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2

Here's another example: $$X=(\mathbb R\times\{0\})\cup(\mathbb Q\times\mathbb Q).$$ Any neighborhood of any point is disconnected, but $\mathbb R\times\{0\}$ is a connected component, so the space is not totally disconnected.


2

How about this space? $$\{(x,y)\in\Bbb R^2\mid x\in\{\frac1n+\frac1m\mid n,m\in\Bbb N\}, y\in[0,1] \}$$


0

You have not died every day since you were born.


-1

One good example of such a sequence is a prime number generating sequence related to the Ulam spiral: $N^2 - N + 41$. Tt generates prime numbers when N = any integer from 0 to 40 inclusive, but obviously fails at 41, returning the result of 41^2. The solutions form a diagonal line on an Ulam spiral that starts at 41 Here is a Youtube video by Numberphile ...


1

In general, the answer is negative. I found an abstract of an article “A new class of spaces with all finite powers Lindelöf” by Natasha May, Santi Spadaro, and Paul Szeptycki. (Topology Appl. 170, 104-118 (2014)): An $\omega$-cover of a set $X$ is a cover with the following property: every finite subset of $X$ is contained in an element of the cover. ...


29

Take $G=\mathbb{Z}/4\times \mathbb{Z}/4$, and $H=Q_8\times \mathbb{Z}/2$ of order $16$, where $Q_8$ denotes the quaternion group. Both groups have exactly $1$ element of order $1$, $3$ elements of order $2$ and $12$ elements of order $4$. Edit: I understood the question as follows: Is there a counterexample where two groups $G$ and $H$ have the same number ...


10

Let $p$ be an odd prime. Let $G$ be the non-abelian group of matrices of the form $$\begin{pmatrix}1&a&b\\0&1&c\\0&0&1\end{pmatrix}\in\operatorname{GL}(3,\mathbb F_p).$$ Then $|G|=p^3$ and each element $g\ne 1$ has order $p$; this follows from the fact that $g-1$ is nilpotent, hence $(g-1)^p=(g-1)^3=0$ and finally ...


18

Here is an example with two groups of order $27$. Consider the group $G$ which is elementary abelian (all elements $x \in G$ satisfy $x^{3} = 1$), of order $27$. And then $H$ the non-abelian group of order $27$ and exponent $3$ (once more, $x^{3} = 1$ for all $x \in H$). Concretely, $$ \left\{\, \begin{bmatrix} 1 & a & b\\ 0 & 1 & c\\ 0 & ...


1

The extension $E/K$ need not be separable. Here is the example I learned from a note by J. Lipman. Consider the rational function field $F=\mathbb{F}_2(y,z)$ and the extension $E=F(x)$, where $x$ is a root of $$f(t)=t^4+yt^2+z\in F[t].$$ If $E/K$ was separable, we would have $f=g^2$, for $g\in K[t]$. We have $g=t^2+\sqrt{y}t+\sqrt{z}$, which means that ...


0

It seems the following. The space $X_0$ of arbitrary functions $f : \mathbb{R} \to \mathbb{R}$ endowed with the topology of pointwise convergence is just the Tychonov product $\mathbb{R}^\frak c$. Thus in order to construct a required counterexample $X\subset X_0$ it suffices to construct a Tychonov (that is completely regular) counterexample $X$ of weight ...


0

The key word here is "operation", not "associative". The associativity of the subset is, indeed, obvious. The fact that it is a binary operation on $M(S)$ is NOT so, and requires demonstration: that is if we call the invertible set of mappings $S \to S,\ A(S)$, that $f,g \in A(S)$ then implies $f \circ g \in A(S)$. The phrase "...with identity $\iota_S$", ...


3

First things first: It's easier for us if you would cite the result by its number in the book, rather than just saying that it comes from the book. This way, those of us who have the book in hardcopy can easily find it. It is Theorem 4.1 in Chapter I. Also, the requirement that $S$ be nonempty is useless. I never understood why some people find it helpful ...


0

Once associativity is established for the larger set, it follows immediately for the smaller set too. The things to worry about here are that composition of invertible mappings is again invertible (so that indeed the operation of composition restricts to the smaller subset) and that the identity mapping is invertible (so that it is present in the smaller ...


