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0

No, it's not! Suppose $R$ is a commutative unitary ring and $M$ a free $R$-module of rank $n$. If $A\subset M$ is a linearly independent set, then $|A|\le n$. We can assume $M=R^n$. A linearly independent set $A\subset R^n$ with $|A|=m$ gives rise to an injective $R$-module homomorphism $\phi:R^m\to R^n$, and then $m\le n$. (You can find a proof here.) ...


3

Off the top of my head, here is something you could try. (Everything you need should be in Fulton's book or Cox–Little–Schenck.) Explain how blowing up a subvariety of a toric variety corresponds to subdivision of the fan. Illustrate with simple examples, like blowing up a point in $\mathbf P^2$ and a line in $\mathbf P^3$. Now use this to ...


1

An example is given in N K Bary's A Treatise on Trigonometric series Volume 1 Chapter 4 section 20. There are other examples of 'similar' nature in this monumental work. It appears that Katznelson's example is countable. The construction is very tedious. It involves Fejer's polynomial. A detail analysis to show that the set of divergence is of second ...


1

Let $\{x_n\}_n$ be the $\pm1$ sequence $$\underbrace{+1}_1,\underbrace{-1,-1}_2,\underbrace{+1,+1,+1}_3,\underbrace{-1,-1,-1,-1}_4,\underbrace{+1,+1,+1,+1,+1}_5,{-1,\ldots} $$ i.e, the $n$th block of constant signs has length $n$ (a more explicit description of a somewhat similar sequence would be $x_n=(-1)^{\lfloor\sqrt n\rfloor}$). Then the sum of the ...


0

Note that in the proper definition of almost convergence we can chose $n=0$ to get $$\forall \epsilon > 0 \exists N_0 \forall N > N_0 \quad \left|\frac1N \sum_{i=1}^N x_i - L\right| <\epsilon$$ wich is precisely the definition of a converging sequence for the sequence of Cesàro means $\sigma_N := \frac1N \sum_{i=1}^N x_i$ For the interesting ...


0

If $A$ is not square, then the determinants of $A^tA$ and $AA^t$ are equal if and only if $A$ has neither full row rank nor full column rank. In this case the common determinant is $0$ Indeed, if the rank of $A$ is the same as the number of rows, then $AA^t$ is invertible, while $A^tA$ isn't. Similarly in the case the rank is the same as the number of ...


0

This is not true even in $\mathbf{R}$, try it with any matrix, it will most likely be false. In particular $A={2 \choose 1}$ works.


4

How about $R = \mathbb R$ and $A = \begin{bmatrix} 2 & 1 \end{bmatrix}$?


4

No. Let $x\neq y$ and let be $ U$ a clopen neighbourhood of $x$ not containing $y$. Then the complement of $U$ is a clopen neighbourhood of $y$, disjoint from $ U$.


0

Define $$ f(x, y) = \begin{cases} \sqrt{x^{2} + y^{2}} & \text{if $x \neq 0$ and $y/x$ is rational,} \\ 0 & \text{otherwise.} \end{cases} $$ Theorem: The function $f$ has (path-)connected graph, but is continuous at precisely one point. Proof: The graph of $f$ is the union of: Lines (through the origin) of irrational slope in the $(x, y)$-plane; ...


0

Let: $$f(x,y)=\sin\left(\frac{1}{x^2+y^2}\right)$$ for $(x,y) \neq (0,0)$ and $f(0,0)=0$ Then graph of this function is: $$\{(x,y,\sin\left(\frac{1}{x^2+y^2}\right))|(x,y) \neq (0,0)\} \cup \{(0,0,0)\}$$ Closure of this graph is: $$\{(x,y,\sin\left(\frac{1}{x^2+y^2}\right))|(x,y) \neq (0,0)\} \cup \{(0,0,x): x \in [-1,1]\}$$ There is a theorem that ...


3

Hint: Choose a function such that the overall limit at $(0,0)$ does not exist (but the limit along some path does). Example: take $$ f(x,y) = \begin{cases} \frac{x^2}{x^2 + y^2} & (x,y) \neq (0,0)\\ 0 & x = y = 0 \end{cases} $$ Verify that $\Gamma_f$ is path-connected, but $f$ is not continuous at $(0,0)$.


