Why may we assume that each interval in $\mathcal{F}$ is contained in $\mathcal{O}$? What warrants this reduction?
Why is statement (4) true? If $x \in E - \bigcup_{k=1}^n I_k$, then $x \in E$ and $x \notin I_k$ for every $k=1,...,n$. Given some $\epsilon > 0$, there exists $I \in \mathcal{...
Corollary 5: Let $\{f_n\}$ be a sequence of nonnegative integrable functions on $E$. Then
$$\lim_{n \to \infty} \int_E f_n = 0 ~~~~~~(5)$$
if and only if
$$f_n \to 0 \mbox{ in measure on } E \mbox{ and } \{f_n\} \mbox{ is uniformly integrable and tight over } E ~~~~~(6)$$
H...
@TobiasKildetoft I am so thankful for this. I was mislead by seeing $\Bbb Z[\alpha]$ just after $\Bbb Z[i]$. So, is there some theorem which lists all $\alpha$, such that $\Bbb Z[\alpha]=\{a+b\alpha :a,b\in \Bbb Z\}$?
@TobiasKildetoft, it seems like $\frac12$ is root of a quadratic polynomial ($4x^2-1$), but set $\{a+b\frac12\}$ contains elements with denominator $2$, while $\Bbb Z[\alpha]$ can have denominator $2^n$, for any $n\ge0$
I'm stuck on the following problem: I want to show that topologies induced by two equivalent norms of a given vector space are the same. So let's denote the vector space with $X$ and for $i=1,2$ the two norms by $\lVert \cdot \rVert_i$, which induce metrics $d_i(x,y)=\lVert x-y \rVert_i$ and those again induce the topologies $T_i$. How can I show $T_1=T_2$?
I tried this: Take an open set $O \in T_1$ which means for every $x \in O$ there is an $\varepsilon$-neighborhood $B_\varepsilon(x)=\{y \in X \mid d_1(x,y) \lt \varepsilon\} \subseteq O$. Then by the equivalence of norms there are scalars $c,C$ s.t. $c d_1(x,y) \le d_2(x,y) \le C d_1(x,y)$.
I want to show that there is a $B_\delta(x) = \{y \in X \mid d_2(x,y) \lt \delta\} \subseteq O$ too.
How do I need to choose $\delta$ so that this is true?
Way 1: Take $U$ open in $T_2$. Then for each $x\in U$ you can fit a $T_2$-ball by definition. Show that you can then fit a $T_1$-ball, so it's $T_1$-open. Repeat with $T_2$ and $T_1$ switched.
Way 2: Show that metric spaces are first countable. Then show they are sequence spaces. Then, since the two metrics have the same convergent sequences, they have the same open sets.
I have a very naive algebraic geometry question: if $f:X\to Y$ is smooth, then $\Omega_X\otimes_{f^{-1}\mathcal{O}_Y}f^{-1}\Omega_Y^{-1}\simeq \Omega_{X/Y}$. Is this still true if one of the fibres of $f$ is singular?
@Secret well, I suspect that we could do so if someone has previously indicated that constantly posting the same problem links here is distracting and irritating
but I frankly don’t care enough to do more than Ignore and move on. ( Perhaps I should, since it’s not a good model for new users, but i just can’t be arsed)
Ok great, somehow I cannot find Ted's telling 193319 to stop spamming question post in the transcript again, guess nothing can be done until some time in the future someone else complains about it
@AkivaWeinberger, In proof that $7+\sqrt[3]{2}$ is algebraic number, how do we derive polynomial $(x-7)^3-2$ when showing $(x-7)^3-2=0$ has root $7+\sqrt[3]{2}$ ?
@AkivaWeinberger I don't think what I did constitutes 'spamming.' You might have noticed from the link Secret furnished us with that I posted those links at most once a day, if at all. Moreover, I recently put a bounty on each question, so I was trying to garner for those questions as much exposure as possible, lest I lose reputations without receiving an answer.
Unfortunately, since I'm not a frequent visitor of either the site or this chat, I don't have much to say on whether user193319 is spamming. That said, I would think that most people who might notice a message in chat promoting a bountied question would also have seen it on the main site anyway.
I'm taking two of them: on one it's broken up into 2 hours on the multiple choice and after a break, 2 hours on the subjective thing. On another it's all at once in 3 hours
(two admission tests for two different unis, i meant)
@BalarkaSen This paper might be too GMT heavy for you, but if you want to read the scalar h principles paper and tell me about it that would be awesome too
@AkivaWeinberger $(1/2i)^n$ lies in $\Bbb Z[\alpha]$ for all $n\ge0$, so, we see that it has arbitrarily small (negative) real (for n even), and arbitrarily small (positive) purely imaginary (for n odd). Since rigs closed under subtraction, and contain 0, we see, it has arbitrarily small real and purely imaginary numbers.
comme pour $z_n=sin(n\pi/4)+\sin(n\pi/2)$ j'ai trouvé dans un livre : on remarques que les sous-suites convergentes sont: $w_{4n}=0, w_{8n+1}=\frac{\sqrt{2}}{2}(1+\sqrt{2}), w_{8n+1}=1, w_{8n+3}=\frac{\sqrt{2}}{2}(1-\sqrt{2})$ d'ou les valeurs d'adhérences @TedShifrin
il faut que je fasse comment l'a fait l'auteur pour cet exemple
(et puis dans le cas des suites périodiques, ben moi je regarderais juste pour quels V est-ce que $\{n \in \Bbb N, w_n \in V\}$ est infini parceque c'est pas bien compliqué)
ben tu peux alors chercher à le montrer pour de vrai
ce qui exclut l'intérieur de l'intervalle de l'ensemble des valeurs d'adhérence
montrer qu'une valeur n'est pas une valeur d'adhérence, c'est facile il suffit de trouver un intervalle autour où $\{n \in \Bbb N, w_n \in I\}$ est fini
la difficulté c'est qu'il faut souvent faire ça pour plein de valeurs en même temps
montrer qu'une valeur est une valeur d'adhérence c'est un peu plus dur puisque qu'il faut donner une sous-suite convergente
mais c'est en faisant un dessin que tu sais qu'est-ce que tu dois prouver pour quelles valeurs
en gros les étapes c'est
1. dessiner
2. observer les valeurs d'adhérence
3. montrer que ce sont bien des valeurs d'adhérence
4. montrer que le reste n'en est pas
mais l'étape 2 est importante pour savoir qui tu traites dans la 3 et qui tu traites dans la 4
c'est quand même un peu bête de faire ça à l'aveuglette
ah mince j'croyais que c'était vrouvrou qu'avait dit "mais c'est pas une preuve"
I tried this: Take an open set $O \in T_1$ which means for every $x \in O$ there is an $\varepsilon$-neighborhood $B_\varepsilon(x)=\{y \in X \mid d_1(x,y) \lt \varepsilon\} \subseteq O$. Then by the equivalence of norms there are scalars $c,C$ s.t. $c d_1(x,y) \le d_2(x,y) \le C d_1(x,y)$.
as math becomes more advanced, you see fewer numbers, in my experience. at least in pure math, not applied math
the benefit of the abstract approach is that then "regular numbers" becomes a consequence of the abstract theorems, but not vice versa. so an abstract result is strong er in that it tells you about numbers and about things more general than numbers
