Paul McKenney
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 Mar 15 comment Constructing sets of certain measure from classes of bijections on the continuum In 2 and 3, do you also want these sets to be Lebesgue measurable, or do you just want them to have positive outer measure? (Measurability seems like it might be a pretty strict requirement in the case of 3.) Feb 29 comment Error bound in the sum of chords approximation to arc length That's not what I'm asking for. I'm asking for the bound on the error using the sum of chords rule. Feb 27 asked Error bound in the sum of chords approximation to arc length Feb 24 comment Problem on infinite cardinal number You should understand without too much difficulty that it reduces to the case $d = 2^e$. Then you need to find a bijection between $(2^e)^e$ and $2^e$. Feb 1 comment Uppercase E notation for sets? An 'E' with brackets commonly denotes an expected value, as from probability theory. Given that they are writing a $u$ in the subscript, they could be defining $L_i$ to be the expected value of $u$ over some event, and similarly for $K$. Jan 22 comment Finding all metrics of set $X=\{1,2,3\}$ You're right that for any metric $d$, a scaling of $d$ by a positive constant is also a metric, so there are infinitely many. I think the point of the question is that you can list them all using only finitely many constants. Jan 22 answered Clopen subspaces of Stonean spaces Jan 20 comment How can it be seen that the Von Neumann universe $V_{\omega+\omega}$ does not model the Fraenkel axiom Thanks Asaf, I knew I was saying something wrong, but I posted anyway. At least now we have some more interesting stuff written here. Jan 20 revised How can it be seen that the Von Neumann universe $V_{\omega+\omega}$ does not model the Fraenkel axiom pointed out a false thing. Jan 20 answered How can it be seen that the Von Neumann universe $V_{\omega+\omega}$ does not model the Fraenkel axiom Jan 20 comment How can it be seen that the Von Neumann universe $V_{\omega+\omega}$ does not model the Fraenkel axiom Look at the function $f : \omega \to \omega + \omega$ defined by $f(n) = \omega + n$. Jan 7 answered There is no Baire bijection between $\mathbb R$ and the set of functions $\mathbb Z\to\mathbb R$ modulo shifts Dec 22 awarded Yearling Nov 7 awarded Nice Answer Sep 15 comment Cardinality of $Def(X)$ Choice probably only appears in showing that the set of finite tuples $(x_0,\ldots,x_m)$ from $M$ has the same cardinality as $M$. Sep 5 comment Every $f\in\omega^\omega$ is bounded by the “increasing enumeration” of the intersection of a countable dense set and a dense open set in $\mathbb{R}$ I'm confused. Isn't it true that if $Y$ and $Z$ are subsets of $X$ and $Y\subseteq Z$, then $f_Z \le f_Y$? Then letting $U_n$ ($n < \omega$) be a countable base for $\mathbb{R}$, we have for every open $V$ that there exists an $n$ such that $f_{V\cap X} \le f_{U_n\cap X}$. Then letting $g$ be a $<^*$-bound on the functions $f_{U_n\cap X}$, we have $f_{V\cap X} <^* g$ for all open $V$. Sep 2 comment Limiting the size of near-coherence classes in $\omega^*$ It's pretty standard, Asaf. Aug 23 answered extending automorphisms in complete boolean algebras Aug 22 comment extending automorphisms in complete boolean algebras @Stefan: Well, every automorphism $f$ of $B$ must permute the atoms, which are just the singletons. So we get a permutation $\pi$ of $\mathbb{N}$. It's then easy to see that $f(A)$ is equal to the image of $A$ under $\pi$, for any set $A$. Aug 22 comment extending automorphisms in complete boolean algebras This won't be true for all CBA's. Here's the example I'm thinking of. Let $B = P(\mathbb{N})$. Let $a_n$ ($n\in\mathbb{N}$) be a partition of $\mathbb{N}$ with $|a_n| = n$, and let $A$ be the set of all possible unions of the $a_n$'s. Then each $a_n$ is an atom in $A$, and you can define an automorphism of $A$ by permuting the $a_n$'s however you like. But every automorphism of $B$ is induced by a permutation, and hence can't map any $a_n$ to $a_m$ for any $n\neq m$.