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1

Differential Forms in Algebraic Topology by Bott and Tu has many examples that you could present if you want. A User's Guide to Spectral Sequences by McCleary has also tons and tons of examples. Here are some examples off the top of my head: The Serre SS for computing the homology of $\Omega S^n$ ("easy"), of $BU_n$ and $U_n$ ("hard"). In homological ...


0

Regarding 2) and 3): The multiplicativity formula for field extension degrees gives $$[F(r_1,r_2):F] = [F(r_1,r_2):F(r_1)][F(r_1):F].$$ The minimal polynomial for $r_2$ over $F(r_1)$ divides the minimal polynomial for $r_2$ over $F$ so $$[F(r_1,r_2):F(r_1)] \leq [F(r_2):F].$$ 2) We know that $[F(r_1,r_2):F(r_1)] \leq [F(r_2):F] =2$. We also know that $r_2 ...


1

So, if we allow the non-measurable set to have infinite measure, this is easy: Take the standard construction of Lebesgue measure (i.e. $R$ is a suitable semiring of intervals) and let $V = [10,\infty) \cup \Bbb{V}$, where $\Bbb{V} \subset (0,1)$ is nonmeasurable. Then $V$ is not measurable, but the inner and outer measure both equal $\infty$. If we ...


4

The matrix $$ T=\pmatrix{0&1\\0&0} $$ has a one-dimensional eigenvector space spanned by $[1,0]^T$, but any 2-vector is an eigenvector of $T^2=0$.


2

This is almost never satisfied since the endomorphisms (automorphisms) of $X$ resp. $Y$ commute when extended to $X \otimes Y$. It is more reasonable to ask if $\mathrm{End}(X \otimes Y) = \mathrm{End}(X) \times \mathrm{End}(Y)$ holds. For example, this holds in the free monoidal category on a category. It fails in most symmetric resp. braided monoidal ...


0

It is still quite trivial but at least... Given the Lebesgue measure $\lambda:[0,1]\to(0,\infty)$. Consider a Vitali set $\mu_*(V)<\mu^*(V)$. Construct the function: $$f:[0,1]\to\mathbb{R}:f(x):=x^2\chi_V(x)$$ This one is not sum of measurable plus step but product.


5

There are many such spaces. One simple Hausdorff example is the Arens-Fort space, which is fairly clearly described in the linked article. In that space let $A$ be everything except the point $\langle 0,0\rangle$; then $\langle 0,0\rangle\in\operatorname{cl}A$, but no sequence in $A$ converges to $X$. An even simpler example, but one that isn’t Hausdorff, ...


0

Ok, I think I got it now... Take a slight variation of the famous example: $$F:(0,1]\to\ell(0,1]:t\mapsto\chi_t$$ This is measurable as: $$\forall\varphi\in\ell(0,1]:\quad\|\varphi-\chi_s\|<\varepsilon<1\implies\varphi(s)>1-\varepsilon>0$$ so that small balls are measurable: ...


1

1) You must see your previus post, I fixed it, and the question turned out to be false, sorry for the mistake. In this case, the question is again false(F(r)=F(r^2)), and the same example works again, the same $\omega$, but instead to take $w^3$ you must take $w^2=e^{\frac{2\pi i}{6}}$ and then $[Q(\omega^2):Q]=\phi(6)=2$. Hint: 2,3,4,5 take $F=Q$. 4 must ...


3

It is false. Take $F=\mathbb Q$ the rationals, and $\omega=e^{\frac{2\pi i}{12}}$. $\omega$ is a primitive $12$-root of unity, and we know that in this case $[\mathbb Q(\omega):\mathbb Q]=\phi(12)=4$. (If you do not know what this means, search for Euler totient.). Now see $\omega^3=e^{\frac{3(2\pi i)}{12}}=e^{\frac{2\pi i}{4}}$, so $[\mathbb ...


2

The function $f(x,y)=y\sqrt{x}$, originally defined in the right halfplane, can be extended as a function differentiable at $(0,0)$, but no extension is differentiable at other points of the $y$-axis.


0

Under the $L^p$ metric, the completion of the space of continuous functions from $\mathbb{R}^n$ to $\mathbb{C}$ is $L^p(\mathbb{R}^n)$ for any $p$ such that $1 \le p < \infty$.


