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We've seen it's not hard to construct explicit counterexamples. But it's at least a little fussy, things have to add up just right. Just for fun we might note that it's easy to use a little functional analysis to avoid that fussiness. Say $X$ is the space of all sequences $a=(a_1,\dots)$ such that $$\lim_{n\to\infty}\frac1n\sum_{j=1}^n|a_j|=0,$$with norm $$|... 3 How about the sequence  a_n = \sqrt[3]{n}, if n is the cube of an integer and 0, otherwise. I think this shows that the claim is not true, as you suspect. Proof Whenever n=m^3 then \frac{1}{n}\sum_{k=1}^n a_k = \frac{m(m+1)}{2m^3}. These are the maximum values of the sequence on the range [m^3, (m+1)^3-1] so this tells us the \limsup_{n\... 1 I only saw this question today; it is indeed hard to find examples of this sort of thing. A somewhat elaborate example, essentially the spectrum of the distributive continuous lattice of lower semicontinuous functions I \to I on the unit interval (where elements of the spectrum are prime elements of the lattice), is given in The Spectral Theory of ... 6 Take f:\mathbb R \rightarrow GL(\mathbb R,2) any bijection. Remember GL(\mathbb R,2) is the group of invertible 2\times 2  matrices. We now define x\star y=f^{-1}(f(x) f(y)) (where f(x) and f(y) are multipled like normal matrices). Then cleary (\mathbb R,\star) is isomorphic to GL(\mathbb R,2) as f(x\star y) = f(f^{-1}(f(x)f(y))=f(x)f(y).... 11 If you already know that matrix multiplication is associative, but not commutative, then you can just choose your favorite bijection f:\mathbb R\rightarrow M_2(\mathbb R), since the two sets are equinumerous. Then, define a\oplus b = f^{-1}(f(a)f(b)) to get an operation on \mathbb R which is associative, but not commutative. If you want to have ... 43 We can define x \oplus y=y. Then (x \oplus y) \oplus z =z= x \oplus (y \oplus z) but y=x \oplus y \neq y \oplus x=x 1 Yes, there are many other bilinear products on a given vector space which are associative but not commutative. For example, real associative algebras (not necessarily commutative), see also this MO-question. Furthermore the quaternion algebra is a real division algebra which is associative but not commutative. Edit: You added that the algebra must be part ... 6 The first counterexample one tries should be the simplest. Let$$X=\{\#\}, \ \ Y = \{*,\&\}.$$Define$$f(\#) = * \ \mbox{ and } \ g(*) = g(\&) = \# $$Note that$$ (g\circ f)(\#) = \#$$Now examine (f\circ g). Do you see the issue? 1 Let the domain of f be the integers and the domain of g be the reals. Then take$$f(n) = n \\ g(x) = \lfloor x \rfloor$$It is easy for see that for all n in the domain of f,$$g(f(n)) = \lfloor n \rfloor = n$$so g\circ f = I_X. Yet for real x (in the domain of g)$$ f(g(x)) = \lfloor x \rfloor \neq x $$unless x happens to be a integer.... 0 Let X=\{0\} and Y=\mathbb{Z}. Now define f:X\to Y by f(0)=0 and g:Y\to X by g(x)=0. Now g\circ f= I_X and f\circ g\neq I_Y since f\circ g(1)=f(0)=0\neq I_Y(1). 1 Let f be the natural inclusion of the real line into the plane, and let g be the natural projection of the plane onto the real line. 7 There is a theorem that if a function is continuous, bounded variation, and has the "Luzin N property" (i.e. maps measure zero sets to measure zero sets) then it is absolutely continuous. One way to get bounded variation is to assume monotonicity but this is not the only way. However, if you don't have bounded variation then you will not have absolute ... 2 The set of 2\times 2 matrices with real entries is non-Abelian when the operator is multiplication, but it has an infinite number of Abelian subgroups. For example consider any subgroup of the form$$\{A | A = \begin{bmatrix} p^n & 0\\ 0 & 1\end{bmatrix} \mbox{ where } n\in \mathbb{Z}\}$$where p is a constant and can be any prime. 