While reading the book by Peter Smith I came across two different definitions of soundness, the general definition
A theory $T$ is sound iff its axioms are true (on the interpretation built
into T’s language), and its proof system is truth-preserving, so all its theorems
are true.
and for theor...
The conclusion I came to I put in the form of a new definition:
Definition 1. Let $C$ be a category with binary products. Let $m : G^2 \to G$, and let $D$ be the diagram in $C$: $$ \require{AMScd} \begin{CD} G^3 @>{m\times \text{id}}>> G^2 \\ @V{\text{id}\times m}VV @VV{m}V \\ G^2 @>{m}>> G \end{CD}$$ Then $G$ is called group object if $\lim D$ exists and $\lim D \simeq G$ itself.
This is a superior definition, for obvious reasons - in math anything expressible with less or other theorems is more elegant, easier to remember and work with
*superior to nLab's group object (3 diagram method)
I'm taking the limit of the square, $G$ by definition of its associated maps and their commutative properties, already forms a limiting cone as shown in the proof
You then take an other cone, glue it in, and if it commutes with limit cone and square (everything commutes) and there is one such morphism $u:X \to G$ that accomplishes that, then $G$ is indeed a limit
See the wikipedia link it goes to a subsection
By definition of limit
So I'm really liking this definition of group object :)
you're observing that a cone with $X$ over the diagram induces a morphism $X\rightarrow G$, but this correspondence is neither injective nor surjective
$(u = i) : X \to G$ 1) always exists, and 2) it's the only such map that $\text{id} \circ u = i$.
I'm saying we're taking $G$ as a usual group object so that the first diagram in my posts commutes. I prove (easily) that it's actually a limit of the square describing associativity
yeah, she cooked the garlic with the spuds. i like both, but i think garlic overpowers the potatoes, especially with a nice roast... but then, i am a minimal sort of fellow
I want to negate definition of uniform continuity: $\forall \epsilon\gt 0, \exists N$ such that $\color{blue}{\text{$\forall n\ge N$ and for all $x\in S$, it follows that $|f_n(x)-f(x)|<\epsilon$}}$. Negation: $\exists \epsilon\gt 0$ such that for all $N$ $\color{blue}{\text{$\forall n\ge N$ and for all $x\in S$, it follows that $|f_n(x)-f(x)|\ge\epsilon$}}$
Because negation of $a\implies b$ is $a \land \not b$ (a and not b)
But what is wrong in considering: $\color{blue}{\text{$\forall n\ge N$ and for all $x\in S$, it follows that $|f_n(x)-f(x)|<\epsilon$}}$ as implication p implies q. And then negating as p and not q ?
i honestly believe that if that had been written in all logical symbols it would have been negated correctly, but people never want to fully commit because they aren't computers
@leslietownes Generally, I advise people to use more English and fewer symbols. But negating moderately complicated theorem statements is a really good example of a situation in which symbols can help.
$\exists \epsilon\gt 0$ such that for all $N$, there exists $n\ge N$ $\color{blue}{\text{or}}$ there exists $x\in S$ such that $|f_n(x)-f(x)|\ge \epsilon$.
koro i'm not a logician but think for a minute about "for all n >= N" what is it really saying. it's not something that can be logically and'ed or'ed with something else because it has no truth value. so the "and" and the "or" in your english translations aren't logical and and or
it might be helpful to think of "forall n >= N, __" as "forall N (n >= N implies __)
whose negation is exists N (n >= N AND (not __))
and maybe try figure out what __ has to be there without any english words in it. no such thats, no follows thats, and no "ands" or "ors" that aren't logical ands or ors
i once had to work with a consultant who insisted on writing ON and OFF in all caps when he meant states of a device, which he insisted had nothing to do with common english on and off, even though his report was written in english and not his alien state language. in context i think he was wrong, and i am not myself an alien, but he had a point
the report made it look like he was yelling
i'm watching my daughter on a baby monitor. after being put to bed, she got up, turned the room light on, sat next to the cat, appeared to almost get in a fight with the cat, lectured the cat about something, turned the light off, and went back to bed
now i can see the cat grooming its paws and her eyes blikning open and shut in night vision
@leslietownes I understood when you nicely said: "for all n>=N" is not a mathematical statement. Because a mathematical statement has a truth value (either the statement is true or false, it's not half and half etc.). With that, we can't "and" or "or" them
it annoyed me when i had to spend about a week teaching students to work through a textbook that did a lot of this at a level that was just not quite formal enough for it to be genuinely helpful
and means AND except when it doesn't mean AND. it's just and. got that?
there's no trick to this, it's just a simple trick
@Ted: Wrote an initial draft of notes. Didn't get to the juicy part about characteristic classes yet, but I will. So far it's mostly definitions, motivations and stable homotopy theory so you won't like it.