0

Claim: $E$ is closed i.e $E=\bar{E}$ suppose not, then $\exists c\in \bar{E}\setminus E$ then the function $f(x)={1\over x-c}$ is unbounded so $E$ must be closed. Claim: E must be bounded. If not then take $f(x)=x$ continous on $E$ but unbounded. So $E$ must be closed bounded subset of reals must be compacto!


1

A space $X$ in which every continuous function from X to the reals is bounded is called pseudocompact. An example of a space that's pseudocompact but not compact is any infinite space with the particular point topology. That's the topology in which you pick some arbitrary point, call it $x$, and then say that the open sets are any set containing $x$, along ...


1

Consider an example similar to one OP used in a now deleted answer, where $M$ is the commutative monoid generated by $x,y$ with the single relation $x^2=y^2.$ Now let $a=x,\ b=y$ so that $\gcd(a,b)=1.$ Let also $t=x.$ Now here's where I have to rely on the OP's statement that, if $\gcd(at,bt)$ exists, then it is necessarily $\gcd(a,b)t$. [That seem right to ...


2

The notation is somewhat strange, but I suppose you define an equivalence relation on $X$ defined by $a_i R a_j$ for all $i,j$ and these are the only non-trivial relations, and set $X / R$ as the quotient. So if $x \notin A:= \{a_i : i \in I\}$, then $\pi(x) = \{x\}$, and otherwise $\pi(x) = A$. $\pi_i := \pi|_{X_i}$ is onto between $X_i$ and $\pi[X_i]$, ...


0

$\pi$-Base is an online encyclopedia of topological spaces inspired by Steen and Seebach's Counterexamples in Topology. It lists the following countable, connected, Hausdorff spaces. You can learn more about any of them by visiting the search result. Gustin's Sequence Space Irrational Slope Topology Prime Integer Topology Relatively Prime Integer ...


2

Actually, $\{ x_A : A \in \mathcal{A} \}$ is always a zero-set in $\Psi(\mathcal{A})$: define the function $f : \Psi(\mathcal{A}) \to [0,1]$ by $$f(n) = \tfrac{1}{n+1} \\ f(x_A) = 0.$$ This is clearly a continuous function (any neighbourhood of $0$ contains all but finitely many of the $\frac{1}{n+1}$, and so it's inverse image under $f$ includes a cofinite ...


2

$\pi$-Base is a database of topological spaces inspired by Steen and Seebach's Counterexamples in Topology. It lists the following fourteen second countable, Hausdorff spaces that are not metrizable. You can learn more about an of them from the search result. Arens Square Double Origin Topology Indiscrete Irrational Extension of $\mathbb{R}$ Indiscrete ...


1

$\pi$-Base is a database of topological spaces inspired by Steen and Seebach's Counterexamples in Topology. It lists the following twenty-two Hausdorff spaces that are not first countable. You can learn more about an of them from the search result. Compact Complement Topology Countable Excluded Point Topology Countable Particular Point Topology Deleted ...


1

The initial semiring is $\mathbb{N}$ (with the usual operations). Hence, the initial boolean semiring is $R = \mathbb{N} / \sim$, where $\sim$ is the smallest congruence relation such that $n \sim n^2$ for all $n \in \mathbb{N}$. Hence, we also have $n_1^2 + \dotsc + n_s^2 \sim n_1 + \dotsc + n_s$. The elements of $R$ are therefore ...


1

Let $V$ be a non-measurable set, it must be uncountable, select a countable points from it $\{x_i\}$, define $f=\chi_V$ except those points, and let $f(x_i)=\frac{1}{n}$. So $f$ is bounded by $1$, unmeasurable, has countable many values.


0

Let $X$ be the real line with the topology in which the usual open sets are open, and in addition $U\setminus A$ is open for any $U$ that is open in the usual topology, where $A=\{\frac1n : n=1,2,...\}$. In other words, every point except the origin has its usual neighborhoods, and the basic neighborhoods of the origin are of the form ...


