New answers tagged

0

Take an operator $T\colon \ell_\infty \to c_0$ given by $$T(\xi_n)_{n=1}^\infty = (\frac{\xi_n}{n})_{n=1}^\infty\quad (\xi_n)_{n=1}^\infty\in \ell_\infty).$$ Certainly $T$ is injective and has dense range. However, it cannot have closed range as it would be an isomorphism which is impossible as $\ell_\infty$ is non-separable. This operator is actually ...


0

This is an example from Booss Bleecker. Let $\mathcal{H}:=l^2(\mathbb{N}_{\geq1})$ and $\{\delta_x\}_{x\in\mathbb{N}_{\geq1}}$ be the orthonormal basis of $\mathcal{H}$. Define $A:\mathcal{H}\to\mathcal{H}$ by: $$ A:=\sum_{x\in\mathbb{N}_{\geq1}}\frac{1}{x}\delta_x<\delta_x,\cdot>$$ Verify that: $A\in\mathcal{B}(\mathcal{H})$ (with $||A||=1$) $v\in ...


0

The idea of maximal solution might come from three body problem. People have very strong interests to know if planets in solar system may eventually collapse onto each other? Without considering the effect of chemical reaction from the sun, the question is roughly equivallent to prove the maximal solution to three body problem exists for $ 0 < t < ...


2

Check the equation $\dot x = x^2$. All the nontrivial solutions have "blowup in finite time" (i.e., a vertical asymptote).


0

Our entire civilization is built on our understanding of classical physics. The fact that Newton's laws give rise to differential equations for which the solutions stretch indefinitely into the future allows us to simulate complex phenomenon and accurately predict what is going to happen. If you want a concrete example then the consider the differential ...


2

The candidate $g(x)=x$ is otherwise ok, but fails the conditions about $g'(\pm1/2)$. A natural idea is to add a term that vanishes at $x=0$ and $x=\pm1$, but fixes the values of those derivatives. The polynomial $r(x)=x^3-x$ is the simplest function vanishing at the critical points $x=0,x=\pm1$. There is no guarantee that using its multiple as a correction ...


0

While your first example is (sort of) fine I'd like you to note that $D_n = D_{n+1}$ for infinitely many $n$'s, namely those for which the $n+1$st decimal place of $\pi$ is $0$. (Actually, I'm not even sure if it has been proven that the decimal representation of $\pi$ contains infinitely many $0$'s, so let's say for many $n$'s and maybe even infinitely many ...


2

Another example (perhaps a bit more familiar with readers of Munkres's text) is $\mathbb R^\omega$ with the box topology.* It is not first countable because you can "diagonalise" through any countable collection of open neighbourhoods of a point. Given $\mathbf x = ( x_n )_n $ and a collection $\{ U_i : i \in \mathbb N \}$ is open neighborhoods of ...


1

Hint: Glue uncountably many copies of $\mathbb R$ together at the origin


1

Perhaps try thinking about matrices. You know from linear algebra that the set of invertible $n \times n$ matrices is a group with group operation matrix multiplication. If you write out the functions that express the entries of the product of two matrices in terms of the entries of the matrices you can see that they are continuous (since they involve just ...


1

For your second paragraph, it might be helpful to see an example of a topological space with a discontinuous group operation. The topological space is $X = \mathbb{R}$. I will pick a discontinuous bijection of $\mathbb{R}$: $f(x) = x$ if $x \ne 0,1$, $f(0)=1$, $f(1)=0$. Using this bijection, I will define a group operation $\oplus$ on $X$, having identity ...


1

It might be easier to think of a topological group as a topological space with a group structure rather than a group with a topology. Let $X$ be a topological space. This means that certain subsets of $X$ are in a family $\tau\subset\mathcal{P}(X)$ such that the usual axioms apply. Now assuming the axiom of choice, there is a group structure on $X$. That ...


0

I don't know if this can be realized as one of your examples, but we can make the polynomial ring $\mathbf{C}[x]$ into a commutative ring $A$ in a bit of a weird way by changing the multiplication: $P+_A Q = P+Q$ $0_A = 0_{\mathbf{C}[x]}$ $1_A = x$ For all $a,b \in \mathbb{N}$, define $x^a \times_A x^b = x^{ab},$ then extend $\times_A$ by ...


