Andreas Blass
Reputation
33,363
373/400 score
 2d comment Correspondence between prime and maximal ideals The formulation of the statement could be misread, so, to be safe, what is wanted is that the natural correspondence restricts to two nice things: (1) a bijection between the prime ideals of $R$ that contain $I$ and the prime ideals of $R/I$. (2) a bijection between the maximal ideals of $R$ that contain $I$ and the maximal ideals of $R/I$. There is no claim about a bijection between prime ideals and maximal ideals. Jan 26 comment Must complete atomless Boolean algebras of the same cardinality be isomorphic? @AsafKaragila To finish translating my answer into the language of forcing, the last sentence says that random is $\omega^\omega$-bounding and Cohen isn't. Jan 25 answered Proof that the minimal inductive set does not contain any more elements than $\{\emptyset, s(\emptyset), s(s(\emptyset)),s(s(s(\emptyset))),…\}$? Jan 25 comment Turn sentence to predicat formula @trolkura No, the formula is true whenever you've read at least two books. For example, if you've read all three books of C. S. Lewis's space trilogy, then BrianO's formula is true because you can instantiate $x$ as "Out of the Silent Planet" and $y$ as "Perelandra"; these instantiations make $C(x)$, $C(y)$, and $x\neq y$ true as required. The fact that you've also read "That Hideous Strength" does no damage to this evaluation of the formula. Jan 22 comment Why do I get one extra wrong solution? Quite generally, when you get a wrong solution as a result of a computation like this, you can usually pinpoint the source of the error by just substituting the wrong value, in this case $x=1$, into each step of the computation. Substituting into the original equation produces a false result ($2-1=-1$); substituting into the final step gives a true result ($1=1$ or $1=4$). Plug $x=1$ into every step and see where the transition from false to true occurs. Jan 21 awarded Enlightened Jan 18 comment Ordered sets $\langle \mathbb{N} \times \mathbb{Q}, \le_{lex} \rangle$ and $\langle \mathbb{Q} \times \mathbb{N}, \le_{lex} \rangle$ not isomorphic The notation for the ordered pairs seems reversed. For example, $\phi(0,0)$ is in $\mathbb N\times\mathbb Q$, so its first component should be in $\mathbb N$ and the second in $\mathbb Q$. Your notation $(q,n)$ suggests (though it doesn't strictly imply) the opposite; later, you use that there's nothing between $(q,n-1)$ and $(q,n)$, which does imply that you meant for $n$ to be in $\mathbb N$, not $\mathbb Q$. Jan 18 comment Prove matrix $X'XA$ equals $X'XB$ iff $XA = XB$ (I see Robert Israel posted the same answer while I was typing mine.) Another way to proceed, once you have $Y'Y=0$ is to add up the diagonal entries of $Y'Y$. Of course the result is $0$, but it's also the sum of the squares of all the entries of $Y$. So $Y=0$ follows without having to introduce the vectors $V$. Jan 18 answered Prove matrix $X'XA$ equals $X'XB$ iff $XA = XB$ Jan 17 comment Proving that a set is not consistent Aha! It seems you're asking only about propositional logic, not first-order logic. In that case, yes, substitute "valuation" for "model". ("Model" would be the appropriate concept in first-order logic, where the same result holds.) Jan 17 comment Proving that a set is not consistent If you don't know what models are, then what do you mean by "$\Sigma\models\psi$ and what do you mean by "satisfiable"? The usual definitions of those concepts involve models. Jan 17 answered Proving that a set is not consistent Jan 16 comment How's it possible for a bounded subset of the reals not to contain it's $\inf$ or $\sup$? If $A$ is the set of numbers $x$ such that \$0