal{U}$ (they are ultrafilters on $\mathbb{N}$) –  Jacob Fox Apr 5 '11 at 19:27
    
Presumably the second one is meant to read $\mathcal{U}\oplus\mathcal{V}\neq \mathcal{V}\oplus\mathcal{U}$? –  joriki Apr 5 '11 at 19:31
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@Jacob: Really? I may be completely off here, but it seems to me that if $\mathcal{U}$ and $\mathcal{V}$ are the ultrafilters that contain the subsets of $\omega$ that contain $0$ and $1$ respectively then $A\in\mathcal{U}\otimes\mathcal{V}\iff (0,1)\in A$ while $A\in\mathcal{V}\otimes\mathcal{U}\iff (1,0)\in A$. What do I fail to see? –  Apostolos Apr 5 '11 at 19:50

2 Answers 2

up vote 3 down vote accepted

For $\mathcal{U}\oplus\mathcal{V}$:

Let $X=\{x_1,x_2,\dots\}$ and $Y=\{y_1,y_2,\dots\}$ be two infinite subsets of $\mathbb{N}$ with this property: $A=\{x_i+y_j;\,i<j\}$ and $B=\{x_i+y_j;\,i>j\}$ satisfy $A\cap B=\emptyset$. For example, $x_i=2^{2i}$ and $y_i=2^{2i+1}$ do satisfy this.

Let now $U=\{C\subset\mathbb{N};\,X\backslash C\text{ is finite}\}$. $U$ is a filter, so let us enlarge it to an ultrafilter $\mathcal{U}$. Similarly we construct $V$ and $\mathcal{V}$ (using $Y$ in place of $X$).

We now have $A\in\mathcal{U}\oplus\mathcal{V}$ (since for every $x_i$, the set $A-x_i$ contains as a subset $\{y_{i+1},y_{i+2},\dots\}$, so $A-x_i\in V\subset\mathcal{V}$). Similary, $B\in\mathcal{V}\oplus\mathcal{U}$. Since $A\cap B=\emptyset$, we have $\mathcal{U}\oplus\mathcal{V}\neq \mathcal{V}\oplus\mathcal{U}$.

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For $\mathcal U\otimes \mathcal V$, let $\mathcal U=(0)$ (principal), $\mathcal V$ -- any other. Then $A=\mathbf N\times \lbrace 0\rbrace$ is not in $\mathcal U\otimes \mathcal V$, (since all vertical sections of $A$ are $\lbrace 0\rbrace\notin \mathcal V$), but is in $\mathcal V\otimes \mathcal U$ (since all vertical sections of $A$ contain $0$).

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