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 Oct 14 comment Why is the intersection of countably many homogeneously Suslin subsets of $\,^{\omega} \omega$ homogeneously Suslin? You are right, that doesn't seem to be the same. I don't know if the proof would be different. Still, in "The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal", Lemma 2.11. on p. 27, the statement appears for homogeneously Suslin sets. Oct 12 comment Show that $\{ a_n \}$ converges. Prove that $a_n$ is increasing. This can be done by using $a_n < 2$, which you already proved. Oct 2 comment There are 4 girls and 3 boys but only 5 seats. How many ways to seat the 3 boys together? Seems correct to me. Sep 18 comment Are there meaningful elementary embeddings of transitive models of set theory without a critical point? Oh, Sorry, I meant that $j$ does not map every element of $M$ to itself, i.e. is not inclusion, but I wrote $j\neq id$. I'm going to change it. Sep 15 comment 1. Measurable sets : uncountable union, 2. null set is disjoint from translated rationals Um, I can't see anything wrong with my post as it is now though. Sep 15 comment 1. Measurable sets : uncountable union, 2. null set is disjoint from translated rationals Note that I was talking about the set $\textbf{remaining}$ after subtracting the obviously open part, as in Both Htob's approach. Sep 15 comment 1. Measurable sets : uncountable union, 2. null set is disjoint from translated rationals We have $A+1 = \{ x+1 \, |\, x\in A\} = \bigcup_{x\in A} \{x+1\}$ and $A-1 = \{ x-1 \, |\, x\in A\} = \bigcup_{x\in A} \{ x-1\}$. These are the definitions of the translations of $A$ by $1$ and $-1$ respectively. Then clearly $\bigcup_{x\in A} \{x+1,x-1\} = \bigcup_{x\in A} \{x+1\} \cup \bigcup_{x\in A} \{x-1\} = (A+1) \cup (A-1)$. Sep 15 comment Sum from $i$ to $\lg(n)$ Well if $m = 2^n$, then by definition of the logarithm $n = log_2(m)$. So if $t_{2^n}$ can be decribed with some formula depending with variable $n$, $t_{2^n} = f(n)$, then $t_m = f(log_2(m))$. So I replaced every instance of $n$ in the formula $f(n)$ by $log_2(n)$ and simplified. Sep 14 comment Intersection of Borel subset with line Yes, that's it. Sep 13 comment Intersection of Borel subset with line The statement in that link is a different one. If $A\subseteq \mathbb{R}^2$ is Borel, then it is not in general true that $\{ x\in \mathbb{R} \, |\, \exists y\in \mathbb{R} (x,y)\in A\}$ is Borel. The proof is not (very) simple. But it is true that for any specific $y\in \mathbb{R}$ the set $\{ x\in \mathbb{R} \, |\, (x,y)\in A\}$ is Borel. Do you see the difference? Sep 13 comment Intersection of Borel subset with line If you prove that (in your first comment), you prove that each Borel set of $\mathbb{R}^2$ is of type $A(y)$ for some Borel set $A$ and $y\in \mathbb{R}$. That's not the same as proving that each $A(y)$ is Borel. The point of my proof is to show that every Borel set $A$ is an element of $M$. Then by definition of $M$, it follows that $A(y)$ is Borel for all $y\in\mathbb{R}$. Here is another proof: Given $A$ Borel and $y\in\mathbb{R}$ consider the function $x\mapsto (x,y)\in\mathbb{R}^2$. It is continuous. Then if you know that the preimage of a Borel set is Borel, you are also done. Sep 13 comment Sum from $i$ to $\lg(n)$ @sonic: You are welcome. marty:The $t_n$ are only defined for $n$ a power of 2. Sep 13 comment Nested interval question involving limits Assume that $a0$ so that $a+\varepsilon < c$. Can you use the definition of convergence to obtain a contradiction? Sep 13 comment Can I determine the groups and homomorphisms? What are your ideas? The first and last lines seem pretty easy to figure out. Sep 13 comment Intersection of Borel subset with line Actually, now I think the version with starting with open cubes is the easiest. The thing about the title is that $A(y)$ is not actually intersection with a line (that would be trivially Borel), but intersection followed by projection to one of the coordinates, which is definitely more subtle. This kind of measurability result is needed for the statement of the Fubini Theorem. Sep 13 comment $\overline{E}\subset \bigcup_{x\in E}\overline{B_x(r_x)}$ if $E\subset \bigcup_{x\in E}\overline{B_x(r_x)}$ In that case $\overline E = [0,1]$ is covered by the union if we choose the $r_x$ for $x\in (0,1)$ in the same way as above. But we could also take an enumeration of $E\cap (0,1) = \{ x_1,x_2,x_3,.. \}$ and let $r_{x_i}' = \frac{r_{x_i}}{2^{i+1}}$ as well as $r_{0} = r_{1} = \frac{1}{8}$ Then the union of the $\overline {B_{x}(r_x')}$ has measure at most $\frac{3}{4}$ and thus cannot cover $[0,1]$. Sep 13 comment Intersection of Borel subset with line Absolutely. Thanks for telling me, I corrected it. Sep 13 comment Intersection of Borel subset with line Right, it is a bit easier to start with closed sets rather than with open for the proof, so I changed that. Sep 13 comment Intersection of Borel subset with line @Carl Mummert: Please have a look at the question again. It isn't asking about the projection of the Borel set, but about the sets $A(y)$ for various $y\in\mathbb{R}$, there is a difference. Sep 11 comment Partial derivative and dependent variables The function $f$ might be given by a formula that technically (mathematically) still makes sense for such $a,b$ that do not satisfy the constraint. Then $\partial f /\partial a$ would be the local rate of change of the value of $f$ if we just increase $a$ independently. That doesn't give us any result for the context of the constraint, but we could compute the directional derivative, which is the gradient of f multiplied with a unit vector indicating a direction. This gives us the rate of change of $f$ if we change $a$ and $b$ along the line with that direction-this makes sense in context.