one time in grad school someone responded to that joke with "zorn's corn?"
they weren't trying to be funny, they didn't know
if you search around on math.SE, the answers to questions like that are usually equivalent to some weak form of choice (maybe choice only up to some specific cardinal, or something)
Given that $\cos(A), \cos(B)$ and $\beta$ are non-zero, show that the
equation $$\alpha\sin(A-B)+\beta\cos(A+B)=\gamma\sin(A+B)$$ reduces to
the from $$(\tan(A)-m)(\tan(B)-n)=0$$
where $m$ and $n$ are independent of $A$ and $B$ if and only if
$\alpha^2=\beta^2+\gamma^2$.
Workings: $$\alpha\sin(...
Given that F is a field and a in F is non zero, if f(x) in F[x] is irreducible over F then af(x) is also irreducible over F.
This assertion is correct, right? For, if not then af(x)= g(x)h(x) where g(x) and h(x) are polynomials of positive degree and have degree less than degree of f(x). This means that $f(x) = (a^{-1}g(x))h(x)$.
where the first factor on RHS is non constant and hence non unit.
This contradicts the assumption that f(x) is irreducible over F.
So in order to prove that P if and only if Q, then we must show that P implies Q and Q implies P. Could someone clarify what P and Q would represent in this question? Would the P be the first trig equation and the Q the second equation in terms of tan(A)? or would P be the first trig equation and Q the condition given at the end?
So @copper.hat , before the all seeing eye appears. I beleive I have the contradiction, but I'm not satisfied by it. We created the convergence subsequence who's (or is it whose?) terms were created specifically to no be in any of the finite union of open sets. So we have created this convergent subsequence which lies in my compact set $X$ but doesn't lie in my infinite union of open sets, but this infinite union of open sets covers my compact set $X$ thus a contradiction.
well convergence to my point means there are an infinite amount of points converging towards the limit point and each of those points is not in an open union of sets.
The usual approach is to put the compact set in a large box. Divide the box into $2^n$ parts (each side by ${ 1\over 2}$). At least one of the little boxes cannot be covered by a finite number of open sets. Keep repeating. Pick $x_k$ from each of the 'no finite cover' boxes. It must converge to some point $x^*$. But some open set covers $x^*$ and so all of the little boxes nearby, hence a contradiction.
let me find what I originally wrote. and the original question.
@copper.hat the question: Suppose $X \subset \mathbb{R}^{n}$ is a compact set. Suppose $U_1, U_2, \dots \subset \mathbb{R}^n$ are open sets whose union contains $X$. Prove that for some $N \in \mathbb{N}$ we have $X \subset U_1, U_2, \dots U_N$. (Hint: if not for each $k$, choose $\mathbf{x}_k \in X$ so that $\mathbf{x}_k \notinx_k \notin \bigcup_{i = 1}^j U_i$.
What I had written: So I will have this convergent subsequence, let's say it converges to $a$ so after a certain $k > K$, $\|x_k - a\| < \epsilon$. So the points that are on the outside of this $\epsilon$ - ball are a finite set of points each of which will have an open set around it.
What I'm concerned about is that will that set of finite open sets be sufficient to cover my compact set?......but I think they will because I can make those open sets as large as I want to
a is in X, isn't it? the union contains X, doesn't it? so there's some N for which a is in U_N, isn't there? and U_N is open, isn't it? so infinitely many x_k in that sequence are in that one set U_N, aren't they? and isn't that a contradiction?
i heard some stuff the other night from somewhere about how we don't know if there are countably many sets. but we do.
but the problem is about countable collections. i don't see the issue.
anything with topology on R^n is going to be sequence based and countable will be enough. i don't know what this would look like for nets.
i'd just have thought by now we could move on to whatever the next problem is, and not circle around this forever wondering about what it would mean if there were uncountably many everything
So the only place we applied $\mathbf{x}_k \ \notin \bigcup_{i = 1}^j U_i$ is in constructing the original sequence from which we could get the convergent subsequence then?
almost any 'construct a sequence via ...' real analysis proof can be redone without this kind of argument. rudin is pretty fond of avoiding sequences outside of series-based arguments. it seems like a waste of time to me if the underlying topology is metrizable.
$x_k$ converges to $a$ along a subsequence, and there is some index $l$ such that $a \in U_l$. Since $U_l$ is open it contains the tail of the subsequence
@leslietownes Can I worry about this when I work through Munkres and not now?....I'm not trying to get anxiety from having to use countable and bijective to the integers right now...
dc3rd: i wouldn't worry about it until you are made to worry about it. you'll know when that happens. it won't be "huh, but what would this exercise mean if i had an arbitrarily indexed collection of sets." you'll actually have one, and a goal in mind, and need to worry about it.