1

Let $Q$ be the space of non-negative rational numbers, $N$ the natural numbers including $0$, and $q:Q\to Y=Q/N$ the quotient map identifying $N$ to a single point. Note that $Y$ is a Hausdorff normal space since $Q$ is such a space and $q$ is a closed map. Now look at $Y\times Q$. This space is Hausdorff. We will show that it's not compactly generated. Let ...


1

Several examples of finite loops (quasigroups with left and right identity) of small order with some additional properties can be found here. Another source showing several examples of finite loops Cayley tables. A third example of very group-like loop which is not a group, answer to a related Math.SE question. Again, in the absence of associativity, ...


1

The Jacobson radical. Take a noncommutative ring $R$ with 1. Any left ideal is either contained in another or is maximal. The elements common to all maximal left ideals, i.e. $$ J = \bigcap_i M_i, $$ is a group in two ways: It is an abelian group because it is an ideal (inherits group additivity from $R$, pretty obvious). It is group under circle ...


1

Joelafrite made a good suggestion: consider the first distributional derivative of a function $f$ that is not absolutely continuous. By definition, this distribution acts as $$\phi\to -\int f\phi'$$ If $f$ is increasing (like Cantor staircase and Minkowski's ?-function), then the distribution $f'$ is a measure. If $f$ has bounded variation, then $f'$ is ...


1

$\mathbb Q[\sqrt{2}]/\mathbb Q$ and $\mathbb Q[\sqrt[4]{2}]/\mathbb Q[\sqrt{2}]$ are normal, but $\mathbb Q[\sqrt[4]{2}]/\mathbb Q$ is not.


1

Take $\sigma=id: F=\mathbb C\to \bar F=\mathbb C$ and choose $E$ and $\alpha$ any way you like: $\sigma$ won't extend to $\mathbb C(\alpha)$.


3

Any example will do, such as $F=\mathbb Q$, $\alpha=\pi$. Since $\overline F$ is algebraic over $F$, $\tau(\alpha)$ would have to be algebraic, say $f(\tau(\alpha))=0$ with $f\in F[X]$. Since $f(\alpha)\ne 0$, $\tau$ fails to be monomorphic.


0

There are good reasons that $2^{\aleph_0}=\aleph_1$ (the continuum hypothesis) is probably true, even if ZFC (+ large cardinals) alone cannot prove it. So people may not be too wrong when naively believing in this proposition. However, we cannot yet be sure, this is still ongoing research. You may want to look at Hugh Woodin's work on "Ultimate-L".


0

Consider the non-linear moving average process $$ y_t = \epsilon_t + \epsilon_{t-1} \epsilon_{t-2},$$ where we take $\epsilon_t \sim N(0,\sigma^2)$ to be i.i.d. random variables. We can easily show that this is a white noise process but $$\mathbb{E} [y_t | y_{t-1},y_{t-2},\ldots] = \mathbb{E} [y_t | \epsilon_{t-1},\epsilon_{t-2},\ldots] = \epsilon_{t-1} ...


0

Essentially, people are confusing $\aleph_1$ with $\beth_1$ (pronounced "beth", see beth numbers). More precisely: People have heard of the $\aleph$ numbers and cardinal exponentiation and think $\aleph_1=2^{\aleph_0}$ which is not true (unless we assume an extra axiom which "by default" we do not). However, people have not heard of the $\beth$ numbers, and ...


8

The cardinality of real numbers is indeed $2^{\aleph_0}$. However $\aleph_1$ is not defined as $2^{\aleph_0}$ (the cardinality of the continuum, often denoted $\mathfrak c$), but it is defined as the smallest cardinality strictly greater than $\aleph_0$. Whether $2^{\aleph_0}=\aleph_1$ (termed the continuum hypothesis) can neither been proven nor disproven ...


-2

$2^{\aleph_0}=\mathcal c$, and it can be $\aleph_1$,$\aleph_2$...,if $\mathcal c=\aleph_1$,this is the continuum hypothesis.


2

To get a broader context for the well above relation you can consult any introductory text on domain theory. However, the context of continuity spaces is a bit different, and I prefer to have the intuition for Flagg's value quantales come directly from their intended role. So, the way I think about the well above relation is that it solves some nasty ...


