# Tag Info

1

This is a slight modification of an example due to K. Alster; it is Example $\mathbf{3.2}$ of J. Chaber, Remarks on open-closed mappings, Fundamenta Mathematicae ($1972$), Vol. $74$, Nr. $3$, $197$-$208$. Let $X=\Bbb R\times\{-1,0,1\}$, and for convenience set $X_i=\Bbb R\times\{i\}$ for $i\in\{-1,0,1\}$. Points of $X_0$ are isolated. For $p=\langle ... 1 @user87690, many thanks for the helpful hint) Maybe, one of the simplest examples is the following. On the set$\mathbb Z$of all integers with left order topology which is obviously$T_0$consider equivalence relation with partition consisting of the set of odd and the set of even integers. This relation is open and closed, since saturation of any nonempty ... 4 For$0 < p < 1$, let $$\ell^p(\mathbb{N}) = \Biggl\{ x \colon \mathbb{N}\to \mathbb{C} : \sum_{n = 0}^\infty \lvert x(n)\rvert^p < +\infty\Biggr\}.$$ We endow it with the$p$-seminorm $$s_p(x) = \sum_{n = 0}^\infty \lvert x(n)\rvert^p$$ and the metric$d_p(x,y) = s_p(x-y)$derived from it. Note the difference from the case$p \geqslant 1$. If ... 1 Given$m,n\ge 1$, the set of all the$m\times n$matrices with entries in the finite field$\Bbb F_2$is, under addition, an abelian group which is also unipotent ($x+x=0$for all$x$). The order of this group is$2^{m\cdot n}$. 4 Let$f:X \to Y$be the desired open and closed continuous surjection. Since$Y$is not$T_0$, it contains two topologically indistinguishable points. If we restrict$f$to such two-point subset of codomain, we still get a desired continuous surjection, so we may assume that$Y$is two-point indiscrete space. Hence, it is enough to find a$T_0$space and its ... 1 As an online resource, you could have a look at this website: http://www.uwyo.edu/moorhouse/pub/bol/. This is about the weaker (not necessarily associative) structure of some finite loops but, among these, you can search the isotopy classes of groups. For example http://www.uwyo.edu/moorhouse/pub/bol/htmlfiles8/8_5_2_0.html or ... 1 The question is about constructing countable union of disjoint sets so dont think about the uncountable interval because any interval is uncountable obviously.Think interval as a unit and then take union of finite units (intervals). UAi where i is from 1 to n and n belongs to set of natural numbers. 0 There is no uncountable collection of open sets in$\Bbb R$. Let$\{A\}$be a collection of disjoint open sets. For each$A$, let$x_A\in A$. Because each$A$is open, there is an$\epsilon_A>0$such that$(x_A-\epsilon_A,x_n+\epsilon_A)\subseteq A$. Because there is always a rational between any two real numbers, there is a rational ... 0 Every collection of disjoint open intervals in$R$is countable because you can choose a rational number (by density theorem) in each of them and rationals are countable. For first your$\bigcup^{\infty}_{n=0}A_n$is a countable collection. It is a union of$\aleph_0$open disjoint intervals. 1 There is no uncountable collection of disjoint open intervals. Suppose there is and call it$\{I_{\alpha} \ \mid \ \alpha \in J\}$. An easy argument is to see that each of these open intervals contains a rational number$r_\alpha$which would disburse an uncountable amount of rational numbers. This is a contradiction since$\mathbb Q$is countable. 1 You can't write$\bigcup \limits_{n = 0}^\infty A_n$, as this is the union of all intervals. You actually want the set that consists of all these intervals, i.e.$\{A_n \;|\; n \in \mathbb{N}\}$. Then your example works. Hint for the second question: Observe that there are always at most countably many disjoint open intervals with length > 1. There are also ... 2 Take odd$f_n$s with $$f_n(x) = \begin{cases} nx, & \text{if 0\le x < 1/n} \\ 2-nx, & \text{if 1/n\le x \le 2/n.} \\ 0, & \text{if x > 2/n} \end{cases}$$$f_n(x)\to 0$pointwisely, since for any$x$there is some$N_x$that$x\in[-2/n,2/n]^c$(namely$f_n(x)=0$) for any$n\ge N_x$. However the ... 3 Just consider e.g. $$(\Omega,\mathcal{F},\mu) := ((1,\infty),\mathcal{B}((1,\infty)),\lambda)$$ where$\lambda$denotes the Lebesgue measure (restricted to$(1,\infty)$) and $$\phi(x) := x^2 \qquad g(x) := \frac{1}{x}.$$ 3 Theorem: Let$X,Y$be real linear spaces and$f\colon X\times Y\to [-\infty,+\infty]$be convex. Then $$\phi(x)=\inf_{y\in Y}f(x,y)$$ is convex. Proof: Let$E$be the image of$\text{epi}(f)$under the projection$(x,y,\alpha)\to (x,\alpha). Then by definition of infimum $$\text{epi}(\phi)=\{(x,\alpha)\in X\times\mathbb{R}\colon \ (x,\beta)\in E,\ ... 1 It is convex! Your first statement that the minimum of convex functions is in general not convex is true, but here you have a lot more structure! In a sense you are projecting onto x. In fact, g is also called the inf-projection of f. Let \lambda \in (0,1) and y_1, y_2 \in J arbitrary:$$ \begin{aligned} g(\lambda x_1 + (1-\lambda) x_2) &= ... 1 First, note that some of\,L^2\,$functions are not differentiable everywhere in$\,\mathbb R$. Second, for the more general case of weak derivatives one can easily come up with such a function. Now, as an example let us take$\ f(x) = \sin\left(1/x^2\right) \in L^2\left(\mathbb R\right)\,$. The derivative$\,f'$looks like$\ f'(x) = ...

