I just started Smullyan's Beginner's Guide to Mathematical Logic
However, by my estimation and what I was able to check via Wolfram Alpha, the very first exercise is asking me to prove that a statement is valid, which is invalid...
(A union B)' inter C = (C inter A') union (C inter B')
Previously he shows an example where A = (1, 2, 3, 4), B = (1, 2, 5, 6), C = (1, 3, 5, 7). My understanding would be that (A union B)' = (7, 8), thus intersection with C is (7).
The other side is clearly not (7) so I'm a bit confounded.
@JakeS: So the universe is the numbers 1 through 8? Anyhow, there is a typo, or you misread the problem. It should be intersection on the right hand side.
I'm a big fan of Smullyan's recreational logic books, hopefully this one isn't a let down, but maybe I should just go straight to Velleman's How to Prove It instead.
My recollection of Velleman's book is that he does way too many things symbolically for my taste. I am very fond of Houston's How to Think Like a Mathematician ...
I'm a first year undergrad with a weak background from high school looking to improve my ability to do proofs, as my university doesn't offer an introductory course in the matter
Internet loves to praise Velleman's book as a successor to Polya's How to Solve It
Yeah, I think it's better to be less symbolic and formal, but some people love symbols.
Amazing, as — at least in the US — now everyone offers "intro to higher math" type courses. And Houston wrote his book in England, if I remember correctly.
If you can find Houston's book in a library, look at it before you make a commitment.
I also made efforts to give guidance on how to think about math and proofs in the introductory (proof-oriented) linear algebra book I wrote a while ago. If you're in Brazil, there's even a Portuguese translation.
I need to get going, but good luck and chat later.
Given a semi-simple Lie algebra, if we wanted to analyze it we could take the largest set of ad-diagonalizable commuting elements (Cartan subalgebra) and ask what happens, great, but why would an abelian ideal spoil this? I know it ends up spoiling the decomposition into su(2)'s/sl(2,C)'s, and it allows for the possibility that a Cartan subalgebra may not exist, but what about cases where it does exist, seems like there is a better reason to ignore solvable Lie algebras from this pov
Let $K$ be a normal subgroup of order 2 in group $G$, show that $K$ lies in the centre of $G$. Describe a surjective homomorphism of the orthogonal group $\mathrm{O}(3)$ onto $C_2$ and another onto the special orthogonal group $\mathrm{SO}(3)$.
@bolbteppa i think the fact that a semisimple lie algebra is really a bunch of sl2 "glued" together in some sense is really crucial to its classification so at least there's that.. i suspect without it things will become much harder - but i dont know anything beyond the semisimple case
@user100000000000000 No, it is not. Pick your favourite nowhere continuous function and restrict it to a point! For a more interesting example consider the indicator function of the rationals, restricted to the irrationals
Let $N$ be a finite subgroup of group $G$, and suppose $G=\langle T\rangle$ and $N=\langle S\rangle$, for some subsets $S$ and $T$ of $G$, then $N$ is normal in $G$ if and only if $tSt^{-1}\subseteq N$ for all $t\in T$. Please give me a counterexample saying that this criterion to check if subgroup is normal fails when $N$ not finite.
Update on the dedekind finite proof: Ok so while it works for the finite case, I cannot be certain whether it works for the dedekind finite case. Better see how Asaf and Noah handle this first, I felt I am missing one more bijection to show that the leftovers is not a singleton
First, you want "union of pairs" to be a "disjoint union of pairs", otherwise any set with two elements or more is even. Now, what if it cannot be expressed as a union of pairs, but can be expressed as a union of sets of size $4$? — Asaf Karagila ♦8 mins ago
hmm... I suspect any finite set of even natural number should be partitionable into pairs so a disjoint union of 4s should be partitionable into a disjoint union of pairs because for each 4s we can split that into two pairs... o wait a minute, then I am making an uncountable number of splittings which it might not exist without choice or without a given bijection, hmmm... — Secret21 secs ago
So you can have strongly even sets that are not weakly even
Or put it in another way, unions are not associative when choice fails
Asaf also answered. Effectively, in trying to map that singleton and assuming bijection, we set up a shift mapping and thus introduced a countable subset, thus blowing up dedekind finiteness
I have a small question regarding the semidirect product. Consider a group $G$ which is the semidirect product $\mathbb{Z}_3 \ltimes (\mathbb{Z}_7 \times \mathbb{Z}_7)$ (internal semidirect product). Let $\phi:\mathbb{Z}_3->Aut(\mathbb{Z}_7 \times \mathbb{Z}_7)$. There will be $\phi_0 ,\phi_1 ,\phi_2$ corresponding to $\bar{0},\bar{1},\bar{2}$ of $\mathbb{Z}_3$, right?