I tend to write little reflection paragraphs after I have trouble with questions like this. Here I think the idea of the union of sets is what was messing me up the most.
shit about like, finite topologies, the long line, etc.
just stuff you come up with when you play with the axioms of topology and wonder what the minimal set of hypotheses is for a T_4 pre-bornological space to be equivalent to a T_{3 and a half} space without the sierpinski property.
none of that math talks to anything except other math about that
when you take a general topology course sometimes it's easy to get the impression that all math requires you to know some minimal set of hypotheses under which a space is metrizable
in the above, i roughly mean, the study of consequences of various axioms for a topology, as distinguished from the use of topology to solve problems that do not arise out of studying the consequences of the axioms for a topology
and there's a ton of this in entry-level books because you eventually run out of stuff to talk about other than that kind of stuff
unless you actually focus on some area of application
you can tell it's garbage because it's described using the same sentence as the previous one, as "a frequently-cited counterexample to many otherwise plausible-sounding conjectures"
[[citation needed]]
what a claim to fame that is
just open 'counterexamples in topology' to a random page, that's what i'm talking about
got you riled up now....starting to behave like my neighbour........turns out she has some issues and doesn't want to take her meds....so the spirals of madness ensue.
I have an exercise way later in Chapter 8 (for anyone who cares) that because $\Bbb R^n$ is separable, you can always reduce arbitrary open coverings to countable open coverings.
@copper.hat yep I'm confused.... Leslie's argument makes sense to me but how is every $U_k$ being used? and apparently there is still something missing in my understanding of it.
not everything? so in my mind i'm picturing your classic open ball and then every point of my sequence inside that ball. But your saying that is not enough?
Well, any limit of a convergence subsequence is called an accumulation point (different authors use different terminology).
The point of the proof is that the point $a$ is a limit of a subsequence, and this is in $U_l$ for some $l$ and hence so are an infinite number of the $x_k$.
But that means there is some $m>l$ such that $x_m \in U_l$ which contradicts the construction.
(In fact there are many such $m$s but we only need one here.)
ah...and that is due to the convergence of the subsequence. Jeez.....I see how it "could" be simple but man....those nuances.....I'm going to have to tighten up my form of arguements and understanding.
@copper.hat My brother lives in London now, just moved there down from Liverpool last week. Get her to bring you a bunch of chocolate as I'm sure you're well aware, there is a huge taste difference between what we get on this side and there.....for the worse
I do not know, so far only manipulation of joints & X-rays. I am not sure that is enough to evaluate, and practitioners tend to jump to certain foregone conclusions.
you already got all the credentials so you don't need to be savy about them, heck you probably helped create the evolution of them somewhere down the line.
Not sure I would go proficient. But I think having a focus that makes the linear algebra a tool rather than the object of study was good for me. I focused on results I could use first and then proofs as backfill.
I learned many things later than most undergrads would, however I have seen many things in an application context which really helps build intuition.
I have one question on the proof that $SO(n)$ is connected. It seems that connectivity of $SO(n)$ can be shown by showing that space is path connected. But the given hint on this problem is slightly different. It says that first imitate the proof of connectivity of $U(n)$ then use a natural isomorphism $U(1)\to SO(2)$.
@anak Forgive me for being pedantic, but it'd be squashing a bug with a tank if I used a theorem or concept which has theorems behind it imo. This is just a restatement of standard facts
everything depends on context but i agree that in general, trying to find the simplest or shortest proof of a result from a list of stated hypotheses "without using any tools" or "without using tool X" is a very separate endeavor from solving a problem
If $(x_n)$ is a sequence with $x_0=x_1=0$ and $x_n=(n-1)/n$ for $n\geq 2$. Let $f$ be defined on $[0,1]$ as $f(0)=0$ and $f(x)=\sum_{\{n\in\mathbb N:x_n<x\}}2^{-n}$ as $x\in (0,1]$. Is $f$ Riemann integrable. I think $f$ is bounded and I was wondering if $f$ is continuous almost everywhere. That would imply $f$ to be integrable. But how to prove it?
Given a dedekind cut $x$, If $x \gt 0$, I want to define the inverse of a cut $x^{-1}= \{ r \in \Bbb Q^{+}\mid \exists p \in \mathbb{Q}-X \ ( pr\lt1)\}$, it’s easy to check that it’s a cut, but how to prove that $xx^{-1}=1^{\ast}$ where $1^{\ast}$ is the usual dedekind cut cut for $1$, it’s easy to see that $xx^{-1}\subseteq 1^{\ast}$ but how to show the other direction?