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A. Br¢ndsted and R. T. Rockafellar,"On the subdifferentiability of convex functions", Bull. Amer. Math. Soc. 16 (1965), 605-611. (see section 5)

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The biggest application that I know for this result is the immediate corollary that a countable metric space is totally disconnected.

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A mildly interesting remark: A comment suggests this result can be used to prove the uncountability of $R$. But the proofs here and elsewhere of this result typically use the existence of a surjective function to an interval $[a,b]$ and the uncountability of that interval. So the argument suggested in the comment would be circular. Here, however, is a ...

1

I think that the only effect that you haven't taken into account is that all trajectories near the equilibrium experience slowdown and as a result the set $Y \cdot \lbrack t, +\infty )$ (at least its left border, $Y \cdot \lbrace t \rbrace$) shaped a little differently than a union of a stripe with a point. I hope that this picture will help illustrate my ...

0

Define $$f(u,v)= \begin{cases} 1 &\hbox{if v=u^2 and (u,v) \neq (0,0)} \\ 0 &\hbox{otherwise}. \end{cases}$$ It is easy to check that the Gateaux derivative at $(0,0)$ exists, but $f$ is not even continuous.

2

This counterexample comes from the Wikipedia page for Frechet Derivative Consider the function $f$ that is $0$ at $(x, y) = (0, 0)$ and $$f(x, y) = \frac{x^3}{x^2+y^2}$$ otherwise. It's Gateaux derivative $g$ as a function of the "direction" $h \in \mathbb{R}^2$ at $(0, 0)$ is $0$ if $h = (0, 0)$ and $$g(h_1, h_2) = \frac{h_1^3}{h_1^2 + h_2^2}$$ otherwise. ...

2

Things are maybe less confusing if we use restricted quantifiers: All $A$ are $B$ - $\forall x\in A\colon x\in B$ Some $A$ are $B$ (there exists an $A$ thatis $B$) - $\exists x\in A\colon x\in B$ However, the correct translations to unrestricted quantifiers are $\forall x(x\in A\to x\in B)$ and $\exists x(x\in A\land x\in B)$. Note that the statement ...

2

$$x = -1 \implies x^4 = 1$$ $$x = -1 \implies x^3 = -1$$ But $$x^4 = 1\not\Rightarrow x^3 = -1$$ because $x^3$ could be equals to $1$.

