Hi all. I have some problems coping with the lack of relationship between limits wrt different absolute values on a number field. Maybe some kind soul could give me a hand?
Fix a prime number $p$. If I see the number $\alpha = \prod_{n=1}^{\infty} n^{1/2^n}$ as an element of $\mathbb{C}_p$, then I can consider $\left|\alpha\right|_p$ and I would like to prove that $\left|\alpha\right|_p \neq 1$.
This seems reasonable if I rewrite $\alpha$ as $\prod_{p} p^{c_p}$ with $c_p = \sum_{n=1}^{\infty} v_p(n) / 2^n$. It would be straightforward to prove if I knew that $\alpha$ is the limit of $\prod_{n=1}^m n^{1/2^n}$ wrt $\left|\cdot\right|_p$, but this seems unlikely to me.
(If it is too complicated I'll post it as an actual question, instead...)
Namely: "If $A$ is in $F$ and $A$ is a subset of $B$, then $B$ is in $F$, for all subsets $B$ of $S$", where $F$ is a collection of subsets of some set $S$.
Since you have only two choices, as you correctly wrote, it must be $\{E, \{a,b\}\}$.
Just out of curiosity, where are you studying on? It might be beneficial to review the part where neighbourhood bases are discussed in your book/notes.
Just for clarity, the definition for neighbourhood base $\mathcal{B}(p)$ at a point $p$ of $(X,\tau)$ should be (piecing it together from memory and the one you linked): $\mathcal{B}(p) \subseteq \tau$ such that:
1. $\varnothing \notin \mathcal{B}(p)$;
2. For every $U,V \in \mathcal{B}(p)$ there is a $W \in \mathcal{B}(p)$ such that $W \subseteq U \cap V$.
3. Any open neighbourhood of $p$ can be written as a union of elements of $\mathcal{B}(p)$.
In particular, your proposed basis failed to satisfy this last property.
@SayanChattopadhyay You know that $$\frac{1}{2\cdot 3\cdot \dotsc \cdot k} < \frac{1}{2^{k-1}}$$ for $k \geqslant 3$, since if you replace all factors in the denominator by $2$, you make the denominator smaller, hence the fraction larger.
@SayanChattopadhyay All terms in that series are positive, so the sum of the series is larger than any of the partial sums. Take a convenient partial sum.
The Online Encyclopedia of Integer Sequences (OEIS) is one of the best tools on the Internet for an amateur to learn mathematics and also perhaps to someday make original discoveries them selves. Currently the site is run by donations. The problem as I see it is that I feel I do not have the right to contribute material there if I not also pay for it.
Stack exchange is also good but it is not as structured as the OEIS. Most of my questions, and I mostly ask questions, are in some form based on what I have learned from surfing the OEIS.
It's a nice doodle. Ascending and descending chain conditions on the bottom, intersection of prime ideals up top, some of her physics stuff on the right (along with some symmetry groups of polygons...)
hi skull ... and hi, mr eyeglasses and @Jasper. I just stopped by before my last advising appointments (well, scheduled ones) start for the last day ever :P
I guess I should say good night, @Mike. It really is night.
But I schedule advising appointments over a 7- to 8-week period, and we're now at the end. Of course, there are students who haven't made appointments, but they're going to need a different adviser next year anyhow ...
I see @Balarka is doing the wonderful Hatcher problem on the homology of the appropriately-identified cube. I loved guiding a grad student through that 10 years ago or whenever it was.
skull, but hopefully not dying for another few months ...
@N3buchadnezzar A Excel spreadsheet formula should do it.
What I mean is that there are programming languages that allows you to write all engineering problems with such symbols/ready-components you have draw above. Then the program will tell you what the outout will be.
But the same problem can be written with code instead of symbols.
@N3buchadnezzar circuitlab.com or similar I searched for pspice which is a program I used to check electric circuits with.
@Vrouvrou You have to only check continuity for preimages of open (or closed) sets. In this case, it means sets of shape $g([0,1]) \cap U$ with $U$ open in $E$.
Since the image of Ted Shifrin's function is $\{a,b\}$, it is enough to check continuity for $\{a\}$ and $\{a,b\}$ (since $g^{-1}(\varnothing) = \varnothing$ is trivial.
@Vrouvrou It is not defined, that's my point!
@Vrouvrou If you find it difficult to understand this in English, I'm sorry. I can more or less understand written French, but I don't know how to speak or write it.
Hum... Ok, I take it back: $g^{-1}(U)$ is a shorthand for the set $\{a \in [0,1] : g(a) \in U\}$. This means that $g^{-1}(U) = E$ for every $U \supseteq g([0,1])$, so it still means that it is enough to check continuity for $g([0,1]) \cap V$ with $V$ open.
i have this set $E=\{a,b,c\}$ with this topology $\tau=\{E,\emptyset, \{a\},\{a,b\}\}$
I have to show if $E$ is path-connected or not ?
I have to construct a continuous function between a and b, b and c , a and c.
1) between a and b: $\varphi_1: [0,1]\rightarrow E, \varphi_1(t)=a, 0\leq t<1, \varphi_1(1)=b$
$\varphi_1^{-1}(\{a\})=\emptyset, \varphi_1^{-1}(\{a,b\})=[0,1[=[0,1]\cap]-1,1[$(is open ), $\varphi_1^{-1}(\emptyset)=\emptyset$ and $\varphi_1(E)=[0,1]$
2) between c and b: if i suppose that $\varphi_2: [0,1]\rightarrow E, \varphi_2(t)=c,0\leq t<1, \varphi(1)=b $ then we have $\va…
@A.P. so i think that the set is not path-connected
user143442
4:55 PM
@Vrouvrou isn't it easier to say that $E$ is not connected and hence not path-connected?
