$\bar{E}$ $\subset$ F for every closed set F $\subset$ X such that E$\subset$F, where $\bar{E}$ is defined as the E$\cup$ E' where E' is the set of limit points E
$E\subset F$. Let $x\in E^\prime$. Then $x$ is a limit point of $E$, hence a limit point of $F$. Since $F$ is closed, $x\in F$. Thus, $\bar{E}\subset F$.
@SimplyBeautifulArt but I used the following: Let x be a limit point of A. If A$\subset$B then x is a limit point of B, hence A' $\subset$ B where A' is the set of limit points of B, I didn't need to use the assumption that x is in B.
i'm showing that that set of limit points of E' are in F. I agree that E' is in F because F is closed, but isn't that a bit circular?
am I incorrect to say that: Let x be a limit point of A. If A⊂B then x is a limit point of B, hence A' ⊂ B where A' is the set of limit points of A ? Does it matter if B is closed or not?
@SimplyBeautifulArt i'm also trying to prove that the closure of a set is closed
may you please tell me if the following proof is correct? please?
Let p$\in$ $(E \cup E′)^c$ which implies that p $\notin$ E and p $\notin$ E'. Since p$\notin$E', $\exists$ $N_r(p)$ : $\forall$ q$\in$ Nr(p) , q $=$ p or q$\notin$ E which is equivalent to Nr(p) $ \cap$ E $=$ $\emptyset$
, therefore Nr(p) ⊂ $E^c$ . Now we would like to show that $N_r(p)$ $\subset$ $E′^c$, so assume that is not the case this implies that $N_r(p)$ $\subset$ E' $\implies$ p $\in$ E' which is a contradiction, therefore $N_r(p)$ $\subset$ $E^c$ $\cup$ E' which means that it is open, therefore the closure is closed.
Let $\Sigma : (\mathbb{N} \rightarrow \mathbb{R}) \rightarrow \mathbb{R}$ be a summation method. What conditions are necessary and/or sufficient for it to be the case that $a < b \Longrightarrow \Sigma(a) < \Sigma(b)$ for all sequences $a,b$? Is this implied by linearity, stability, and/or any other similar properties?
Well the contradiction showed that $N_r(p)$ $\subset$ $E'^c$ and also since $N_r(p)$ $\subset$ $E'^c$ then the neighborhood is contained within the union, thats my argument
$a < b \Longrightarrow b - a > 0 \Longrightarrow \Sigma(b - a) > 0 \Longrightarrow \Sigma(b) - \Sigma(a) > 0 \Longrightarrow \Sigma(a) < \Sigma(b)$. The problem is how do I justify $c > 0 \Longrightarrow \Sigma(c) > 0$?
@SimplyBeautifulArt okay so first since p is not a lp of E, there is a neighborhood that does not intersect E so the neighborhood is contained within the complement of E. Is it so far so good?
@SimplyBeautifulArt since the neighborhood does not intersect E, it therefore must lie in the complement of E. So then I want to show that the neighborhood is contained within $E^c$ $\cup$ $E'^c$ because then that means that it is open. So I say seek a contradiction by assuming that the nbhd is not a subset of $E'^c$ so $N_r(p)$ $\subset$ E'
but that yields a contradiction because then p$\in$ E'
Well, zeta function regularization doesn't respect linearity, for example. What I mean to ask is what's the largest class of summation methods for which this property holds.
Let $A=B\cup B'$ be the closure of $B$. Let $x$ be a limit point of $A$. For all $\epsilon>0$, we can find another point $y$ in $A$ such that $|x-y|<\epsilon/2$.
Lol yeah, that and I remember once in analysis, we had a problem to the effect of, show that if a space is reflexive or has separable dual, then there's a sequence in the unit sphere converging weakly to 0
I found myself having to defend teaching it several times in my career, Mike.
I mean, I've never been a Moore-style point-set nut, but certain results/definitions are needed, and I like the example/counterexample flavor for a good course.
So where's the content of this problem and where are these hypotheses coming from? We realized that the key word was sequence, since you could in principle have a net
And if a space is reflexive, you take a separable closed subspace, then that guy is separable + reflexive, so the dual is then separable, and then you apply the first part of the problem
@TedShifrin yeah, I realised that as well, and i'm trying to minimise it too, but out of curiosity, will using contradiction alot yield other bad habits, in your opinion?
@maths: I think contradiction is (a) necessary occasionally (b) useful for figuring out intuition. But in the latter case, you can often then be more succinct with a direct proof. For example, suppose I ask you to prove $x\cdot y = 0$ for all $y$ implies $x=0$. How do you do it?
@TedShifrin there are a few ways i could think of i could consider the contrapositive of the statement: if a$\neq$ 0 then $\exists$ b $\in$ $\mathbb{R}$ such that ab$\neq$ 0, well if I take b=a then i'm done
I'm stuck at FTA on your abstract algebra book; I started topology and when things were getting too complicated I decided to do it right, from the easier to the harder.
@CaptainAmerica: Well, OK. Just don't overdo. I already said that contradiction sometimes gives intuition for how a direct argument should go. But for some things (like $\sqrt2$ irrational) you have NO options.
@CaptainAmerica16 it sure is. But proving that there's a nontrivial solution to every Pell equation is cooler. And stuff about differential geometry, manifolds and group symmetries is even cooler!
