This set is due April 3 at the beginning of lecture. Details of the homework policy can be found on the syllabus and here.

1. Find where and determine all its subfields. Make sure you justify your answer. For example, if you state that two subfields and are different, you need to prove that this is indeed the case.

2. Do the same for

[Updated, April 2: I guess the hint I gave for problem 2 makes no sense, sorry about that. Rather, you may want to begin by looking at how factors. Then, to compute it may be helpful to look at a triangle with angles and ]

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[…] Homework 6, due April 3, at the beginning of lecture. Possibly related posts: (automatically generated)117b – Homework 5Buffalostyle Forges OnHomework battles and the biggest genius in the school, part IThe Myth About Homework […]

Concerning HW assignment 6, I am wondering if there is a typo in the hint you gave us for problem number 2. I may be wrong but I believe the denominator under the radical on the left hand side should be 18. Thanks!

Hi,
Okay…I am struggling with the second problem. I have solved the quartic down to w^6 and found my value for w. Then, when I plug everything back in, I cannot get any given u to solve the equation where u^3 + ….=0. Therefore I am second guessing everything I have done. And I appreciate the new hint, but I’m not sure how to apply it. I have worked the equation many times, each time hoping my calculations are wrong…unfortunately so far, they are not.
Any help would be greatly appreciated!

(As I pointed out in a comment) yes, partial Woodinness is common in arguments in inner model theory. Accordingly, you obtain determinacy results addressing specific pointclasses (typically, well beyond projective). To illustrate this, let me "randomly" highlight two examples: See here for $\Sigma^1_2$-Woodin cardinals and, more generally, the noti […]

I am not sure which statement you heard as the "Ultimate $L$ axiom," but I will assume it is the following version: There is a proper class of Woodin cardinals, and for all sentences $\varphi$ that hold in $V$, there is a universally Baire set $A\subseteq{\mathbb R}$ such that, letting $\theta=\Theta^{L(A,{\mathbb R})}$, we have that $HOD^{L(A,{\ma […]

A Wadge initial segment (of $\mathcal P(\mathbb R)$) is a subset $\Gamma$ of $\mathcal P(\mathbb R)$ such that whenever $A\in\Gamma$ and $B\le_W A$, where $\le_W$ denotes Wadge reducibility, then $B\in\Gamma$. Note that if $\Gamma\subseteq\mathcal P(\mathbb R)$ and $L(\Gamma,\mathbb R)\models \Gamma=\mathcal P(\mathbb R)$, then $\Gamma$ is a Wadge initial se […]

Craig: For a while, there was some research on improving bounds on the number of variables or degree of unsolvable Diophantine equations. Unfortunately, I never got around to cataloging the known results in any systematic way, so all I can offer is some pointers to relevant references, but I am not sure of what the current records are. Perhaps the first pape […]

Yes. Consider, for instance, Conway's base 13 function $c$, or any function that is everywhere discontinuous and has range $\mathbb R$ in every interval. Pick continuous bijections $f_n:\mathbb R\to(-1/n,1/n)$ for $n\in\mathbb N^+$. Pick a strictly decreasing sequence $(x_n)_{n\ge1}$ converging to $0$. Define $f$ by setting $f(x)=0$ if $x=0$ or $\pm x_n […]

All proofs of the Bernstein-Cantor-Schroeder theorem that I know either directly or with very little work produce an explicit bijection from any given pair of injections. There is an obvious injection from $[0,1]$ to $C[0,1]$ mapping each $t$ to the function constantly equal to $t$, so the question reduces to finding an explicit injection from $C[0,1]$ to $[ […]

One way we formalize this "limitation" idea is via interpretative power. John Steel describes this approach carefully in several places, so you may want to read what he says, in particular at Solomon Feferman, Harvey M. Friedman, Penelope Maddy, and John R. Steel. Does mathematics need new axioms?, The Bulletin of Symbolic Logic, 6 (4), (2000), 401 […]

"There are" examples of discontinuous homomorphisms between Banach algebras. However, the quotes are there because the question is independent of the usual axioms of set theory. I quote from the introduction to W. Hugh Woodin, "A discontinuous homomorphism from $C(X)$ without CH", J. London Math. Soc. (2) 48 (1993), no. 2, 299-315, MR1231 […]

This is Hausdorff's formula. Recall that $\tau^\lambda$ is the cardinality of the set ${}^\lambda\tau$ of functions $f\!:\lambda\to\tau$, and that $\kappa^+$ is regular for all $\kappa$. Now, there are two possibilities: If $\alpha\ge\tau$, then $2^\alpha\le\tau^\alpha\le(2^\alpha)^\alpha=2^\alpha$, so $\tau^\alpha=2^\alpha$. In particular, if $\alpha\g […]

Fix a model $M$ of a theory for which it makes sense to talk about $\omega$ ($M$ does not need to be a model of set theory, it could even be simply an ordered set with a minimum in which every element has an immediate successor and every element other than the minimum has an immediate predecessor; in this case we could identify $\omega^M$ with $M$ itself). W […]

[…] Homework 6, due April 3, at the beginning of lecture. Possibly related posts: (automatically generated)117b – Homework 5Buffalostyle Forges OnHomework battles and the biggest genius in the school, part IThe Myth About Homework […]

Concerning HW assignment 6, I am wondering if there is a typo in the hint you gave us for problem number 2. I may be wrong but I believe the denominator under the radical on the left hand side should be 18. Thanks!

Hi Tommy,

Hmm, yeah. I’m just about convinced now that the hint is nonsense. So, I have added another hint.

Hi,

Okay…I am struggling with the second problem. I have solved the quartic down to w^6 and found my value for w. Then, when I plug everything back in, I cannot get any given u to solve the equation where u^3 + ….=0. Therefore I am second guessing everything I have done. And I appreciate the new hint, but I’m not sure how to apply it. I have worked the equation many times, each time hoping my calculations are wrong…unfortunately so far, they are not.

Any help would be greatly appreciated!