0

Instead of "$a_\lambda^j\cdot a_{\lambda'}^{j'}=a_\lambda^{j+j'}$ if $|a_\lambda|$ is odd", it should just say "$a_\lambda^j\cdot a_{\lambda}^{j'}=a_\lambda^{j+j'}$ (hopefully this makes more sense). Strictly speaking, you should also clarify that "$a_\lambda^j$" is a shorthand for the formal expression $a_1^{i_1}\dots a_n^{i_n}$ where $i_\lambda=j$ and ...


10

Posets (viewed as categories) provide many examples of categories without certain limits. A product of two elements in a poset is simply a greatest lower bound, so you just need to make a poset where some pair of elements has no greatest lower bound. For example, the poset $\{ a, b, c, d \}$ where $a \leq c, a \leq d, b \leq c, b \leq d$. Then both $a$ and ...


10

Let $K$ be a field. In the category of fields, let's prove $P = K \times K$ (categorical product) doesn't exist in general. If it did, it would come equipped with two projection morphisms $\pi_1, \pi_2 \colon P \to K$. If $\operatorname{id} \colon K \to K$ is the identity morphism, the pair $(\operatorname{id},\operatorname{id})$ must factor through $P$, ...


11

For fields the most basic problem is fields of different characteristic. If $K$ and $L$ are fields, then the product $K\times L$ (if it existed) would have to be a field that can map into both $K$ and $L$, which means in particular it has the same characteristic as both $K$ and $L$, which is impossible if $K$ and $L$ have different characteristic. Even if ...


0

The Wikipedia page on semi-locally simply connected has a good example of a space that is semi-locally simply connected but not locally simply connected: the cone over the Hawaiian earring. The thrust of this example is that small open neighborhoods will generally contain non-contractible circles—but these circles can be contracted in $X$ by collapsing them ...


1

A nice example is Helly space $H$. This space is compact Hausdorff (so $T_4$, countably compact/pseudocompact), and not metrisable (e.g. as it has a discrete subspace of size $\mathfrak{c}$, see the linked answer). So it is also not submetrisable (a strictly coarser topology cannot be Hausdorff, by standard arguments). It is also a convex subset of a ...


3

$x_n=\sin \frac{n\pi}2$ has limit points $0,1,-1$. For any $m\in\Bbb N$, $x_n = n\bmod m$ has $m$ limit points $0,1,\ldots,m-1$. $x_n=n \bmod {\lfloor \sqrt n\rfloor^2}$ has countably many limit points (namely $\Bbb N_0$). $x_n=\sin n$ has uncountably many limit points, namely $[-1,1]$ We know that $\Bbb Q$ is countable. If $x_n$ is an enumaration of ...


1

A very simple example: $$a_{m,n}=\begin{cases} 1,&\text{if }m\le n\\ 0,&\text{if }m>n\;. \end{cases}$$ If you write out the double sequence as an infinite array, it’s very easy to see what happens: $$\begin{array}{ccc} 1&1&1&1&1&\ldots&\to&1\\ 0&1&1&1&1&\ldots&\to&1\\ ...


0

Here are some general remarks. First, some model theory. The claim that $(R, r, s)$ is a tuple consisting of a ring $R$ and two elements $r, s \in R$ such that $rs \neq 0$ but $sr = 0$ can be formalized in first-order logic. The completeness theorem then guarantees that such a statement has a model (namely, a ring exists with these properties) if and only ...


7

In the ring of $2\times 2$ matrices, say with real entries, look at $$\begin{pmatrix}1&0\\0&0\end{pmatrix}$$ and $$\begin{pmatrix}0&0\\1&0\end{pmatrix}$$


1

Let $C\subset [0,1]$ be the standard Cantor set. Define $$f(x)=(\operatorname{dist}(x,C))^{\alpha}$$ with $\alpha\in (0,1)$ to be chosen later. This is an $\alpha$-Hölder continuous function since it's a composition of the Lipschitz function $x\mapsto \operatorname{dist}(x,C)$ with the $\alpha$-Hölder function $t\mapsto t^\alpha$. The complement of Cantor ...


0

No such graph exists. Every edge can be part of a Hamiltonian cycle, and the graph isn't a cycle. From these, we can deduce that: At least two Hamiltonian cycles exist. All vertices have valence > 2. All vertices have valence > 3. All vertices have even valence. The connectivity > 3. For 2, we merely need to use the two Hamiltonian cycles. They ...