Hi. Question here. I have three skew lines in projective space $\mathbb{P}^4$. I want to show that there is a projective transformation such they can be described in this way: $\{(x_0,x_1,x_2,x_3):x_1=x_0=0 \}$, $\{(x_0,x_1,x_2,x_3):x_2=x_3=0 \}$ and $\{(x_0,x_1,x_2,x_3):x_0=x_2 \text{ and } x_1=x_3\}$. Is the best way to go just trying to work in $\mathbb{R}^4$?
@MatheinBoulomenos I can't figure out why $\langle S\rangle$ is not normal. Shall I try $g\langle S\rangle g^{-1}$ is proper subgroup of $\langle S\rangle$, or that $gsg^{-1}\notin \langle S\rangle$ for some $s\in \langle S\rangle$?
I am reading a small section in Herrlich's Axiom of Choice, and in the proofs of the various properties of dedekind finite sets, it is quite similar in that often restrictions of bijections are considered and then finding some reference point to make the choices
Also what you said is correct @Secret partitions of 4 require choice to implies partitions of 2(we choose bijective from a four elements to {0,1,2,3} and make a partition of $\bigcup_{x\in P_4,n\in\{0,2\}}\{f_x(n),f_x(n+1)\}$)
Well, trying to define them without natural objects (e.g. natural numbers) is pretty hard and I don't have enough time to read in detail how it is done.
here's an overview. Basically, they like to release games such that it is 1, 2 and something else, but never 3 appeared in the title. Half life in particular it goes 1, 2 episode 1, episode 2
@TobiasKildetoft You mean $x^{-1}(y^nxy^{-n})x$ or $y^{-1}(y^nxy^{-n})y$? i think latter is indeed in $\langle S\rangle$. and so is $x^{-1}(y^nxy^{-n})x=y^0x^{-1}y^0(y^nxy^{-n})y^0xy^0$.
@Silent The only substantial book I have read more than 2 chapters is Munkres. That is largely motivated by trying to understand infinite objects and weird spacetime topologies
I have a huge list of books which I still don't have time to read more than 10 pages
That is almost nothing compared to weird things called rank into rank cardinals which I will need a much stronger understanding of model theory to understood them
@Secret Whenever i have tried to start learning mathematical logic, i have to end that endeavor because, unlike other math disciplines, in logic, i can't even formulate my questions!
My thinking is actually more closer to an artist than a scientist, with weird jumps in logic in my thinking. I actually find formal logic quite painful to follow until I just treat the whole thing as a computer program
My personality is one that loves to ponder what happens if certain rules will be broken, but logic is a world that you must follow all the rules without question
I normally don't follow rules if I don't understand them
Recently, I was wondering how division of aleph numbers would work. First, I thought about how finite cardinality division would work. What I came up with was that the result of $A/B$ where $A$ and $B$ are both cardinalities, is the number of times that each element of $B$ had to be mapped to an ...
@Silent Back in my undergrad, he is one of the professors I consulted to share about their research. His approach of putting everything into surds can be very useful if crystallography parameters follow that standard, but ultimately his method is equivalent to the usual sin cos approach as there is a correspondance between them
I have not check with him about his ultrafinite and real number argument though, as at that time I was not as focused on infinity as today
It is known that the Dedekind-finite cardinals are closed under addition and multiplication, so one may do arithmetic in them, as opposed to only natural numbers.
How much can those two arithmetics be different? For example, can there be a Diophantine equation which is not solvable in the natura...