2 Your question: Let f_n:X\rightarrow \mathbb{C} be a sequence of measurable functions such that f_n\rightarrow f in measure. Let f_{n_k} be a subsequence of f_n such that f_{n_k}\rightarrow g pointwise a.e.. Then, is it necessary that f=g a.e.?What would be a counterexample? The answer is YES. It is true that f=g\: a.e.. First, ... 12 Consider the subgroups of \mathrm{SO}(3) (visualized as the rotational symmetries of the 2-sphere) representing rotations about a fixed axis through the center of this sphere. There are infinitely many choices of this axis, each of which specifies an (abelian) subgroup isomorphic to \mathrm{U}(1). 43 Any infinite group G must have infinitely many abelian subgroups. Note that for each x \in G, there is a cyclic subgroup \langle x \rangle, which is abelian. If there is an x such that \langle x \rangle is infinite, then \langle x \rangle has infinitely many abelian subgroups. If no such x exists, there must be infinitely many distinct finite ... 30 Take the product G = S_3 \times \Bbb Z, which is non abelian since it has a non-abelian subgroup, namely S_3. However, \{1\} \times n\Bbb Z are abelian subgroups of G for every n \geq 0. 0 The other answers find counterexample for f(A\cap B) =f(A)\cap f(B), but it is easy to show that f(A\cap B)\subseteq f(A)\cap f(B): If A\subseteq B, then f(A)\subseteq f(B). Now, A\cap B\subseteq A, B, and thus f(A\cap B)\subseteq f(A), f(B)\implies f(A\cap B)\subseteq f(A)\cap f(B). Actually, equality holds if and only if f is injective. ... 2 Just to give a "minimalist" counterexample that clearly makes use of the hints in the comments: consider X = \{0,1\}, let A = \{0\} and B = \{1\}. Take f: X \to X: x \mapsto 0. Then$$f(A\cap B) = f(\emptyset) = \emptyset$$But$$f(A) \cap f(B) = \{0\} \cap \{0\} = \{0\}$$0$$A=\{0,-1\} \,\, \, , \,\, \, B=\{0,1,-2\}\,\,, \,\, C=\{0,1,-1,-2\}\,\,,D=\{1,2,3\}$$let$$f:C\to D$$such that$$f=\{(0,1),(1,1),(-1,3),(-2,3)\}$$we have f(A)=\{1,3\} , f(B)=\{1,3\} , f(A\cap B)=\{1\} whereas f(A)\cap f(B)=\{1,3\} Generally we have$$f(A\cap B)\subseteq f(A)\cap f(B)$$7 f(x) = x^2, A = [-1,0],B = [0,1]. f(A\cap B) = f(\{0\}) = \{0\} But f(A)\cap f(B) = [0,1] 1 The sets X_s = X \times \{s\} \subset X \times S are all open since S is discrete, and \pi_X|_{X_s} \colon X_s \to X is a homeomorphism (the inverse is x \mapsto (x,s)). Since \{X_s\}_{s\in S} form an open cover of X\times S, we have shown \pi_X is a local homeomorphism. 3 You have given an argument that \pi_X|_{U_x} is continuous and open, but you are missing the most important part: a homeomorphism needs to be a bijection! In fact, \pi_X|_{U_x} won't be a bijection if you choose U_x and V_x arbitrarily (for instance, you might choose U_x to be all of X\times S and V_x to be all of X, and then \pi_x is not ... 2 As mickep observes in the comments, the spectra are not the same! The Dirichlet spectrum is$$ \lambda_{kl} = \bigg(\frac{\pi k}{a}\bigg)^2 + \bigg(\frac{\pi l}{b}\bigg)^2,\ k,l = 1,2,3,\ldots $$while the Neumann spectrum is$$ \nu_{kl} = \bigg(\frac{\pi k}{a}\bigg)^2 + \bigg(\frac{\pi l}{b}\bigg)^2,\ k,l = 0,1,2,3,\ldots $$The intuition is that the ... 2 You can consider [0,1] and \mathbb{R}. 1 Let X_1 and X_2 have the discrete topology. Then X=X_1\times X_2 has the discrete topology, so \mathcal{B}_X=P(X). But if X_1 and X_2 are large enough \mathcal{B}_{X_1}\otimes\mathcal{B}_{X_2} will not be all of P(X). Here's one way to prove this. For S\subset X and x\in X_1, let S_x=\{y\in X_2:(x,y)\in S\}. Now let$$\mathcal{A}=\{...