4

Symmetric difference is associative. Therefore, $(A\mathop{\triangle}B)\mathop{\triangle} C=A\mathop{\triangle}(B\mathop{\triangle}C)$. Moreover, $B\mathop{\triangle}B=\varnothing$, and $B\mathop{\triangle}\varnothing=B$, for any set $B$. Combine these and we have that: ...


4

It is countable. $A \bigtriangleup C = ( A \setminus C ) \cup ( C \setminus A )$. Then $A \setminus C = (A \setminus C \cap B) \cup ( A \setminus C \cap B^c)$. We see that $A \setminus C \cap B \subset B \setminus C \subset B \bigtriangleup C $ and $A \setminus C \cap B^c \subset A \cap B^c \subset A \bigtriangleup B $. Hence $A \setminus C \cap B$ is ...


1

The matrix $$\pmatrix{1&0&1&0\cr0&1&1&0\cr0&1&0&1\cr}$$ has nullspace generated by $(1,1,-1,-1)$, so every nonzero vector in the nullspace has at least 2 negative entries.


1

In $\mathbb{R}^{2}$ take two disjoint infinite lines $L_{1},L_{2}$ connected by vertical lines $B_{1},...$ (like a ladder). Then define $$A_{n}=L_{1}\cup L_{2}\cup \bigcup_{k\geq n}B_{k}.$$ These form a descending chain and their intersection will be $L_{1}\cup L_{2}$.


2

Define $T:L^1(\Bbb R)\to L^1(\Bbb R)$ by $Tf = gf$, where $$g(t)=\frac{1}{1+t^2}.$$Functions with compact support are dense in $L^1$, hence $T(L^1)$ is dense in $L^1$. But you can easily find an example showing that $T(L^1)\ne L^1$; hence $T(L^1)$ is not closed in $L^1$.


3

Your proof is not correct as it stands, as $U_a$ is not equal to $\{a\}$ but we know that $U_a \cap A \subseteq \{a\}$, in general. We also don't really need the fact that $A$ is closed (though this is true). Pick for every $x \in X$ some open $U_x$ that contains $x$ and such that $U_x \cap A \subseteq \{x\}$. This is, by definition almost, an open cover of ...


1

John Ma's answer, together with my calculation for $\sqrt{x}$, settles the issue: the notes are wrong (on any reasonable interpretation of the supremum that makes the function $\Delta(x)$ well-defined), and lower semicontinuity is the best that can be concluded in general. But even this weaker result can be used to prove the fact—call it Heine's theorem—that ...


1

You are applying the root-finding Newton's method $x_{n+1}=x_n-f(x_n)/f'(x_n)$, which in fact converges to $0$ in this example (albeit slowly). Generally, this method works okay for $C^1$-smooth convex functions, such as yours. But the problem was to minimize $f$. The minimizing Newton's method is $x_{n+1}=x_n-f'(x_n)/f''(x_n)$ (i.e., the root-finding ...


5

I'm not sure why you say the definitions seem to agree "in most cases"; exceptions seem far more common than examples to me. For a simple counterexample, in the category of groups, no object is discrete in the first sense, and the trivial group is discrete in the second sense. More generally, any category of algebraic structures which include distinguished ...


3

Let $$f(x,y) = x_1y_2\qquad \text{and}\qquad g(x,y)=y_1x_2\qquad \forall x=(x_1,x_2),y=(y_1,y_2)\in\Bbb R^2$$ Then $f(x,x)=g(x,x)$ for all $x\in\Bbb R^2$ but $f\neq g$.


0

Provided $\varepsilon < \frac{\max{f}-\min{f}}{2}$, the function is continuous: let $x\in[a,b]$. Assume, for a contradiction, that there is some sequence $x_n \rightarrow x$ s.t. $\Delta(x_n)\geq \Delta(x)+ \alpha$ for some $\alpha>0$. But then, for large enough $n$, we have for all $y\in (x-\Delta(x)-\alpha, x + \Delta(x) + \alpha)$, for large enough ...


1

Avoiding the trivial case, assume that $f$ is nonconstant and $\epsilon <\sup f - \inf f$. So $\Delta$ is finite valued. One can show that Theorem !@: The function $\Delta$ is lower-semicontinuous. That is, $$\liminf_{z\to x} \Delta (z) \ge \Delta (x).$$ To see this, let $x\in [a, b]$ and $\delta <\Delta(x)$ be arbitrary. By definition, ...