2

\begin{align} \mathcal A & = \{5,6\} \\ \mathcal B & = \{1,2,3,4,5,6\} \\ \mathcal C & = \{3,4,5,6,7,8,9,10\} \\[12pt] A & = [x\in\mathcal A] \\ B & = [x\in\mathcal B] \\ C & = [x\in\mathcal C] \end{align} The $A\to B$ and $A\to C$ are true but $B\to C$ is false. Second example: \begin{align} A & = [x\in\{1,2,3,4,\ldots\}] \\ B ...

0

$A$: "I do not exist" $B$: "I work for a living" $C$: "I am a billionaire" In other words, make $A$ false, $B$ true, $C$ false.

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If $(X,d)$ is a metric space then $$d'(x,y)=\min\{d(x,y),1\}$$ is a metric on $d$, as well. Moreover, the metrics $d$ and $d'$ generate the same topology (hence the same subsets of $X$ are closed in $(X,d)$ and in $(X,d')$). Every subset is bounded in $(X,d')$. See also: Proof that every metric space is homeomorphic to a bounded metric space So it ...

2

The first thing I thought of was $\displaystyle \bigcap_{n=0}^\infty [n,+\infty)$. If the object called $+\infty$ were included in the space, with the appropriate topology so that $n\to\infty$, then these sets would not be closed, but would become closed if one added $+\infty$ to them as a new member, and then the intersection would not be empty because ...

4

Another simple example is to look at the "punctured line": $(-\infty, 0) \cup (0, \infty)$, which is just the real numbers with $0$ removed. The sets $A_n = \{ x \ \in \Bbb R \,|\, |x| \le 1/n \text{ and } x \ne 0 \}$ are closed and bounded in the punctured line, but their intersection is empty.

1

Do you know a theorem about nested bounded closed sets having non-empty intersection? If you do then you would need to find a metric space that does not satisfy the conditions of that theorem. $\mathbb{R}$ satisfies that theorem and hence you're not going to find a counter-example there. But there is a smaller metric space sitting inside it, namely ...

7

Let $\mathbb N$ be endowed with the discrete metric. In this metric space, every subset is bounded (although not necessarily totally bounded) and closed. Moreover, the subsets \begin{align*} A_1\equiv&\,\{1,2,3,4,\ldots\},\\ A_2\equiv&\,\{\phantom{1,\,}2,3,4,\ldots\},\\ A_3\equiv&\,\{\phantom{1,2,\,}3,4,\ldots\},\\ \vdots&\, \end{align*} are ...

1

Since $0$ is the additive identity, you have to have $0$. You can have groups without $1$. for example {..., -6, -4, -2, 0, 2, 4, 6, ...}

1

It seems to be implicit in the question that $\mathbb Z_n$ is an additive group of integers, i.e. a subgroup of $\mathbb Z$. This is not the case. $\mathbb Z_n=\mathbb Z/n\mathbb Z$ is the group of residue classes modulo $n$. Its elements are often represented by integers, but they are not integers (else $1$ would generate all of $\mathbb Z$), but classes of ...

3

Though the question has effectively been answered by the combined comments of Martin Brandenburg and Derek Holt, I thought I'd write up a complete answer for completeness' sake. Let $N:=\Bbb{F}_{11}$ the finite field of $11$ elements and $H:=\Bbb{F}_{11}^{\times}$ its unit group. Let $G:=N\rtimes H$ the semidirect product of $N$ and $H$ given by the ...

1

In a topological space $(X,\tau)$ a subset $A \subset X$ is dense, iff its closure is the whole space, i.e. $\overline{A} = X$, while a subset $B \subset X$ is closed iff it is its own closure, i.e. $\overline{B} = B$. To recite the example of $\mathbb{R}$ with euclidian topology: $\mathbb{Q} \subset \mathbb{R}$ is dense but not closed, since: ...