2

I’ve not dealt much with the concept, but an example that I find helpful is $\Bbb R^2$ with the product partial order, $\langle x_0,y_0\rangle\le\langle x_1,y_1\rangle$ iff $x_0\le x_1$ and $y_0\le y_1$. Then it’s not hard to check that $\langle x_0,y_0\rangle\prec\langle x_1,y_1\rangle$ iff $x_0<x_1$ and $y_0<y_1$. Consider, for instance, the set ...


1

I cannot in good faith answer question 1, because I didn't see enough examples of that relation to get an informal idea of it. As for 2, first note that $q \succ p$ implies $q \geq p$ (as stated in the linked article, too), and here's an example where the two relations differ: Consider the set $S = \{a,b,c,d\}$, so that its power set $V$ is a complete ...


1

You should have five subgroups of $D_4/N$ and five subgroups of $D_4$ which contain $N$. Of the subgroups of $D_4/N$, you're missing $\{N,rN\}$. Of the subgroups of $D_4$ which contain $N$, you're missing two: $\langle r^2,s\rangle$ and $\langle r^2,rs\rangle$. (As Nishant said.) Contrary to what you claim in the comments, $\langle r^2,rs\rangle$ is not ...


1

You're missing $\langle r^2, s\rangle$ and $\langle r^2, rs\rangle$.


2

Note that the right hand side is empty if $A'=\emptyset$. So it is tempting to try and find an $A$ without limit points such that $A+A$ does have limit points. The set $A$ must be closed and discrete. This could be achieved by picking one point from every interval $[n,n+1]$ for integer $n$. Can you see a way of doing this such that the sum $A+A$ has $0$ as a ...


0

There are a lot of counter examples. (a+b)/2=x a+b=2x As long a and b add up to a even number, its possible. Assume 0 < a < 1 (odd number+a)+(even number+[1-a])=odd number+1=even number You also can change this a little bit and you can get way more equations for counter examples.


1

Is there any better way? Since all you have to do is give a counterexample, there is not much need to try to improve your "disproof"; however, a simpler and more all-encompassing consideration would lend itself to multiple counterexamples. Let $\mathbb{I}$ denote the set of irrational numbers. Suppose that $a,b\in\mathbb{I}$, where $b=-a$. Then $$ ...


-1

every finite dimensional Banach space is a reflexive bot it need not be uniformly convex .$X=\mathbb{R}^n , n\geq2 $with the norm $\Vat x\Vat_{1}=\sigma {i=1}^\n\vert x{i}\vert


2

Define $\mu\{1\} =\mu\{4\}=1$, $\mu\{2\}=\mu\{3\}=0$ and $\mu\{1\} =\mu\{4\}=0$, $\mu\{2\}=\mu\{3\}=1$. Then $\mu\{1,2\}=1=\nu\{1,2\}$; $\mu\{1,3\}=1=\nu\{1,3\}$; $\mu\{2,4\}=1=\nu\{2,4\}$ and $\mu\{3,4\}=1=\nu\{3,4\}$, but $\mu$ and $\nu$ are far from agreeing on the $\sigma$-algebra $\mathcal A$.


1

One counter example to you question is : $$1^2\equiv 2^2\mod 3 ,\text{ and } 5\equiv 2\mod 3 \text{ and } 2^5 \equiv 2 \not \equiv 1^5\mod 3$$ Note that if we modify a little your statement, if$a,b$ are coprime then we have : $$\left(a^k\equiv b^k\mod n\ \text{ and }\ \ k\equiv j\mod \varphi(n)\right)\Rightarrow a^j\equiv b^j\mod n $$ with $\varphi(n)$ is ...


1

This is NOT transitive because even though (1,2) and (2,3) are in the relation but (1,3) is not in the relation. Hence this is NOT a partial order.


2

Couldn't you take a set of the form $\{0\} \times A$, where $A$ is non-measurable? This is measurable of measure zero, being a subset of the the measure zero set $\{0\} \times \mathbb{R}$.


3

consider $X = \mathbb{R}$ with discrete metric space...then $X$ cannot be embedded in $\mathbb{R^n}$ for all $n$...since $f(X)$ would be an discrete set of $\mathbb{R^n}$..but any discrete set can atmost be countable in $\mathbb{R^n}$.