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The preimage of a open set is still open. And the preimage of a singleton under a bijection is still a singleton. So the singletons is open in both spaces. A space with all singletons open can only have discrete topology.

6

If all singletons are open, then the space is discrete, because an arbitrary union of open sets is an open set, and any set is a union of singletons. So the second space has the discrete topology. A discrete space is only homeomorphic to another discrete space whose underlying set has the same cardinality. So there are no "nontrivial" examples.

5

No. Consider $a_n=1/n$ while $n=2^k$ for some $k$, and $a_n=1/n^2$ while $2^k<n<2^{k+1}$.

1

Summarising the comments: The lower limit topology being strictly finer than $\tau_{\text{usual}}$ means it's not equal to $\tau_{\text{usual}}$. It's essentially the difference between $<$ and $\le$: $x < y$ means exactly $x \le y$ and $x \neq y$. Or really, the difference between $\tau_1 \subsetneq \tau_2$ and $\tau_1 \subseteq \tau_2$. The only ...

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An easy metric construction is $d'(x,y)=d(g(x),g(y))$ where $g$ is a bijection and $d$ is another metric on the base set. In the new metric the open sets are the $g$-images of open sets in the original metric. So all we have to do is come up with a bijection that maps $f$ to a discontinuous function. For example, let $g$ swap $0$ and $1$, that is $g(0)=1$, $... 0 Sure. Let$d |_{[1, + \infty)^{2}}, d |_{(- \infty, 1]^{2}} $be equal to the standard metric, but if$x < 1 \leq y$or$x \leq 1 < y$, then let$d(x, y) = |x| + |y| + 7$. You can check this is a metric. But you'll have that if$\epsilon < 7$, then we get weird behavior around$1$. Namely, we have discontinuity at$-1$. Refer to the definition of ... 1 Let$M = \{1, e, a, b, 0\}$where$e^2 = e$,$aa = ab = 0$,$ae = b$,$ba = bb = 0$,$be = b$,$ea = eb = 0$,$1x = x1 = x$and$0x = x0 = 0$for all$x \in M$. Take$N = \{e\}$. Then$N$is a submonoid of$M$(it is even a subgroup) and one has$xy \in N \iff yx \in N$. However,$aN = \{ae\} = \{0\}$but$Na = \{ea\}$. 5 I would say that$f(x)=-x^2$is anyway defined on all$\mathbb R$. Moreover, I think that the usual definition of a convex function (see https://en.wikipedia.org/wiki/Convex_function) is the opposite with respect to the one you adopted here, so$f(x)=x^2$is indeed convex. Having said that, you can by sure find a strictly convex function$f(x)$defined on ... 1 The usual property says that$\phi(N)$is normal in$\phi(G)$, so actually it's not necessary in this case that$\phi(N) \vartriangleleft H$. Although the property is false in the general case as Devlin said, if we consider that$\ker \phi \subseteq N$, then the result follows. Indeed, consider the map$\psi: G \longrightarrow \phi(G)/\phi(N)$such that$\...

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It is true when $N$ contains $\ker \phi$. That is part of the isomorphism theorems.

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It is true that $\phi (N)$ will be normal in $\phi(G)$ (though not in $H$ itself!), and thus $\phi(G)/\phi(N)$ will be a group (to see this, use the fact that $\phi$ is a group homomorphism). However, the claim itself is false; take $N \leq G$ and $\phi$ to be the trivial map, so $\phi(G)/\phi(N)$ is trivial but $G / N$ is not.