0

One explanation is in terms of continuous functions. A continuous function (on a topological space $X$) is determined by its values on a dense subset of $X$. This is a defining characteristic of dense sets; any non-dense subset does not have this property. A prototype of a closed subset of $X$ is the solution set of a collection of equations defined by ...

0

Every commutative (symmetric) quasigroup operation. In particular, for example, the one associated to a Steiner Triple System $S_{\nu}$: $$a\circ b:=\left\{ \begin{array}{ccc} a & , & b=a\\ c\colon\left\{a,b,c\right\}\in S_{\nu} & , & b\ne a \end{array} \right.$$ over a set $X$ with $\nu > 3$ elements.

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Take $\mathbb R$ with the usual topology. Then the set of integers is closed (any converging series of integers converges to an integer), but not dense (you get nowhere close to $1/2$). On the other hand, the set of rational numbers is dense, but not closed (the limit of a sequence of rational numbers can be irrational). IMHO a good intuition for a closed ...

16

I want to add one thing. The only closed, dense set in a topological space is the space itself! So these two concepts are pretty far apart. So far that in most situations, they are mutually exclusive!

3

I'd say the most glaring difference between the two definitions is that dense sets need not contain their limit points. Closed sets necessarily do. There are plenty of examples of "dense not closed" and "closed not dense" scattered throughout this question now. I'd also argue that your intuition for a dense set only makes sense in spaces where distance ...

3

Loosely speaking: Suppose $X$ is a topological space (with some topology, obviously). If $A \subseteq X$ and we say $A$ is dense in $X$, we mean that for each point $x \in X$, we can keep finding points from $A$ "closer" and "closer" to $x$. If $B \subseteq X$ and we say $B$ is closed in $X$, we mean if you can find points in $X$ that are really really ...

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$[0,1]$ is a closed subset of $\mathbb R$ that is not dense. It contains all of its limit points, so it is closed. Some points in $\mathbb R$, for example $2$, are not limit points of this set, so the set is not dense. $\mathbb Q$ is a dense subset of $\mathbb R$ that is not closed. It is not closed because it does not contain all of its limit points. ...

4

Let the metric space be $\mathbb{R}$. Then $\{0\}$ is closed but not dense in $\mathbb{R}$. While $\mathbb{Q}$ is dense but not closed in $\mathbb{R}$.

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A set is dense/closed in a given topological space. $[0,1]$ is closed in $\mathbb{R}$ but it is not dense in $\mathbb{R}$ since there are real numbers that can not be approached arbitrarily close by elements of $[0,1]$. $[0,1]\setminus\{\frac{1}{2}\}$ is dense in $[0,1]$ but it is not closed in it.

1

You can show that $H_1(S^2\setminus C_1\cap C_2)\ne 0$ using Mayer–Vietoris sequence, and then use Mayer–Vietoris sequence again for the pair $S^2\setminus C_1$, $S^2\setminus C_2$ and show that rank of $H_0(S^2\setminus(C_1\cup C_2))$ is at least two.

0

Take $\mathbb{G}=\mathbb{Z}_q$; take $e(1,1)$ to be any non-trivial element of $\mathbb{G}_T$, and define $e(P,Q)=e(1,1)^{PQ}$.

2

It doesn't satisfy the intermediate value theorem : $$f(\frac{9}{10}) = \frac{9}{10}$$ $$f(\frac{11}{10}) = \frac{6}{10}$$ But there is no $x \in [\frac{9}{10}, \frac{11}{10}]$ such that $f(x) = \frac{8}{10}$

4

Examples aren't in mathematics. They are tools for learning mathematics, but are exterior to mathematical systems. There couldn't be a "mathematical" symbol for it for just that reason. Granted, you could use a symbol, but it would be just a symbol that represents a word, not an acceptable character in a well-formed formula.

3

As a first thought, I would say $f^{-1}$ is always injective, since a function is invertible if and only if it is injective and surjective. $f^{-1}$ is certainly invertible since its inverse is $f$, thus it is also injective. For the bounded question, consider the function  f(x) = \left\{ \begin{aligned} &\dfrac{1}{1+(\frac{1}{x})^2} &&: x ...

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