2

To add to the answers above, to get a Hausdorff space of continuum cardinality you can take a disjoint union of countably many simplexes of unbound degree (i.e. a $1$-simplex, a $2$ simplex etc.). Since an $n$ simplex can only be embedded in $\mathbb{R}^m$ for $n\le m$ you get that their union fits the goal.


6

Since each $\mathbb{R}^n$ is Hausdorff, and as all subspaces of Hausdorff spaces are Hausdorff, any non-Hausdorff space would suffice. A simple example would be $X = \{ 0 , 1 \}$ with the trivial topology.


5

Note that for any $n \in \mathbf N$, we have $|\mathbf R^n| = |\mathbf R|$. If we equip $X:= \mathcal P(\mathbf R)$ with any, to be concrete say the discrete, topology, there isn't even any one-to-one map $X \to \mathbf R^n$, hence no embedding.


2

Here is a useful result in Folland's Modern Real Analysis. Let $G$ be a locally compact group that is homeomorphic to an open subset $U$ of $\Bbb R^n$ in such a way that, if we identify $G$ with $U$, left translation is an affine map, i.e. $xy = A_xy+b_x$, where $A_x$ is a linear map on $\Bbb R^n$ and $b_x\in\Bbb R^n$. Then $|\det A_x|^{-1}\,dx$ is a ...


1

Note that $$ [1+\gamma e_{xy}, 1+\delta e_{yz}] = (1-\gamma e_{xy})(1-\delta e_{yz})(1+\gamma e_{xy})(1+\delta e_{yz}) = 1 + \gamma\delta e_{xz}. $$ It is stated in the question that if $L$ is a normal subgroup of $G$ then $L$ contains some $1+\alpha e_{xy}$. By taking commutators with $1+\beta e_{wx}$ and $1+\beta e_{yz}$ for $w < x$ and $z > y$, it ...


0

When I was in kinder-garden I was sick one day so I missed class and the next day the teacher gave us a quiz on basic multiplication. But I didn't know what multiplication was so when I got the test paper I thought that she wrote the + signs really badly. I didn't want to make her feel bad by pointing it out (not that my writing was any good either). So I ...


1

If you set $X = Y = \mathbb{R}$, you can construct examples fairly quickly. One example would be: $f(x) = \left\{\begin{array}{ll} 1/x & x \ne 0\\ 0 & x = 0\end{array}\right.$


0

Let us consider this finite group of order $8$, call it $G$. Then $H=\{0,1\}$ is a subgroup of $G$, but $H$ is not normal. And, there, we have: $$3H=\{3,6\}=6H,$$ but $$H3=\{2,3\}\ne\{4,6\}=H6.$$ Similarly, in the same environment, we have: $$H2=\{2,3\}=H3,$$ but $$2H=\{2,4\}\ne\{3,6\}=3H.$$


0

Just take the union of $0$ and a sequence of disjoint closed annuli around $0$ of decreasing sizes. It's still closed and absorbing, but not convex, and it includes no ball around 0. Edit: the above is wrong. The definition of absorbing set is equivalent to be a neighbor of 0 in every intersection with a one-dimensional subspace. An example of a radial set ...


0

The only quasigroups of orders $1$ and $2$ are the (unique) groups of those orders. So, a minimal example must have order $\geq 3$; indeed, (up to isomorphism) there are five quasigroups of order $3$ but but only one of these (the unique group with three elements) has an identity. Probably the most familiar of the four quasigroups of order $3$ without ...


1

There are examples that are simple to state (in $n$ dimensions): $$f(x) = |x|^{\alpha-n}\chi_{|x|\le 1},\qquad \frac{n}{2}<\alpha\le \frac{n+1}{2}$$ But proving that $f\notin H^{1/2}$ is a bit awkward; it's best done by using the divided-difference-integral characterization of $H^{1/2}$ or a related, simplified formulation in Exercise 33 here. Then ...


4

If you don’t find a counterexample using the approaches already suggested, try to prove the theorem, and see where you get stuck: this may well give you a better handle on the properties that a counterexample will have to have, and that’s invaluable information when you’re searching for a counterexample or trying to construct one from scratch. This doesn’t ...



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