First, two exercises to work some with the notion of ultrapower: Check that whenever is a nonprincipal ultrafilter on the natural numbers, and

for all or

Our argument for compactness required the existence of nonprincipal ultrafilters. One might wonder whether this is a necessity or just an artifact of the proof. It is actually necessary. To see this, I will in fact show the following result as a corollary of compactness:

Theorem. If is a nonprincipal filter on a set then there is a nonprincipal ultrafilter on that extends

(Of course, this is a consequence of Zorn’s lemma. The point is that all we need is the compactness theorem.)

Proof. Consider the language Here, each is a constant symbol, is another constant symbol, and is a symbol for a binary relation (which we will interpret below as membership).

In this language, consider the theory A model of this theory would look a lot like except that the natural interpretation of in namely, is no longer nonprincipal in , because is a common element of all these sets.

Note that there are indeed models of thanks to the compactness theorem.

If let and note that is a nonprincipal ultrafilter over that contains

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A database of number fields, by Jürgen Klüners and Gunter Malle. (Note this is not the same as the one mentioned in this answer.) The site also provides links to similar databases.

As the other answer indicates, the yes answer to your question is known as the De Bruijn-Erdős theorem. This holds regardless of the size of the graph. The De Bruijn–Erdős theorem is a particular instance of what in combinatorics we call a compactness argument or Rado's selection principle, and its truth can be seen as a consequence of the topological c […]

Every $P_c$ has the size of the reals. For instance, suppose $\sum_n a_n=c$ and start by writing $\mathbb N=A\cup B$ where $\sum_{n\in A}a_n$ converges absolutely (to $a$, say). This is possible because $a_n\to 0$: Let $m_0

Consider a subset $\Omega$ of $\mathbb R$ of size $\aleph_1$ and ordered in type $\omega_1$. (This uses the axiom of choice.) Let $\mathcal F$ be the $\sigma$-algebra generated by the initial segments of $\Omega$ under the well-ordering (so all sets in $\mathcal F$ are countable or co-countable), with the measure that assigns $0$ to the countable sets and $1 […]

You assume $\omega_\alpha\subseteq M$ and $X\in M$ so that $X$ belongs to the transitive collapse of $M$ (because if $\pi$ is the collapsing map, $\pi(X)=\pi[X]=X$. You assume $|M|=\aleph_\alpha$ so that the transitive collapse of $M$ has size $\aleph_\alpha$. Since you also have that this transitive collapse is of the form $L_\beta$ for some $\beta$, it fol […]

No, this is not possible. Dave L. Renfro wrote an excellent historical Essay on nowhere analytic $C^\infty$ functions in two parts (with numerous references). See here: 1 (dated May 9, 2002 6:18 PM), and 2 (dated May 19, 2002 8:29 PM). As indicated in part 1, in Zygmunt Zahorski. Sur l'ensemble des points singuliers d'une fonction d'une variab […]

I don't think you need too much in terms of prerequisites. An excellent reference is MR3616119. Tomkowicz, Grzegorz(PL-CEG2); Wagon, Stan(1-MACA-NDM). The Banach-Tarski paradox. Second edition. With a foreword by Jan Mycielski. Encyclopedia of Mathematics and its Applications, 163. Cambridge University Press, New York, 2016. xviii+348 pp. ISBN: 978-1-10 […]

For the second problem, write $x=-3+x'$ and so on. You have $x'+y'+z'=17$ and $x',\dots$ are nonnegative, a case you know how to solve. You can also solve the first problem this way; now you would set $x=1+x'$, etc.

Ha! Yes, sorry. (It is fixed now.)

I think there is a typo in the assignment. You say that is an ultrapower. Did you mean ultrafilter here?