3

Here is another simple example: Write $n$ in binary (or any other basis) and define $\eta$ to be the erasing of the first digit. This has the desired property because for all integers $x$, as long as $2^m >x$ we have $$\eta(2^m+x) =x$$


3

Let $A_{1}$ be the set of all positive integer whose lowest prime factor is 2 , $A_{2}$ be the set of all positive integer whose lowest prime factor is 3 , $A_ {3} $ be the set of all positive integer whose lowest prime factor is 5 and so on....clearly $A_{i}$ are infinite and mutually disjoint. Now define $\eta \colon \mathbb{N} \rightarrow\mathbb{N}$ by $ ...


4

Let $x\in\mathbb{N}$, and let $\eta(x)$ be the number of times the digit 1 appears in the binary expansion of $x$. For any $n$, the equation $\eta(x)=n$ has infinitely many solutions. For instance, if $n=4$, then the numbers with the binary expansions $1111$, $11110$, $\ldots$, $111100\ldots 00$ are all mapped to $4$. Notice that in this example, we have ...


0

An answer to the first question is Counterexample. Let $\mathbb T^2$ (the 2-torus) endowed with periodic coordinates $(\theta,\phi)$ and with the symplectic form $\Omega = \mbox{d}\theta \wedge \mbox{d}\phi$. Then a costant vector field on $\mathbb T^2$ is locally Hamiltonian but not Hamiltonian (Compare [1], chapter 5). This can be important, e.g, in ...


0

@Adayah's answer provides the proof of this lemma: Lemma: Let ${(u_n)}_{n \geq 0}$ and ${(v_n)}_{n \geq 0}$ be two sequences of positive numbers such that $v_n \searrow 0$. If $\sum u_n v_n < \infty$ then $\epsilon \sum_{i=0}^{n(\epsilon)} u_i \to 0$ when $\epsilon \to 0^+$, where $n(\epsilon)=\min\{n \mid v_{n+1} < \epsilon\}$. Surely, @Adayah's ...


0

Another example. If $\mathbb O$ is an algebra of octonions with a standard basis $e_0,\ldots,e_7$, then $L=\{\pm e_0,\pm e_1,\ldots,\pm e_7\}$ is a finite non-associative and non-abelian loop with elegant law of composition (Fano plane).


4

$\eta(x) = \lfloor |\tan(x)| \rfloor$ will do, because (1) $\tan$ is periodic with period $\pi$ and maps the set $D = [0, \pi/2) \cup (\pi/2, \pi)$ continuously onto $\mathbb{R}$ and (2) for any $n\in\mathbb{N}$, reduction modulo $\pi$ maps $\{m \in \mathbb{N} \mathrel{|} m > n\}$ onto a dense subset of $D$. Statements (1) and (2) imply that for any real ...


8


2

$\eta\left(x + \frac{n(n+1)}{2}\right) = x$ where $x \leq n$. Equivalently, $\eta(x) = x - \frac{n(n+1)}{2}$ where $n$ is such that $\frac{n(n+1)}{2} \leq x < \frac{(n+1)(n+2)}{2}$, $\therefore n = \lfloor\frac{1}{2}(\sqrt{8x+1} - 1)\rfloor$ This is a nonrecursive formulation of Farnight's answer.


8

Let for any $x\in \mathbb N$, $x>1$ $\eta(x)$ be the number of factors in factorization of $x$ in the product of prime numbers. We define $\eta(1)=1$. Then $\eta$ satisfies your conditions. Here $\mathbb N=\mathbb{Z}_{>0}$. Another variant: Infinite Hotel. Third variant. For $x\in\mathbb N$ let $\eta(x)=1+\max\{k\in\mathbb{Z}_{\geq 0}:2^k\mid x\}$, ...


6

You could choose $$\eta(0,1,2,3,4,5,6\dots)= (\color{red}0,\color{blue}{0,1},\color{orange}{0,1,2},\color{green}{0,1,2,3}\dots)$$ i.e: $\eta(0)=\eta(1)=0$ and $$ \eta(n+1) = \begin{cases} \eta(n)+1 & \text{if } \eta(n)\leq \max\{\eta(0),\dots,\eta(n-1)\} \\ 0 & \text{otherwise} \end{cases} $$ It's easy to see that $\eta(x)=n$ has ...


1

I think the following works. Let me know if anything is unclear; I've been trying for a while to also find an elementary proof of this fact, and I'll be very glad if this ends up being correct! We follow Eisenbud's hint. Suppose $\alpha\colon A \twoheadrightarrow B$ and $\beta\colon B \rightarrowtail C$ form such a factorization. Let $C\alpha$, $C\beta$ be ...


