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I'm concerned that not all mathematical models are time based. For example you might want to model the frequency domain response of a filter specified in the S-Domain. This is a Deterministic-Static-Continuous model, but neither static nor continuous in time. It's static and continuous in the frequency domain (because the equations are analytic). I ...


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The Volterra algebra is an example of a commutative Banach algebra without maximal ideals (hence with empty character space). See also Definition 4.7.38 in H. G. Dales, Banach algebras and automatic continuity, London Math. Soc. Monographs, Volume 24, Clarendon Press, Oxford, 2000.


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You're right to be skeptical. As you've already noted at your other question, Artinian rings are ruled out from consideration since the Krull-Schmidt theorem can be used to prove they have the aforementioned property. Happily, I can demonstrate a ring without the property that is even von Neumann regular. I was inspired by a theorem in Goodearl's book von ...


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Hint: We can let simply $f:=1$ then the values of $g$ can be either $+1$ or $-1$, it neither has to be continuous.


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"Continuous" Borel measures is a jargon used to denote measures not charging any point. Most times is synonymous of "non-atomic". Discrete measures is also a jargon, used for measures which are sum of Dirac masses - for any finite measure this sum is at most countable. This kind of measures are also referred as "atomic" or "purely atomic" measures. ...


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Another way of seeing this about cancellation is to notice that free semigroups/monoids are cancellative, and every semigroup/monoid is a quotient of one.


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Here is a silly example: Let $A$ be any non-discrete topological abelian group. Endow it with the zero multiplication. Then $A$ is a non-unital topological ring and $UV=\{0\}$ is not open for all non-empty subsets $U,V$. In order to get a unital example, consider the unitalization $\widetilde{A}$. The underlying abelian group is just $\mathbb{Z} \times A$, ...


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1) Think of monoids as categories with a single object: any homomorphism $f : M \to N$ that is surjective but not an isomorphism is an example of a functor that is full but not faithful. 2) Pick a prime $p$ and let $\mathcal{P}$ be the category whose objects are the $\mathbb{F}_p^n$ and whose morphisms are polynomials in several variables. Let $\mathcal{F}$ ...


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Let X be a mapping cylinder of the obvious map between the subspaces of the real line $\{n | n \in N\} \rightarrow {0}\cup\{1/n |n \in N^+\}$. Then the obvious map $X \rightarrow ({0}\cup\{1/n |n \in N^+\})\times I$ meets the requirement. Denote $({0}\cup\{1/n |n \in N^+\})\times I$ by $Y$, and ${0}\cup\{1/n |n \in N^+\}$ by $Z$. So we have $f : X ...


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I think that 2) holds if you are referring to property P as one of the convergence theorems like monotone convergence or dominated convergence theorem. The point is that Fatou's lemma remains valid if we replace a.e. convergence with convergence in measure. Try to prove this.


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The example that comes immediately to my mind is the obvious functor from a category $\mathcal{C}$ to its preorder-reflection $\mathcal{D}$, i.e. the category $\mathcal{D}$ whose objects are the same as in $\mathcal{C}$ and with a unique morphism $x \to y$ in $\mathcal{D}$ if and only if there is at least one morphism $x \to y$ in $\mathcal{C}$. It is full ...


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The convex subdifferential of a proper convex lower semicontinuous function defined in a Banach space is a maximal monotone operator (that is a result of Rockafellar). There are several examples of subdifferentials without a convex domain, the one written below being taken from Rockafellar's book "Convex Analysis" on page 218. Let $f(\xi_1,\xi_2)=\max ...


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Let $f(x) = \begin{cases} -1,&x < 0,\\1,& x \geq 0, \end{cases}$ with $x_0 = 0$.


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Also notice that since Borel Set contains every interval and since sigma algebra it is closed under countable intersection we have that for $x\in\mathbb{R}$ $$\bigcap_{i=1}^{\infty}\left[x-\frac{1}{i},x+\frac{1}{i}\right]=\{x\}$$ Thus Borel Sigma Algebra contains every singleton


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For a more funky example, look at the Cantor set.


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How about the union of two disjoint intervals? Any countable set also works.


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Let $G$ be the free group on two elements $a,b$. Let $H$ be the group generated by $g_n = a^nba^{-n}$. H is countably infinitely generated. To get a handle on these arguments the nicest way is to look at covering maps of graphs, and study the fundamental groups (I realize this may be a bit beyond what you'd like to use).