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Let $a\in R$ be such an element. Then there exists in particular an element $e\in R$ with $ae=a$, and an element $\bar a$ with $a\bar a=e$. If the multiplication is commutative, this implies that $e$ is unity: For $b\in R$ with $b=ac=ca$, say, we find $$be=cae=ca=b.$$

0

Let $S$ be any semigroup with the property that $xS = S$. e.g.the one with two elements $x$ and $y$ satisfying $x^2 = yx = x$ and $y^2 = xy = y$. Then, the semigroup rng $R = \mathbb{Z}[S]$ will satisfy $xR = R$; e.g. if $$\sum_i m_i s_i$$ is an element of $R$, then we can select $t_i$ so that $x t_i = s_i$ and get $$x \sum_i m_i t_i = \sum_i m_i (x ... 1 The simplest counter example for this question is f(x)=x^2. This function is concave up on [1,\infty], but most real analysis textbooks will show that it is not uniformly continuous. 5 What follows is not so much an answer to the question (it was already fully answered in comments) as an essay intended to bring consolation to, perhaps even cheer up, the inequality stated in the question, which must feel somewhat mistreated, poor thing, and justifiably so. The counterexample provided by "Macavity" is elegant (albeit not fully spelled out... 1 Your reasoning is correct. Your argument is not quite correct. One can use the following, for a commutative ring R with 1 and A an R-algebra -$$M_n(R) \otimes A \simeq M_n A$$Note that the above is in fact an R-algebra isomorphism. Now let A = R^2 and use the commutativity of tensor products with direct sums. Coupled with (2), this ... 0 associative but not commutative String concatenation: "This is my interesting sentence." = "This" + "is my" + "interesting" + "sentence." = ( "This" + "is my" ) + ("interesting" + "sentence.") but not "This" + "interesting" + "sentence." + "is my" = "This interesting sentence. is my" 3 What about f(x, y) = \frac{x y}{x^2 + y^2} if (x, y) \neq (0, 0), and f(0, 0) = 0? It's clear that x \mapsto f(x, y) is smooth if y \neq 0, and of course for y = 0 as well since f(x, 0) is identically zero. By symmetry, y \mapsto f(x, y) is smooth for each fixed x. This function is discontinuous at (0, 0) since the limit as (x, y) ... 0 The product fact is indeed enough: if we have X and Y that are limit point compact Hausdorff and such that X \times Y is not limit point compact, we know that neither X nor Y is compact (as the product of a limit point compact space and a limit point compact k-space (which includes all compact Hausdorff spaces) is limit point compact. But this is ... 0 \omega (the first uncountable ordinal) is limit point compact and Hausdorff with the order topology, but not compact. 4 If you want to prove that such a homomorphism exists, you only need to find one. And because any homomorphism will do, there is nothing wrong with choosing one that is as simple as possible. So, what is the most simple homomorphism that you can think of...? In the comments, you suggested \mathbb Z/2\mathbb Z\times \mathbb Z\to \mathbb Z/2\mathbb Z. I ... 1 Let me elaborate a bit on the suggestion I made in the above comment. Let \varphi \colon \mathbb R\to\mathbb R be any smooth function such that \varphi(x)=0 for x\le 0 and \varphi(x)=1 for x\ge1. I.e., something similar to one half of bump function. Let us define$$f(x,y)= \begin{cases} 0,&x,y\le0\\ \varphi(\frac xy),& 0<x\le y\\ \...

1

For proving $f(x)= x^3\sin\frac1x$ is not uniformly continuous on $(0,+\infty)$ using the definition, note that if $x-y>\delta/6>0$ then we have \begin{align}f(x)-f(y)&>x^3\left(\frac1x-\frac{1}{6x^3}\right)-y^2 \\ &=x^2-y^2-1/6> \frac16 (\delta(x+y)-1).\end{align} By the way, even if we assume $f(x),g(x)\to+\infty$ your general claim ...

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I assume you mean "uniformly continuous on $(0,\infty)$, since otherwise, your variable $M$ is overloaded. For your particular function, the easiest way to show it is not uniformly continuous is to show that $f'(x)\to\infty$ as $x\to\infty$. Such a function cannot be uniformly continuous because of the mean value theorem. That said, for your general ...

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