0

If $A$ is a coherent ring, any ring of power series $A[[X_1,\dots,X_n]]$ is faithfully flat over $A$. Also, the direct sum of a flat $A$-module and a faithfully flat $A$-module is faithfully flat. The direct sum $\bigoplus\limits_{\mathfrak m\in\operatorname{Max}A} A_{\mathfrak m}$ is faithfully flat over $A$.


3

Here is another example involving closed (but not bounded) sets. Let $A_n = (-1, 1) \times (-n, n)$. Let $B_n = \mathbb R^2 - A_n$. Each $B_n$ is connected. We have $$ \bigcap_n B_n = ((-\infty, -1] \times \mathbb R) \cup ([1, \infty) \times \mathbb R). $$ Thus, $\bigcap_n B_n$ is disconnected.


1

$\Delta \implies \Theta$ Let $\varepsilon > 0$ and take such $N_1$ that $$\sum_{i=n+1}^m \frac{r_i \log r_i}{q_{i+1}} < \varepsilon$$ whenever $N_1 \leqslant n \leqslant m$. There is such $N_2$ that $$\frac{ \displaystyle \sum_{i=0}^{N_1} r_i \log r_i }{ q_{m+1} } < \varepsilon$$ if $m \geqslant N_2$. Let $N = \max \{ N_1, N_2 \}$. For $h < ...


1

So you can regard any abstract group as a topological group by just giving it the discrete topology, but this usually defeats the purpose of topological groups, which is to exploit a meaningful topology on the group in order to do something. Most of the answers here are saying something along the lines of: "If you try to do the analogous stupid thing for ...


0

For example, consider x = NU{0} under discrete topology and Y = {0}U{1/n|n€N} as subspace of real line R. Difine f:x------Y by f(0)=0,f(n)=1/n,n€N.then f is a bijection. and f is continues. Therefore,y is continous image of x. Note that x is being a discrete space is locally connected but y is not locally connected.


3

What you describe is in fact known as a topological group, rather than an algebraic group. And yes, any group can be made topological by giving it the discrete topology (though this is usually not interesting. An algebraic group, however, is slightly different (but of the same flavour). It is a group which is also an algebraic variety (usually assumed ...


1

I suppose you could always endow any group with the discrete topology (every subset is open). Then the group operations are trivially continuous.


5

What you've described is a topological group. The definition of an algebraic group is somewhat more complicated. And algebraic groups aren't topological groups (although in certain contexts they give rise to topological groups). There isn't really a sense in which any group $G$ can be viewed as an algebraic group, unless you drop a finiteness condition ...


15

The silly counterexample is this: if $H$ is not normal in $G$, then we have $$H \not\lhd G\quad G\lhd G$$ Indeed, this need not even be true if $K$ itself is normal in $H$. For example, in $S_4$, we have $$C_2 \lhd V_4\lhd S_4$$ but $C_2\not\lhd S_4$. (Here, $V_4 = \{(1), (12)(34),(13)(24),(14)(23)\}$ and $C_2 = \{(1), (12)(34)\}$) The flaw in your ...


8

G is a normal subgroup of itself, but it might have subgroups that are not normal.


0

Here is one I found myself (those are complete prime factorizations): $$\begin{align}38&=\textbf {2}\cdot 19\\ 38\, 111&=\textbf{23}\cdot \color{#0bc}{1657}\\ 38\, 111\, 111 &=\textbf{233}\cdot \color{#0bc}{163567}\\ 38\, 111\, 111\, 111 &=\textbf{2333}\cdot 31\cdot 526957\\ 38\, 111\, 111\, 111\, 111 ...


12

The following proposition provides only a partial response to the latest edit. Proposition. There does not exist an $ f: \Bbb{R} \to \Bbb{R} $ that fits the stronger requirement reflected in the OP’s latest edit if we assume $ g $ to be continuously differentiable. Proof As $ g $ is, by assumption, surjective and differentiable everywhere, there ...


2

One family of functions that works is $f(x)=|x|$ and $g(x)$ any monotonic, surjective, differentiable function with $g(0)=g'(0)=0$. Suppose $g$ is monotonically increasing. Then $g(x)<0$ for all $x<0$ and $g(x)>0$ for all $x>0$. Thus, $f(g(x))=-g(x)$ for $x<0$ and $f(g(x))=g(x)$ for $x>0$. It follows that $f(g(x))$ is continuous and ...