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I think I have an example, but I can't prove it. Let $G$ be the free group generated by two elements $a$ and $b$, and $H$ be its subgroup consisting of all elements which are a product of an even number of elements among $a$, $b$, $a^{-1}$ and $b^{-1}$. Then I'm pretty sure that $R_H = 3$.


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Here are some examples coming from linear algebra. Let $R$ be a ring ( commutative or not). Let $n$ and $1\le k \le n-1$ natural numbers. Consider the ring $N_{n,k}(R)$ of $n\times n$ matrices with elements in $R$ that have zero entries below the diagonal $\{ (i,j)\ | \ j-i =k\}$. Then $N_{n,k}(R)$ has nilpotency index $\le (n-k+1)$, and if $R$ has a ...


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All the topological atlases will be topologically compatible ( that is, the coordinate changes will be homeomorphisms) because the charts are local homeomorphisms with open subsets of $\mathbb{R}$ ( so the topology is already taken care of). However, if you just consider a chart as a bijection, without regard for the topology of $\mathbb{R}$ then it is ...


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A good example is the average of $f\in C[0,1]$ on the Cantor set. If $C=\bigcap_{n\in\mathbb N}C_n$, where $C_1=[0,1/3]\cup [2/3,1]$, $C_2=[0,1/9]\cup[2/9,1/3]\cup[6/9,7/9]\cup[8/9,1]$, etc. As $\mu(C_n)=2^n/3^n$, we define $$ \varphi_n(f)=\frac{1}{m(C_n)}\int_{C_n}f(x)\,dx, $$ and show that the sequence $\varphi_n(f)$ converges, for every such $f$, to ...


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My favourite examples are the Riesz products, that is, weak*-limits of $$\prod_{k=1}^n \big(1+\cos(3^kt)\big)$$ in $C(\mathbb{T})^*$. Colin C. Graham used the Riesz products to give a slick proof of the Wiener–Pitt phenomenon in $M(\mathbb{T})$.


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The spectral theorem for bounded selfadjoint operators is a surprising consequence of the Riesz Representation Theorem for continuous functions $C[a,b]$ on a finite interval $[a,b]\subset\mathbb{R}$. If $A$ is a bounded selfadjoint operator on a Hilbert space $X$ with spectrum $\sigma(A)$, it is not so difficult to show that $\|p(A)\| \le ...


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My favorite example is harmonic measure. Let $\Omega$ be a domain in $\mathbb R^n$ (with smooth boundary $\Gamma=\partial \Omega$, to avoid technicalities). For every $\phi\in C(\Gamma)$ there is a unique harmonic function $u$ on $\Omega$ such that $u(x)\to \phi(y)$ as $x\to y$, for every $y\in \partial\Omega$. Fix $z\in\Omega$. By the maximum ...


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I think I have figured it out. The converse doesn't hold. A counterexample is pretty intuitive. My idea was right, but one just need to tune the parameters a little to get satisfy all the conditions.


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No. $$ \left( \begin{array}{cc} 0 & 1 \\ 2 & 0 \end{array} \right) \left( \begin{array}{cc} 0 & 1 \\ 2 & 0 \end{array} \right) = \left( \begin{array}{cc} 2 & 0 \\ 0 & 2 \end{array} \right) $$ which is symmetric.


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Hint: No, for instance some nilpotent matrices.


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HINT: No, for instance take $A$ antisymmetric.


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I agree that, if you make the direct sum of a finite number of Hilbert spaces, then you do not need to complete anything. Indeed, the finite product of complete metric spaces is also complete. However, things change quite a little bit when you switch to infinite products: indeed, the algebraic direct sum of an infinite family of vector spaces consists not on ...


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I need a couple of preliminary results. Lemma $\mathbf 1$: Suppose that $X$ is $T_1$ and weak Hausdorff. Let $K$ be a compact Hausdorff spaces and $f:K\to X$ continuous; then $f[K]$ is not just compact, but also Hausdorff. Proof: Let $x$ and $y$ be distinct points of $f[K]$, and let $H_x=f^{-1}[\{x\}]$ and $H_y=f^{-1}[\{y\}]$; $H_x$ and $H_y$ are ...


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Take $A = \{a\}$ to be any one-point set, take $B$ to be any set whatsoever, take $f : A \to B$ to be any function whatsoever, and take $g : B \to A$ to be any function whatsoever. Then $g \circ f(a)=a$ and therefore $g \circ f$ is the identity map on $A$. But if $B$ has more than one element then $f$ is not surjective and $g$ is not injective.


