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overset
▪ I. ˈoverset, n. [f. overset v.] The act or fact of oversetting, in various senses of the vb.: † a. Overthrow, defeat. Obs. b. Overturn, upsetting, upset. † c. Putting off, postponement. Obs. † d. Overload, excess. Obs. e. Printing. Matter set up in excess of space.1456 Sc. Acts Jas. II (1814) 45/2...
Oxford English Dictionary
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$X=\overset{\circ}{(X\setminus U)}\cup \overset{\circ}{B}$ if $\bar U\subseteq \overset{\circ}{B}$? Let $X$ be a topological space and $U,B\subset X$ two subspaces of $X$ with the property that the closure of $U$ is c...
Yes, it is. The interior of the complement is the complement of the closure.
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$X_n\overset{\mathcal{D}}{\rightarrow}X$, $Y_n\overset{\mathbb{P}}{\rightarrow}Y \implies X_n\cdot Y_n\overset{\mathcal{D}}{\rightarrow}X\cdot Y\ ?$ The title says it. I know that if limiting variable $Y$ is constant ...
The premise is fulfilled, but $X_n\cdot Y_n = 0\overset{\mathcal{D}}{\nrightarrow}Y = X\cdot Y$.
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$K \overset{\subset}{\to} NK$ subset homorphism of groups? If $K$ and $N$ are subgroups of a group $G$ with $N$ normal in $G$, then $N \trianglelefteq NK$. What is the definition of the homomorphism $K \overset{\subse...
$K$ is a subgroup of $NK$, so the inclusion $k\in K\mapsto k= 1\cdot k\in NK$ is a group homomorphism (this works for any subgroup of any group). This is what the author calls $K\stackrel{\subset }\to NK$. Another frequent notation is $K\hookrightarrow NK$.
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Proving $ \overset{\circ}{E} \cap \overset{\circ}{F} =\overset{\circ}{( E \cap F )}$? I'm trying to prove $ \overset{\circ}{E} \cap \overset{\circ}{F} =\overset{\circ}{( E \cap F )}$. For clarification, I take $\overs...
If $p\in \overset{\circ}{E\cap F}$ then there is some $R>0$ so that $N_R(p)\subset E\cap F$.
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Does $\xi_{ni}\overset{p}\to0$ imply $\frac1n\sum_{i=1}^n\xi_{ni}\overset{p}\to0?$ Suppose for each $i=1,\cdots,n$, we have $$\xi_{ni}\overset{p}\to0.$$ Can we claim that $$\frac1n\sum_{i=1}^n\xi_{ni}\overset{p}\to0?$...
You deleted a similar question where there was only one subscript. But anyway, the result is false. Let $\xi_{n,i}=n 1_{[\frac{i-1}n,\frac{i}n)}$ be a sequence of random variables on the usual $[0,1)$. Then for each $i$ (we can even allow it to depend on $n$), $\xi_{n,i}\xrightarrow{\mathbb{P}}_n0$ ...
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Fluent Overset 网格案例 - 知乎 - 知乎专栏
1 案例描述 本次计算的案例模型如图所示。 计算模型长3000mm,宽200mm,模型中心有一个障碍物方块,边长100mm。 重叠网格需要包含两套网格:背景网格及前景网格。 如上图中的网格模型(a)作为背景网格,而模型(b)则作为前景网格。 2 模型准备 两套网格模型需要各自准备。 不过需要注意的是,建模过程中必须确保几何位置正确。 案例几何在SCDM中创建。 2.1 创建前景网格 启动Workbench,建立如下图所示的流程 双击A2单元格进入SCDM,在XY平面上绘制如下图所示草图 点击Pull按钮生成下图所示平面 入B3单元格,采用网格尺寸8 mm进行网格划分 生成网格如图所示。 创建边界命名,左侧为入口,命名为inlet,右侧命名为outlet
zhuanlan.zhihu.com
Show that $X_n \overset{p}{\to} X$ and $Y_n \overset{p}{\to} Y$ implies $(X_n, Y_n) \overset{p}{\to} (X,Y)$. How can I show that if $X_n \overset{p}{\to} X$ and $Y_n \overset{p}{\to} Y$, then $(X_n, Y_n) \overset{p}{\...
For any $\epsilon>0$ $$P\left\\{d\left((X_n,Y_n),(X,Y)\right)>\epsilon\right\\}\le P\\{|X_n-X|>\epsilon/\sqrt{2}\\}+P\\{|Y_n-Y|>\epsilon/\sqrt{2}\\}\to 0.$$ Here, $d(z, w)=\|z-w\|_2$.