The most concrete application that I have in mind right now, which is the reason I made the nosedive into this stuff in the first place, is $S^{2n}$ has no almost complex structures other than $n = 1, 3$. There are other proofs, apparently, but I like the one I could come up with.
it uses the relation between these Steenrod powers (mod p Steenrod squares) and Chern classes modulo p that I mention at the end, and will elaborate on
If row vector of matrix is linearly dependent then is column vector of matrix linearly dependent?
I made row vector of 3 by 3 and 2 by 2 linearly dependent and found that determinant is 0 so I was wondering if it is true whether column vector of matrix linearly dependent if rows are dependent. I then tried to prove that for 3 by 3 by fixing the first row of matrix as linearly dependent out of other two.
https://i.stack.imgur.com/w2Q2H.jpg
(Here is the illustration of the matrix)
But this is only true if $b_1=k_1b_2+k_2b_3 $ and $c_1=k_1c_2+k_2c_3$ where $k_1,k_2\in \Bbb{R}$.
So my question is if row vector is linearly dependent will at least one of the column also be linearly dep…
Right now I don't know anything about rank and other stuff but just learning some basic linear algebra for multivariable calculus.
Like only dot product matrix multiplication and vector properties and vector arithmetic.
Is there a way I can prove by very basic stuff that I learn? Or do I need very advance math?
Some people pointed out that column rank= row rank.
Okay If I know row rank or 3 by3 matrices and column rank then I will be satisfied.
@MethNoob I wrote something wrong. But from linear dependence of the rows, you deduce that there is a nontrivial solutiion of $Ax=0$, and hence this gives a nontrivial relation among the columns.
Let $f(z)$ be a conformal map from the open unit disk onto $D$, which is a domain.
I would like to show that the distance from $f(0)$ to the boundary of $D$, denoted $\partial D$, is given by $\mathrm{dist}(f(0), \partial D) \le |f'(0)|$, that is, it is bounded by $|f'(0)|$.
What is a good way ...
@TedShifrin Sorry. Had to run off for a second. A function $f : G \to \Bbb{C}$ is said to be conformal if it has the angle preserving property and also has $\lim_{z \to a} \frac{|f(z)-f(a)|}{|z-a|}$ existing.
I mean, maybe it follows from that definition that $f$ is injective, at least...
The problem states that $f$ maps the unit disc ONTO $D$, so it is definitely surjective. I see some definitions include injectivity in the definition of being conformal. Is this problem solvable/true if one doesn't include injectivity?
I was trying to find the CFG for the language below. However, I couldn't do that. Can anyone help with this problem?
$$\{1^n 0^m 1^k 0^p | n \geq 2, m,k,p \geq 1, n+k = m+p\}$$
yes, I was able to build design a PDA that can recognize the language, however, I'm not able to derive the context-free grammar, I would like some help
Yes, but, assuming this is a mathematical notation, I don't know what it means or how it relates to the formal grammar in terms of alphabet and syntax.
Problem $L = \{ w \in \{a, b\}^* : n(a) \neq 2 \cdot n(b)\}$
This can be done easily with NPDA, but I couldn't find a way to make it work with CFG. My idea was to break it into 2 cases: $n(a) > 2 \cdot n(b)$ or $n(a) < 2 \cdot n(b)$. I first try to generate a language which makes them equal, the...
Well the problem isn't my understanding of your needs. I learned about these things informally, i.e. not through university or college. The issue here is only my understanding of these formal constructs of communication. Once I know what they are, then I can help you quite easily. I will look at the question.
if i'm not mistaken, a "language" here is just a set of strings of 1s and 0s, and something is in the "language" if it has the given form for integers n, m, k, p satisfying the relation
the part you asked about is defining the general shape of the language, the language will have a set of 1s followed by a set of 0s followed by a set of 1s followed by a set of 0s where the sum of 1s and 0s is equal
also, the string should at least have two 1s at the beginning
I'm not sure if this will help you, but the way I intend to implement the inputs and outputs of my metacompiler would represent that set of strings as a tree, namely, a binary decision tree (as a directed graph).
You have one input (the decision tree) describing all possible outputs, and another input (the predicate tree) describing all possible decisions. The third is the input to operate on which is passed to the predicate tree which, starting at the root node, computes a decision on which path, either left or right, to take in the decision tree.
In the case of a formal grammar, the predicate tree would identify what the current element is, and the decision tree would give what the next expected element is and would do nothing but begin or end a context, where a context here means a symbol, word, clause, etc.
This describes one possible implementation of a parser on said metacompiler
You can chain different trees together to form a complete compiler or even software.