@TobiasKildetoft oh! Thanks for heads up. I was going to listen his online lectures on differential geometry and algebraic topology. But I do not find any other online lectures on those topics :(
@Silent He might actually know something about those topics. It is just hard to imagine that he does due to his way of addressing anything to do with infinite sets
O that reminds me: whatever you do, avoid his differential geometry lecture. It is universally confirmed to be a disaster in my uni when he was assigned to teach it in 2016, which is why the maths department send him away from teaching that course
The issue is that differential geometry needs the concept of infinitesimals, which cannot exists in an ultrafinitist framework. I have many peers who done that course that year and it is a disaster
well tbh, I don't know of any use of Hamel functions outside of set theory
But they are relatively tame compared to the $\beth_2$ many functions out there
there, fix it for you
I think it is safe to say that nonmeaurable sets have close relationships with geometry since their existence allow the duplication of things by partitioning them
Compared to nuking choice, the weirdness of allowing choice is much less and mostly concentrated in the nonmeasurable sets and certain nonisomorphic structures
The reason why there is a lot more weirdness in choiceless universe mostly because the 5 notions of finiteness are no longer equivalent
The following list comprehensively shows Fields Medal winners by university affiliations since 1936 (as of 2018, 60 winners in total). This list considers Fields medalists as equal individuals, regardless of the total number of winners who received the medal each time at an International Congress of Mathematicians (ICM). It does not include affiliations with research institutes such as IAS and MSRI in the United States, as well as IHES and CNRS in France. In this list, universities are presented in descending order starting from those affiliated with most Fields medal winners.
The universit...
@TobiasKildetoft, Dummit and Foote says "the property of being normal is an embedding property, that is, it depends on the relation of N to G, not on the internal structure of N itself (the same group N may be a normal subgroup of G but not be normal in a larger group containing G). "
Is this same as saying that being normal subgroup is not transitive?
Oh, i think embedding is a little more general than transitivity.
"(a) Define a $\mathcal{V}_G$-sentence $\varphi$ such that $\varphi$ has arbitrarily large finite models and, for any model $G$, $G$ is a connected graph."
"(b) Find a connected graph that does not model the sentence $\varphi$ you found in part (a)."
I'm thinking that if $E(x,y)$ denotes that "$x$ and $y$ have an edge in common", then creating $K_n$ is just $\forall x_i\forall x_j(E(x_i,x_j)=1)$ where $i\neq j\in\{1,2,\dots,n\}$ for $n\in\Bbb{N}$. I'm not really sure what it would mean that a graph doesn't model the sentence.
Okay. So, that $A$ models $F$, denoted $A\models F$, means that the assignment function $A:S\to\{0,1\}$ for some sentence $S$ gives each atomic formula a truth value of 0 or 1, and that the formula $F$ fulfills that $\mathcal{A}(F)=1$ for some assignment. Seeing this as a truth table, this corresponds to a row in that table, where we have all the assignments of the atomic formulas, and then the formula $F$ which is then fulfilled.
I don't really know what we would have here. An example of a connected graph is $K_n$. I think that something like $\forall x_1,\dots,x_n(E(x_i,x_j)=1)$ construc…
@OskarTegby maybe explaining it to someone who doesn't know a lot of math will help you understanding the problem yourself. If that's possible, of course.
I'm looking at this closed question and trying to understand the answer if I could post a similar one myself. I mean, the original one was closed due to a lack on context. If I could give that, then it should be a valid question.
Hi. I'm trying to calculate the modulo inverse of a 10x10 matrix A. I have determined that rank(A) == 10 so theoretically the inverse should exist. I've tried multiple computer programs like NumPy and Wolfram Mathematica but all of them says the matrix is "invalid for modulo 64"
@iBug One thing you could do--I think--is compute the eigenvalues of the matrix, and then take them modulo 64; if any of them are zero, then no such inverse exists.
Let $(X_1,\tau_1)$ and $(X_2,\tau_2)$ be separable spaces, and let $\tau$ denote the product topology on $X=X_1 \times X_2$. Then $(X,\tau)$ is a separable space.
Proof. $(X_1,\tau_1)$ and $(X_2,\tau_2)$ be separable spaces. So, there exists countable dense subsets $D_1$ of $(X_1,\tau_1)$ and $D...
(unrelated curiosity though) What does it mean geometrically to have a zero divisor as a determinant. how is it differ from a zero determinant matrix...?
Let $K$ be a normal subgroup of order 2 in group $G$, show that $K$ lies in the centre of $G$. Describe a surjective homomorphism of the orthogonal group $\mathrm{O}(3)$ onto $C_2$ and another onto the special orthogonal group $\mathrm{SO}(3)$.