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Whenever $g\circ f$ is injective, we have that $f$ is injective. For suppose that $f(x)=f(y)$. Then $g(f(x))=g(f(y))$. Since $g\circ f$ is injective, $x=y$. But, if $g\circ f$ is surjective then only $g$ need be surjective. The hint I'll leave is that you can get a counterexample with finite sets where $A$ is smaller than $B$.


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Your counterexample doesn't make sense; the function $f(x) = x$ is certainly continuous at each point of $A \cup B = (-1/2, 1/2)$. You are, however, correct that statement a) is false, but for a different reason. Consider $M = [0,1]$, and let $A = [0, 1/2)$ and $B = [1/2, 1]$. Then $A \cup B = M$. What happens if we define $f : A \to \mathbb{R}$ to be ...


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Just consider $\mathbb{R}\times\mathbb{R}$ and the sequenece $\{(a_n,b_n)\}_{n\in\mathbb{N}}$ where $$ a_n=\begin{cases} n&\text{if } n\text{ is odd}\\ 0& \text{else} \end{cases} \quad\text{and}\quad b_n=\begin{cases} n&\text{if } n\text{ is even}\\ 0& \text{else.} \end{cases} $$


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I use Sobolev injection Which requires some assumptions on the set $\Omega$. Certainly, being connected is one of them, but also the boundary should not be too irregular: cusps are a problem, as are other features that obstruct passage to the boundary. The usual assumption in the Sobolev embedding theorem is that the domain is of Lipschitz class. (My ...


2

The union of two ideals will possess the absorption property, but not necessarily be closed under addition and subtraction. For example, in $\mathbb Z$, we have $2\mathbb Z \cup 3\mathbb Z$, which contains $2$ and $3$ but not $1=3-2$.


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For the second example start with $\Bbb R$ with the usual topology, and let $X$ be the quotient space obtained by identifying the set of negative reals to a point $p$, so that $X=\{p\}\cup[0,\to)$. Then $X$ isn’t even $T_1$, since every open nbhd of $0$ in $X$ contains $p$. I don’t know a really simple example of the first type. Engelking has the following ...


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Hint. Work out what $s_n$ needs to be so that $\sigma_n = (-1)^n/\sqrt{n}$. The reason this seemed reasonable is that for any sudden swing in $\sigma_n$, there needs to be a corresponding value of $s_n$ about $n$ times larger than the swing.


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Let $\mathscr{E}$ be the usual topology on $\mathbb{Q}$, and let $\mathscr{T}$ be the topology obtained by isolating each point of $\mathbb{Q}$ except $0$. Clearly $\langle\mathbb{Q},\mathscr{T}\rangle$ is Hausdorff, but not locally compact at $0$. Let $U_0=\{0\}\cup\{2^{-n}:n\in\mathbb{N}\}$, and let $U_1=\{0\}\cup\{-2^{-n}:n\in\mathbb{N}\}$. For ...


2

I claim that $I=J=0$ already yields a family of counterexamples to $R\otimes_R R \cong R$. We must have $R\cdot R \neq R$ for this to work (I think this condition might also be sufficient for the natural map to not be an isomorphism, but I haven't worked out all the details). $R=n\mathbb{Z}$ is such a rng: $n\mathbb{Z} \otimes_{n\mathbb{Z}} n\mathbb{Z} ...


0

It's clear that $G3$ always holds. If $A,B,C,D$ are four distinct simple submodules of $M$ such that $(A+B)\cap (C+D)\neq\{0\}$, then $A+B+C+D$ has length $3$. (You can see this by noting that $(A+B+C+D)/(A+B)\cong (C+D)/((A+B)\cap(C+D))$ and that the length of the right hand side is $1$ and then it must be that $len(A+B+C+D)-2=1$. Then it is impossible for ...


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As you have remarked one possible solution is $$ f(x)=\frac{x}{2}+x^2\sin\left(\frac{1}{x}\right) $$ Note that, for $n\geq 1$, $$ f'\left(\frac{1}{ \pi n}\right)=\frac{1}{2}-(-1)^n $$ This means that $f'$ changes its sign on each interval $I_n=\left(\dfrac{1}{\pi(n+1)},\dfrac{1}{\pi n}\right)$, so there is some $s_n\in I_n$ such that $f'(s_n)=0$. Moreover, ...


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Something like $|x|^{3/2} \sin(1/x)$ will work.



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