So that is my informal understanding of languages in the context of CS
@bazzinga Ok, so the only thing I'm not understanding then is this use of exponentiation notation. I assume it has something to do with the number of 0s or 1s in that position, e.g. 1^2 0^3 1^1 0 ^2 would be the sequence 11000100.
Also that just gave me an idea for a compact encoding format. Thanks lol
Ok, got it. So, how is the CFG intended to be described formally in CS? I'm pretty sure you just take the constraints (the predicate) to the right describing the relative sequence and just rewrite it formally in some other language.
Well the rewrite rules in the case of a CFG are just one level above Type-3. The process I would take is to make a Type-3 grammar, then convert upwards from there to Type-2 or even Type-1. The process of this conversion is essentially a compaction of the grammar that can make it more terse. That is how I see it.
Unfortunately, I don't have an intuition for rewrite rules, so I can't help you there.
But I believe that you can figure it out on your own if you put your mind to it, and if you do, you can always answer your own question. :)
@Semiclassical Ok that makes sense, the problems with educators here is that they just use formula for everything without telling the essential logic which causes confusion, Also I don't know why I am studying this thing at night after being exhausted completely
"Commutative algebra" is a very broad field, and there is active research going on. It is a very long way away from anything that I know (or care) much about, but there are certainly people who do care.
i mean, i think a lot of departments wouldn't hire someone who couldn't talk that language for an AG-focused postdoc or tenure track position.
not so much self-identification.
but again, a lot of departments, not all departments.
one semester a couple of my friends took commutative algebra from eisenbud and i went in a different direction. we all wound up in happy places but i am not an algebraic geometer.
i mean, if you refer to my comments which may still be on the screen.
"if you want to hitch yourself to the AG gravy train," "if you want to style yourself as an algebraic geometer," applying "for an AG-focused postdoc or tenure track position."
if you have specific jobs and connections in mind that don't require these, then OK, but i don't know why you'd be asking the question about commutative algebra, if you do.
@Asinomás (1) Thinking about coursework as a means to employability is the wrong mindset. You are learning a body of theory, and you are learning a way of thinking. You should pursue what interests you. While it doesn't hurt to think about what kinds of jobs you may end up pursuing, it shouldn't be the primary focus.
(2) There are active fields where "speaking" commutative algebra is going to be an assumed prerequisite. Algebraic geometry is one of those (probably; depending on the institution / program / research group). So, if you are interested in algebraic geometry, you probably want to know commutative algebra.
i don't think people take graduate courses in commutative algebra for fun. it's not something to take for general mathematical culture, as far as i know.
i was TAing calculus once, and after the first midterm someone came and asked about how to get a good grade. i asked what their expectations or requirements. they were an english major who had not spoken to advisors, and assumed that there was some general requirement to take calculus in college and get some minimum grade. i said, "uh, oops, i'm not the authority, but do go back and talk to your major advisors again. i know no reason for you to take this." they dropped the class.
if you can drop a commutative algebra class that you're not interested in, i'd say, in the same spirit, do it
Generally speaking, you should not be taking very many (if any) graduate classes which you don't find interesting or worthwhile. Graduate school is when you narrow down your focus to a research topic.
You are probably going to have to take one or two courses which are not that interesting to you (beyond the "general education" courses)---I had to take a quarter of Lie theory because there was literally nothing else offered that semester, and I was trying to fast track my advancement to candidacy---but you generally are going to be taking classes which interest you.
If something is not interesting, don't do it, and focus your research elsewhere.
i think my anecdote actually has more relevance than i first realized. a lot of people in my grad class took courses relating algebraic curves or more or less by default, if they had come from place where that was the thing. ditto PDE, ditto harmonic analysis, ditto set theory.
@XanderHenderson I didn't realize that there was any aggression at work there. At some point, the microabuse flags will accumulate and I will be in trouble.
While reading the book by Peter Smith I came across two different definitions of soundness, the general definition
A theory $T$ is sound iff its axioms are true (on the interpretation built
into T’s language), and its proof system is truth-preserving, so all its theorems
are true.
and for theor...
Let $f(z)$ be a conformal map from the open unit disk onto $D$, which is a domain.
I would like to show that the distance from $f(0)$ to the boundary of $D$, denoted $\partial D$, is given by $\mathrm{dist}(f(0), \partial D) \le |f'(0)|$, that is, it is bounded by $|f'(0)|$.
What is a good way ...
I was trying to find the CFG for the language below. However, I couldn't do that. Can anyone help with this problem?
$$\{1^n 0^m 1^k 0^p | n \geq 2, m,k,p \geq 1, n+k = m+p\}$$