2

Tomek's answer is very comprehensive. However, I thought I would add a "lower level" explanation. Proposition: $\{0\} \oplus c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty \oplus \ldots$ is not complemented in $c_0 \oplus \ell_\infty \oplus \ell_\infty \oplus \ell_\infty\oplus \ldots$. The idea is to use the following simple lemma: ...


2

Here is one of the simpler examples. Let $$A=\left\{\left\langle\frac1m,\frac1n\right\rangle:m,n\in\Bbb Z^+\right\}$$ and $$L=\left\{\left\langle\frac1m,0\right\rangle:m\in\Bbb Z^+\right\}\;.$$ Let $X=\{\langle 0,0\rangle\}\cup L\cup A$. These ordered pairs are a bit clumsy, so let me introduce some abbreviations: for $m,n\in\Bbb Z^+$ let ...


5

Each separable Banach space embeds into $C[0,1]$–in particular $C[0,1]\oplus \ell_2$. This space however is not complemented in $C[0,1]$ because $\ell_2$ is reflexive and $C[0,1]$ has the Dunford-Pettis property. Okay, embed now $C[0,1]\oplus \ell_2$ into $C[0,1]$ and add $\ell_2$ to the latter space. Regarding your second question, these two spaces are ...


4

As mathmax points out, first countability doesn’t imply even the weakest separation axiom, $T_0$. Moreover, adding some separation doesn’t help: first countability doesn’t imply Hausdorffness even for $T_1$ spaces, since the cofinite topology on $\Bbb N$ is first countable and $T_1$ but does not have any disjoint non-empty open sets.


6

Consider any topological space with at least two points and the indiscrete topology: It is first countable but not Hausdorff.


1

How about the following: Consider a torus $$\bigl(\sqrt{\xi^2+\eta^2}-a\bigr)^2 +\zeta^2=b^2,\qquad 0<b<a,$$ in $(\xi,\eta,\zeta)$-coordinates. After a suitable squaring this can be written in the form $$\Phi(\xi,\eta,\zeta)=0\ ,\tag{1}$$ where $\Phi$ is some polynomial of degree $4$ in the variables $\xi$, $\eta$, $\zeta$. Then let the $\zeta$-axis ...


2

A $C^{0}$ semigroup has to do with the parameter of time in a time-dependent equation such as the heat equation $$ \frac{\partial}{\partial t}h(x,t)=\Delta h(x,t),\\ h(x,0) = h_{0}(x) $$ The solution operator $S(t)$ evolves the heat distribution $h_{0}$ at time $0$ to that at time $t$. That is ...


3

The Taylor series of a bounded function that converges on all of $\mathbb{R}$ never converges uniformly to that function because the difference between every partial sum of the series and some point in its limit is as large as you want to make it. Polynomials increase in absolute value without bound as we stray far enough from 0.


2

One of the most important places where non-associativity plays an important role is in computations. A lot of useful algorithms rely on the fact the computational complexity of evaluations is non-associative. For instance, consider a rank $1$ matrix-vector product. Let the matrix $A = uv^T$ and the vector be $x$. Let $u,v,x \in \mathbb{R}^{n\times 1}$. To ...


5

This is taken from Inside Interesting Integrals by Paul J. Nahin. ...


1

What about $$ S(x,y,z)=4-x^2(y^2+z^2)+yz^2(7x-y) $$ We have $S(1,2,0)=S(2,0,1)=S(0,1,2)=0$ but $S(t,t,t)=4(t^4+1)\neq 0$


1

If $R[X]$ is integral, the $R$ is integral. The units in $R[X]$ are the units in $R$ (by lack of zero divisors). The irreducible elements of $R$ are the degree zero irreducible elements of $R[X]$. Consequently, for $0\ne r\in R$, the unique factorization in $R[X]$ consists of degree zero factors and is a factorization in $R$ iand is still unique.


4

It is even possible that $(a)(a) \neq (a^2)$, even in unital rings. Take the ring of $2 \times 2$-matrices over any field. Consider the matrix $a = \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}$ and define $b = \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}$. Then one computes $a=a \cdot ba$, and this lies in $(a)(a)$. But it doesn't lie in ...