Just wondering, how can you describe a vector field to a non trivial tangent bundle in a global way? When the tangent bundle is trivial you can always just take a vector valued function. But what if it's not? I think one can describe a complete flow and then take the corresponding vector field, that feels more "global" in the sense that it doesn't have to rely on coordinates. Is there an obvious choice for a "global" way to describe a vector field?
@DanielFischer Hi, I would like to prove that given $F\in C(\overline{D(0,1)})$ and holomorphic on $D(0,1)$ such that $\vert F(z) \vert\le 1$ for $\vert z\vert=1$ then $\vert F(z)\vert\le 1$ holds as well on the $(\overline{D(0,1)})$
I tried to take a converging subsequence $z_{n_k}\rightarrow \alpha_0$. I have $\vert z_0 -\alpha_0\vert=\lim_{k\rightarrow +\infty}\vert z_0-z_{n_k}\vert$. I denote $z_n$ an element of $D(z_0,1+1/n)\cap D(0,1)$. @DanielFischer
@user The closed subsets of that topological space are $\{\varnothing, \{c\}, \{b,c\},E\}$. Since every two non-empty closed subsets intersect, it follows that $E$ must be connected.
@Vrouvrou In 1) you write $\phi_1^{-1}(\{a\}) = \varnothing$, but this is not correct. Since you defined $\phi_1$ as $\phi_1(t) = a$ for $t \in [0,1)$ and $\phi_1(1) = b$, it follows that $\phi_1^{-1}(\{a\}) = [0,1)$.
@Vrouvrou Fix an $r \in (0,1]$ and define $I_0 = [0,r)$, $I_1 = [r,1]$, so that $[0,1] = I_0 \cup I_1$ with $I_0$ open. Now consider the following functions $[0,1] \to E$: $\phi_{ab}(t) = a$ if $t \in I_0$, $\phi_{ab}(t) = b$ if $t \in I_1$; $\phi_{ac}(t) = a$ if $t \in I_0$, $\phi_{ac}(t) = c$ if $t \in I_1$; $\phi_{bc}(t) = b$ if $t \in I_0$, $\phi_{bc}(t) = c$ if $t \in I_1$. For each of these, the preimage of $\varnothing$ is $\varnothing$ and the preimage of $E$ is $[0,1]$ (this is a general fact about functions). As for the remaining two open subsets of $E$:
@nerdy My topology is (very) rusty. What do you need?
I heard that the concept of topology is always more abstract than the concept of metric. But, could this notion en.wikipedia.org/wiki/Helly_metric be carried to the topology environment ? Im thinking not. So in a sense, metric is not a subset of topology, they are simply two overlapping sets
I don't understand what you want to achieve, @nerdy. Once you define a metric on a space, by definition you have a metric space...
Also, note that topological spaces are not more abstract than metric spaces but they are strictly more general, in the sense that a metric always induces a topology, but not every topological space is metrizable (i.e. allows the definition of a metric).
For example, topology gives you the minimal framework to work with limits (at least in the more or less classical sense), but to do some calculus (derivatives, integrals, ...) you need more than a generic topological space.
Find all the solutions of the form $y(x)= x^m \sum_{n=0}^{\infty} a_nx^n, \ x>0 (m \in \mathbb{R})$ of the differential equation $3x^2y''+5xy'+3xy=0$.
That's what I have tried:
Since $x>0$ the differential equation can be written as follows.
$$y''+ \frac{5}{3x}y'+ \frac{1}{x}y=0$$
$$p(x)=\fra...
And world cat library. I would expect to pay a couple of hundred dollars though, when I check amazon volume 1 is about $260 (still incredibly expensive @javra
I feel like it will rely on a different trick, try integration by parts, with $u = x\ln(1+x^2)$ and $dv = \frac{1}{x}dx$ or some other variation like that
what context did you encounter it? If it was an exercise from the book, the section you found it in should provide some sort of hint on how to proceed.
@javra I am not sure, but it does look like you could check it out from a library (through world cat if a local one doesn't have a copy). I am sure you already know of it, but there is also Topos Theory, which is a dover book and at around $20
I acknowledged that it was not in terms of u. But you made it sound like I had somehow performed the substitution that led to that 'incorrect' state wrong. So I went back and redid it.
The final trick will involve changing $\frac{2x^2}{1+x^2}$ into something a bit more convenient. We are always allowed to "add or subtract zero", just like we are always allowed to "multiply by one." Its what we "call" zero that makes a difference
Yes., so, using that to simplify your integral, $\int \frac{1}{x^2+1} dx = \int \frac{\sec^2(u)}{1+\tan^2(u)}du = \int \frac{\sec^2(u)}{\sec^2(u)}du = \int du = u+C$
But, remembering that $x=\tan(u)$, and $u=\arctan(x)$
Find all the solutions of the form $y(x)= x^m \sum_{n=0}^{\infty} a_nx^n, \ x>0 (m \in \mathbb{R})$ of the differential equation $3x^2y''+5xy'+3xy=0$.
That's what I have tried:
Since $x>0$ the differential equation can be written as follows.
$$y''+ \frac{5}{3x}y'+ \frac{1}{x}y=0$$
$$p(x)=\fra...