I suppose it can be shown that there exists $a_{n}$ in $x$ such that $1/(a_{n}+1/n)$ is in $x^{-1}$ and then using the fact that $a_{n}/(a_{n}+1/n)$ eventually approaches $1$, but is there an easier proof especially one that avoided the Archimedean property and does my proof hold water?
it isn't clear to me that "a_n/(a_n + 1/n) eventually approaches 1" this seems to rely on statements about the growth of a_n that are not explicit above.
well a_n/(a_n + 1/n) = (a_n + 1/n - 1/n)/(a_n + 1/n) = 1 - 1/(n(a_n + 1/n)) = 1 - 1/(n a_n + 1). i think we're still using something about the choice of a_n here to conclude that this fraction goes to 1? this would not be true for any sequence of bounded numbers a_n (for example a_n = 0)
i'm going to bed now but maybe look at the above links
all of the verifications of the stuff about arithmetic of dedekind cuts are horrible proofs, they involve arbitrary choices and being clever in ways that do not reflect much more than the definition of dedekind cuts
you will need the archimedean property
this is why many analysis books introduce the real numbers via cauchy sequences :)
this is maximal pedantry, but on rereading some parts of hatcher AT, i came across his usual definition of covering map where he allows it to be non-surjective, but im lapsing and dont see how this is allowed: A covering space of a space $X$ is a space $\tilde{X}$ together with a map $p : \tilde{X} \rightarrow X$ satisfying: Each point $x \in X$ has an open nbhood $U$ in $X$ such that $p^{-1}(U)$ is a union of disjoint open sets in $\tilde{X}$
and then bla bla bla these project homeomorphically onto $U$
my concern is how can non-surjective maps be allowed if for every $x \in X$, we need an open nbhood of $x$ evenly covered by $p$?
@Koro it should because the cases in which that holds are kind of dumb, basically the only way $f$ can be reducible in $\mathbb{Z}[X]$ but not $\mathbb{Q}[X]$ is when $f$ is divisible by an irreducible of $\mathbb{Z}$
so if you restrict to primitive polynomials this annoyance dissapears
the discrepancy arises because irreducibles of $\mathbb{Z}$ are units in $\mathbb{Q}$
whereas they are clearly non units in $\mathbb{Z}$
yeah anyway the point is the only bad cases are the ones you just brought up
every other polynomial behaves normally
so just restrict to monic polynomials or polynomials whose coefficients have gcd 1 and everything is normal
oh also what i said is wrong, you dont need this for the proof of gauss lemma, you do need it as well as the gauss lemma to show $R $ UFD implies $R[X_1,...,X_n]$ UFD
oh , im being dumb, if $x \in X \setminus p(\tilde{X})$ then one must have $p^{-1}(U) = \emptyset$ for any open $U$ containing $x$ and $p$ maps $\emptyset$ homeomorphically onto $U$ vacuously...
you’re right. There’s a theorem that says: if f(x) is in Z[x] and degree f(x)= degree (f(x) mod p) for some prime p such that f(x) mod p is irreducible over Z_p then f(x) is irreducible over Q.
@porridgemathematics but I don’t understand the reason you have given for why it is so.
f(x)= $x^4+1$ is given. f(x+1) is irreducible. Suppose that f(x) is reducible then f(x)= g(x) h(x) and this gives f(x+1)= g(x+1) h(x+1), which means that either g(x+1) or h(x+1) has degree 0.
If g(x+1) is of degree 0 then g(x+1) is a constant and hence g(x) is a constant.
and this is a contradiction to the assumption that f is reducible.
Incidentally, $x^4\equiv-1\pmod p$ can only have solutions if $x^2\equiv-1\pmod p$ does, and it's an important well-known result that the latter is true for $p\not\equiv3\pmod4$.
@Koro im pretty sure ive done an exercise from somewhere where they ask you to use eisenstein on to show it, and it cant be done as is so the next best bet is to use a simple isomorphism to convert it into something eisensteinable
so i.e. i would also like to know if this sort of thing is a special case of some deeper approach to seeing why its irreducible
@copper.hat I learned Git about 10 years ago, but I haven't used it much, so most of my Git knowledge has evaporated, and the great tutorial I used is no longer freely available. :( There's an online game that teaches Git: learngitbranching.js.org It's highly recommended by several of the Python room regulars. I guess I should check it out...
About the cheapest thing I can buy anywhere where tipping exists is a short of espresso for three dollars. When I pay for it, the tablet they use for processing credit card transactions asks how much I want to tip the barista.
Americans simply don't frown on tipping at all (as a broad generality).