1

Verify directly that $A$ is closed. Assume $\{ x_n \}$ and $\{ Ax_{n}\}$ converge to $x$, $y$ respectively. Then, for all $z \in \mathcal{H}$, $$ (Ax_n, z) = (x_n, A^{\star}z) \\ \implies (y,z)=(x,A^{\star}z)=(Ax,z) \\ \implies y = Ax $$


1

You're saying that for every $\psi \in \mathcal H$, the linear functionals $\phi \to \langle \psi, A \phi \rangle$ are bounded. Use the uniform boundedness principle...


1

So if, say, $X=\{1,2\}$, then the free Lie algebra $L(X)$, with $x\mapsto e_x$, consists of all linear combinations of the elements $e_1,e_2$, $[e_1,e_2]$, $[e_1,[e_1,e_2]]$, $[e_2,[e_1,e_2]]$, $[e_1,[e_1,[e_1,e_2]]$, $[e_1,[e_2,[e_1,e_2]]$, $[e_2,[e_1,[e_1,e_2]]$, $[e_2,[e_2,[e_1,e_2]]$, and so forth, such that skew-symmetry and the Jacobi identity holds. ...


1

You can modify your example a little bit to get the general one (I like to use $\chi_D$ to stand for characteristic function of a set $D$): for any $d\in [0, c]$, let $$f_n = (c-d) \chi_{[n-1, n]}+ d\chi_{[-2,-1]}.$$


3

Take $$a_n=\frac{(-1)^n}{\sqrt{n}}.$$ Then $$ \lvert a_{n+1}-a_n\rvert=\left|\frac{(-1)^{n+1}}{\sqrt{n+1}}-\frac{(-1)^n}{\sqrt{n}}\right|= \left|\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n}}\right|\ge\frac{1}{\sqrt{n}}, $$ and hence $$ n\lvert a_{n+1}-a_n\rvert\ge\sqrt{n}\to\infty. $$


1

Define $$d:\Bbb R\times\Bbb R\to\Bbb R:\langle x,y\rangle\mapsto\begin{cases} x-y,&\text{if }x\ge y\\ 1,&\text{if }x<y\;. \end{cases}$$ For $\epsilon\le 1$ we have $$\{y\in\Bbb R:d(x,y)<\epsilon\}=[x,x+\epsilon)\;,$$ so this generates the topology of the Sorgenfrey line, also known as the lower-limit topology. This space is separable, ...


0

My main use for modular arithmetic at the age was for this: http://everything2.com/title/Cracking+a+Master+Lock It's interesting and practical (e.g. when you forget a combination for an old lock), though naturally judgment should be exercised before volunteering this kind of information to Jr. High students.


0

You probably know this, but the fact that the remainder of a number modulo $9$ is the same as that of the sum of its digits is a simple calculation involving powers of $10 = 1$. There's a similar formula modulo $11$, where $10 = -1$. Also, the fact that if $2^n - 1$ is prime (a Mersenne prime), then $n$ is prime. If $n = ab$, then reduce $2^n - 1$ modulo ...


1

Consider the free algebra $k\langle x,y\rangle $. The ideals $I=(x)$ and $J=(y)$ are such that $IJ\neq JI$.


4

The term you want is concretizable (meaning admits a faithful functor to $\text{Set}$). The category of representations of a quiver does not work: if $Q$ is such a quiver with vertex set $Q_0$ and $V$ is a representation, then $$V \mapsto \prod_{q \in Q_0} V(q)$$ is a concretization. More generally, a category $C$ is concretizable if it has a small ...


8

It is true, which can be readily shown in ${\mathbb{R}^2}$. Let $U = \left\{ {\left( {x,0} \right):x \in \mathbb{R}} \right\}$ and $V = \left\{ {\left( {x,y} \right):x - y = 0} \right\}$. Let ${P_U}\left( {x,y} \right) = \left( {x + y,0} \right)$ and ${P_V}\left( {x,y} \right) = \left( {x,x} \right)$. Note that both are projections, since $P_U^2 = {P_U}$ ...


3

It's true for any topology, and the proof depends on the form of your definition of a set's interior. For example, one definition of the interior of a set $A$ is as the union of all open sets $U$ such that $U\subseteq A.$ Another (equivalent) definition is that the interior of a set $A$ is the unique open set $U$ such that for every open set $V$ with ...



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