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一只瓢虫在瓶子里总重量为5,此时瓢虫起飞悬停在瓶子正中央,总重量会有什么变化?
于是,我们得到系统A所受到的合外力为: \overset{\rightharpoonup }{F}_合=\overset{\rightharpoonup }{F}_支+\overset{\rightharpoonup }{G}=-\overset{\rightharpoonup }{F}_压+\overset \overset{\rightharpoonup }{a} 为系统A的质心加速度。 \overset{\rightharpoonup }{G}=m\overset{\rightharpoonup }{g} 为系统A所受到的重力,它也是一个恒量。
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If $X_n\overset{d}{\to}c$ for some constant $c$, does that implies that $X_n\overset{\text{a.s.}}{\to}c$? I know that if $X_n\overset{d}{\to}c$ for some constant $c$, that implies that $X_n\overset{p}{\to}c$, and that...
It's false. Suppose we have a sequence of random variables $X_n$ along with $Y$ such that $X_n - Y \to 0$ in probability. Then if your statement was correct, this would imply that $X_n - Y \to 0$ almost surely. But this implies $P(\lim |X_n - Y - 0| > \epsilon )=0$ Which implies $P(\lim |X_n - Y|> \...
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Playing with closure and interior to get the maximum number of sets Can you find $A \subset \mathbb R^2$ such that $A, \overline{A}, \overset{\circ}{A}, \overset{\circ}{\overline{A}}, \overline{\overset{\circ}{A}}$ ar...
According to Kuratowski's closure-complement problem, the monoid generated by the complement operator $a$ and the closure operator $b$ has $14$ elements and is presented by the relations $a^2 = 1$, $b^2 = b$ and $(ba)^3b = bab$. Now you are interested by the submonoid generated by the **closure oper...
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Convergence in probability to a sequence converging in distribution Consider two sequences of real-valued random variables, $\\{X_n\\}_n$ and $\\{Y_n\\}_n$, and a real-valued random variable $Y$. Suppose that $X_n\ove...
Start with the identity $P(U\le a) \le P(V\le a+\epsilon)+P(|U-V|>\epsilon)$ for $\epsilon>0$. Let $a$ be a continuity point of $F_Y$. Then \begin{align} P(X_n\le a) & \le P(Y_n\le a+\epsilon)+P(|X_n-Y_n|>\epsilon) \\\ & \le P(Y\le a+\epsilon+\epsilon')+P(|Y_n-Y|>\epsilon')+P(|X_n-Y_n|>\epsilon) \en...
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Kernel of an arrow that factors through a monic? Suppose an arrow $A\overset{f}{\rightarrow}B$ factors as $A\overset{q}{\rightarrow} J \overset{j}{\rightarrowtail}B$. When does $\ker f=\ker q$ and how can I prove it?
Always. The kernel of $f$ represents maps $z$ into $A$ which are killed by composition with $f$; but if $fz=jqz=0,$ then $qz=0$ and $z$ factors also through the kernel of $q$. In the other direction, if $qz=0$ then certainly $fz=0$.
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Prove $x<y \land z>0 \Rightarrow x\cdot z < y \cdot z$ for all$x,y,z\in \mathbb{K}$ is it possible to prove it like that: $\begin{gather*} x\cdot z < y\cdot z \quad | \cdot z^{-1} \\\ x\cdot \underbrace{(z \cdot z^{-...
If $x<y$ then $0<y-x$. Since $0<z$. Then, by multiplicative axiom, you have $0<(y-x)(z)$. So, $0<yz-xz$. Hence $xz<yz$.
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Very general object-free categories? Is there a name for categories $\mathcal C$ such that $\text{Obj}(\mathcal C)$ coinside with $\text{Mor}(\mathcal C)$? In the diagram below the morphisms are $a\overset{a}{\to}a$,...
Interpreted in a certain way what you are asking is standard. A category $C$ is a collection $C_1$ together with two maps $s,t:C_1\to C_1$ and a map $m$ from $C_1\times_{\langle s,t\rangle} C_1 =\\{(f,g)\,|\,s(f)=t(g)\\}$ to $C_1$ satisfying: $st=t$, $ts=s$, $m(m(f,g),h))=m(f,m(g,h))$ and $m(f,s(f))...
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