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 Yearling
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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 $n\neq m$.