I accidentally saw a proof of L'hopital today. I think it said it used Taylor series, but I tried not to read too much just in case I want to do it on my own.
@Rithaniel: Here is an example, "Prove that the identity element $e$ for a group is unique.", Suppose otherwise, then $\exists e'$ that is also an identity element. This implies that $ee' = e'e = e = e'$, hence it must be true."
I actually got stumped once when asked to prove that the identity element for a group had to be unique, because I forgot the part of the definition which specified that $ea=ae=a,\forall a\in A$. I ended up thinking in circles because I was allowing for distinct right identity and left identity.
Proposition: There is no finite number such that multiplying it with some finite number gives an infinite number:
Proof: Suppose there exists a pair of finite numbers cd such that cd is infinite. We knew that there exists some n,m finite such that n<c<m. Now multiply this both sides by d, we get: nd<cd<md. But this is a contradiction because cd is infinite and nd,md are both finite
Ah, I am using the usual definition: Anything that bijects with an element of the naturals
But I think the proof is actually using something more general: the property that operations between finite numbers formed a closed ring and that there always exists at least a pair of finite number larger than or smaller than any given finite number
i'm trying to minimize complexity for searching a sum in an array, the sum can be any series of consequent numbers. meanwhile i only managed to minimize the complexity to _O( n * ((n+1)/2) )_ do you know computer science?
@Shimmy I don't know enough about array summing algorithm to push that down even further
@Rithaniel Usind the finite fields as examples, since you can pick a representative for each equivalence class as its name and define operations on them so they combine these classes in a structurally preserving manner, it follows that $[c]\in \Bbb{Z}_n$ for fixed n
Hi all, if $F$ is a tempered distribution and if let $\hat{F}$ denote its Fourier transform, what is the precise meaning of $\hat{F} \in L^{\infty}$?
Im doing an exercise in Folland and it asks me to show that the space of tempered distributions satisfying $||F \ast \phi||_2 \leq C||\phi||_2$ for some $C > 0$ is actually equal to the set $\{F \in S' : \hat{F} \in L^{\infty}\}$
Im just confued by what it means for a distribution to be in L infinity... He provides a hint saying to use Plancherel theorem.
Hi, for the sequence $(S_n)$ defined by $S_n = \sum_{k=1}^n (-1)^k \frac{1}{k}$, exercise want me to prove that $(S_{2n})$ and $(S_{2n+1})$ are adjacent sequences, meaning that one of them is increasing, the other is decreasing and that the difference of them converges to 0. I proved this. And the exercise want me to infer that $(S_n)$ is convergent because of the fact I showed.
@Holo I have not thought too much into set theory level when I made algebraic proofs. So multiplication is merely defined to be a map from $\Bbb{N}^2$ to $\Bbb{N}$ that preserves the ordering <
@Holo and yes, that's what I originally want to say, but I keep making typos after typos after typos while on mobile previouly
I can intuitlvely see why but I don't know which theorem can be used to show that proving that $S_{2n}$ and $S_{2n+1}$ will imply all subsequence of $S_n$ is convergent
yeah, clearly I need a lot more logic to think through this properly
But my suspicion is that the "interfinite sets" that I really want to construct, may be already ruled out in some subtheory of PA-, so that product of finite sets can only be finite may be pinned down by something much weaker than PA-
This diagram is, however suggestive that such thing may be able to be defined, probably via some kind of partial ordering instead of total ordering
(you can imagine those bunch of dots are converging to some infinite object. Thus going via the bottom route, you can only get there in infinite steps, but the top route, you can reach it in 9 steps)
(If this diagram is a portion of $\Bbb{R}^2$, and the dots are points in this space, then a topology can be defined such that convergent sequences will behave exactly as suggested by the above, namely there is only one sequence (and its subsequence) can converge to the point at the lower right corner)
This is Theorem 8.2 from baby Rudin, where it is assumed that series given has radius of convergence $1$. :
I can't understand last couple of lines in proof fully. In last line we have to derive $$(1-x)\sum_{n=0}^{N}|s_n-s||x|^n<\frac{\varepsilon}{2}$$ for some $\delta>0$ with $x>1-\...
Is there a standard presentation for the group of units modulo $n$? It'd save me some time if someone could let me know it before I have a deeper look into the literature.
It's actually not too bad, let $J$ be the all 1s matrix, then this guy is $bJ + (a-b)I$
Call the dimension $n$, then $J$ is rank 1 so $0$ is an eigenvalue of multiplicity $n-1$ and $n$ is an eigenvalue of multiplicity $1$
In this case, it'd be $0$ has multiplicity $n-1$ and $bn$ of multiplicity $1$. But then if the eigenvalues of $A$ are $\mu_1,\ldots,\mu_n$, then the eigenvalues of $A - \lambda I$ are $\mu_1 - \lambda, \ldots, \mu_n - \lambda$
This is Theorem 8.2 from baby Rudin, where it is assumed that series given has radius of convergence $1$. :
I can't understand last couple of lines in proof fully. In last line we have to derive $$(1-x)\sum_{n=0}^{N}|s_n-s||x|^n<\frac{\varepsilon}{2}$$ for some $\delta>0$ with $x>1-\...