1

Consider $$A = \begin{pmatrix} 46 & 7 & 8 \\ 7 & 36 & 16 \\ 8 & 16 & 30\end{pmatrix}\qquad \text{ and }\qquad B=\begin{pmatrix} 34 & 7 & 12 \\ 7 & 48 & 10 \\ 12 & 10 & 44\end{pmatrix}.$$ Then $$\frac{1}{2}\|A+B\|_1 =\frac{1}{2}\|A+B\|_{\infty}=62$$ however $$ \|A^{1/2}B^{1/2}\|_1\approx 62.1527 \qquad \text{ ...


3

The function $g,$ which can actually be defined for $x\in \mathbb {R},$ is uniformly continuous on every compact subset of $\mathbb {R},$ and is differentiable everywhere. Proof: $f$ is continuous everywhere, hence is bounded on compact sets. Thus the sum converges uniformly to $g$ on compact sets by the Weierstrass M-test. As the summands are all ...


2

Theorem $\,\ [a,b]\,$ exists $\,\Rightarrow\, (a,b) = ab/[a,b],\: $ for $\ (x,y)=\gcd(x,y),\,$ $\,[x,y] = {\rm lcm}(x,y)$ Proof $\ \ c\mid a,b \iff a,b\mid ab/c \iff [a,b]\mid ab/c\iff c\mid ab/[a,b]$


5

Let $a,b \in M$ and $m \in M$ be a least common multiple of $a$ and $b$. As $ab$ is a common multiple of $a$ and $b$, we must have $m \mid ab$, say $md = ab$ for some $d \in M$. Now let $t \in M$ be any common divisor of $a$ and $b$, say $a = st$, $b = rt$, define $m' = rst$. Then $m'$ is a common multiple of $a$ and $b$, so for some $d' \in M$ we have $m' = ...


0

Let $d=-1$, so that our field is $\mathbb Q[i]$ and the ring of integers is $\mathbb Z[i]$. Then for fixed $\beta$, the numbers $\gamma\beta$ with $\gamma\in\mathbb Z[i]$ form a square lattice. The trick is now to translate this lattice in such a manner that it has two distinct points of minimal norm. So for example we can let $\beta=2$ and translate it by ...


0

For $n=1$ we have $$\sum_{i=0}^ni^4=1 $$ and $$\left(\sum_{i=0}^ni\right)^3=1$$ so that $n=1$ is a counter-example to the (therefore wrong) claim that for all $n$ $$\sum_{i=0}^ni^4\ne\left(\sum_{i=0}^ni\right)^3.$$


3

No need of finding particular $n$s such that the LHS differs from the RHS. The LHS is a fifth-degree polynomial in $n$, the RHS is a sixth-degree polynomial in $n$, hence they cannot be the same.


0

$$\sum\limits_{i=0}^n i^4 =\frac{1}{30}n(n+1)(2n+1)(3n^2+3n-1)$$ And: $$\left(\sum\limits_{i=0}^n i\right)^3 =\frac{1}{8}n^3(n+1)^3$$


3

The first counter-example is $n=2$, since $$\sum\limits_{i=0}^2 i^4 = 0^4+ 1^4 +2^4 = 0 + 1 + 16 = 17,$$ and $$\left(\sum\limits_{i=0}^2 i\right)^3 =(0+1+2)^3= 3^3= 27,$$ and $17\neq 27.$


15

The first condition suggests $f(x)=|x|$ as a relatively simple example. Then $g(x)=x-|x|$ is also not differentiable at $x=f(0)=0$, and $(g\circ f)(x)=|x|-|x|$ is identically zero, so it is certainly differentiable at $x=0$.


4

$$ f(x) = g(x) = \begin{cases} \frac{1}{x} & x \neq 0 \\ 0 & x=0 \end{cases}, $$ if you don't mind lack of continuity.


12

Borrowing from this answer: $$f(x)=\begin{cases} x+1 & x\in\mathbb Q\\ x& x\notin\mathbb Q\end{cases}$$ $$g(x)=\begin{cases} x-1 & x\in\mathbb Q\\ x& x\notin\mathbb Q\end{cases}$$ These functions are inverses of each other and nowhere continuous, ergo nowhere differentiable. Their composition is the identity which is differentiable ...



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