1

For (1), take $A,B$ disjoints with $A\cup B=\Omega$. Then $P(A)=P(B)=1/2$ and $P(A\Delta B)=P(\Omega)=1$. For (2), $P(A\Delta B)=P(A\cup B)-P(A\cap B)=P(A)+P(B)-P(A\cap B)-P(A\cap B)=2p-2p=0$ if $P(A)=P(B)=P(A\cap B)$ (please note that we don't use $0<p<1$). For (3), we only have $P(A\Delta B)=P(A)+P(B)-2P(A\cap B)$. Thus, we need ...


1

If $A$ is an abelian group, then the new operation $a * b = - (a+b)$ is commutative but is typically not associative. (This comes up naturally in the classical chord-tangent law on a cubic curve $C$ in the projective plane: if $a$ and $b$ are two points on the curve, then $a*b$ is the third point of intersection of the line through $a$ and $b$ with $c$.)


0

Adding a little to @DietrichBurde's answer, Jordan algebras (characterized by $xy=yx$ and a limited associativity $(xy)(xx)=x(y(xx))$ are a reasonably natural not-fully-associative class of algebras. The standard Jordan algebras are made from associative algebras with multiplication $\cdot$ by defining $xy=x\cdot y+y\cdot x$. For non-commutative algebras, ...


5

Write down a blank $n \times n$ matrix, and randomly fill in the entries on and above the diagonal with values from $\{1,\ldots,n\}$. Then mirror these into the below-diagonal half to make it a symmetric matrix. Reading this as a multiplication table, it determines a commutative operation on $\{1,\ldots,n\}$. The probability of this being associative gets ...


2

A smaller example is $\mathbb{CP}^2$. It's homogeneous since $GL_3(\mathbb{C})$ acts transitively on it and has the fixed point property by the Lefschetz fixed point theorem. More generally, if $G$ is a Lie group and $H$ a closed subgroup of it, then $G/H$ is a homogeneous smooth manifold. This is a rich class of spaces which includes, for example, the ...


11

The simplest example of a nonassociative commutative binary operation (but lacking an identity element) is the two-element structure $\{a,b\}$ with $aa=b$ and $ab=ba=bb=a;$ note that $a=bb=(aa)b\ne a(ab)=aa=b.$ This is the NAND (or NOR) operation of propositional logic, where $a=\text{true}$ and $b=\text{false}$ (or vice versa). See this answer or this one.


6

I asked the SMT solver Z3 to provide a function that is not associative but commutative. The way to do this is to assert commutativity for all possible inputs and ask for a counter-example for the associativity property. The Z3 script in SMTlib format is: (declare-fun f (Int Int) Int) ;We are looking for a function f (assert (forall ((a Int) (b Int)) (= ...


50

Here is an example with identity element and inverses. On $\mathbb{R}_{\geq 0}$, define $a*b = \left| a-b \right|$. Then $*$ is clearly commutative, $0$ is its identity and the inverse of any $a$ is itself. Yet, it is not associative, since for instance $2*(1*1) = 2$ while $(2*1)*1 = 0$.


38

Consider the "rock-paper-scissors" operation $\ast$ on the set $\{R, P, S\}$ where we declare each element to be idempotent and declare the product of two distinct elements to be the winner according to the usual rules of Rock, Paper, Scissors: \begin{array}{r|ccc} \ast & R & P & S\\ \hline R & R & P & R\\ P & P & P ...


21

Consider $\Bbb R$ endowed with the arithmetic mean operation $$a \oplus b := \frac{a + b}{2}.$$ It is clearly commutative but not associative, as $$(a \oplus b) \oplus c = \frac{a + b + 2c}{4}$$ but $$a \oplus (b \oplus c) = \frac{2a + b + c}{4} .$$ (It's easy to see that $\oplus$ has no identity, and hence no inverse, though it is a quasigroup, which means ...


7

Convolution of distributions is commutative, but not necessarily associative. You have the identity $R*(S*T)=(R*S)*T$ in the case where at least two of these distributions are of compact support. If this condition is violated, there is a standard counterexample $$(1*\delta_0')*H = 1'*H=0*H=0,\quad 1*(\delta'_0*H)=1*H'=1*\delta_0=1.$$


6

Consider the class of non-associative Jacobi-Jordan algebras. They are in general not associative, for dimension $n\ge 5$, but they are always commutative Jordan algebras, hence commutative. Proposition $4.1$ gives examples of such algebras: Proposition: Any Jacobi-Jordan algebra of dimension $5$ over an algebraically closed field of characteristic ...



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