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First, you should know that, for any $B\subseteq X$, $X\setminus\overline{B}=(X\setminus B)^\circ$ and that $X\setminus B^\circ=\overline{X\setminus B}$. Now \begin{align*} A\text{ nowhere dense }&...

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Just as you mentioned, given $\varepsilon>0$, we can find $x_1,\ldots,x_n\in M$ such that $M\subseteq \bigcup_i B_{\varepsilon/2}(x_i)$. Let's show that $\operatorname{cl}(M)\subseteq\bigcup_i B_\... View answer 6 answers 24 votes 9k views 23 votes Since$1000=2^3\cdot5^3$, the even divisors of$1000$have the form$2^i5^j$, where$1\leq i\leq 3$and$0\leq j\leq 3$. There are only 12 of them, so you can do this calculation directly. ... View answer 4 answers 23 votes 25k views Accepted answer 20 votes Suppose that$E$is connected. Let$A,B\subseteq X$be separated sets (that is,$\overline{A}\cap B=A\cap\overline{B}=\varnothing$) such that$\overline{E}=A\cup B$, and suppose that$A\neq\varnothing$... View answer 12 answers 34 votes 2k views Accepted answer 19 votes This is done so that classical propositional calculus follows some natural rules. Let's try to motivate this, without getting into technical details: The expression "$P\Rightarrow Q$" should be read "... View answer 6 answers 1 votes 1k views 13 votes As you said,$A$is a square matrix. Since$A^2=0$, then $$0=\det 0=\det(A^2)=\det(A)^2$$ so$\det A=0$, and this means that the rows and columns of$A$are LD. View answer 1 answers 11 votes 3k views Accepted answer 13 votes Let$f:[a,b]\rightarrow\mathbb{R}$be a function with countable discontinuities, and let$c\in\mathbb{R}$. We must prove that$f^{-1}(-\infty,c)$is measurable. Let$A=\text{int}f^{-1}(-\infty,c)$(... View answer 4 answers 18 votes 2k views 12 votes This is basically a long comment: The thing is that when we construct sequences in the manner you described ("Having picked$x_1,x_2,\ldots,x_n\in S$, pick$x_{k+1}$,...") is much simpler to grasp ... View answer 3 answers 12 votes 1k views Accepted answer 12 votes Let$X$be a (non-empty) c.c.c. metric space. Let us temporarily fix a positive integer$n\geq 1$. Let$\mathscr{F}_n$be the family of classes of open balls of radius$1/n$(in$X$) which are ... View answer 1 answers 9 votes 2k views Accepted answer 10 votes Theorem Let$(M,d)$be a metric space. The following are equivalent: (a)$M$is compact; (b)$M$is sequentially compact; (c)$M$is complete and totally bounded. Proof: (a$\Rightarrow$b) ... View answer 2 answers 16 votes 2k views 10 votes No. First, recall that the product measure is not unique for non-$\sigma$-finite spaces. Let$(X,\mathscr{M},\mu)=(\mathbb{R},2^\mathbb{R},\#)$and$f(x)=|x|. We consider the product measure $$\nu(A)... View answer 4 answers 20 votes 18k views 10 votes This is not true in general. Let X=Y=[0,1]. Take X with the usual topology. For Y, take the topology$$\tau=\left\{\varnothing,Y,(1/2,1]\right\}.$$Then id:x\in X\mapsto x\in Y is continuous, ... View answer 1 answers 5 votes 2k views Accepted answer 10 votes It is obvious that \Vert Tx\Vert_\infty\leq\Vert x\Vert_\infty, hence \Vert T\Vert\leq 1. Since \Vert T(1,0,0,\ldots)\Vert_\infty=\Vert (1,0,\ldots)\Vert_\infty, we conclude that \Vert T\Vert=1... View answer 5 answers 6 votes 2k views Accepted answer 9 votes The exponential funtion t\mapsto \exp(t) is convex, so$$\begin{align} xy&=\exp(\log(xy))\\ &=\exp(\log(x)+\log(y))\\ &=\exp((1/p)\log(x^p)+(1/q)\log(y^q))\\ &\leq (1/p)\exp(\log(x^p)... View answer 2 answers 7 votes 1k views Accepted answer 9 votes SupposeX$wasn't connected. Let$A$be a non-trivial clopen in$X$, and let$B=X\setminus A$. Then$A$and$B$are closed, hence compact, subsets of$X$, so there exist$a\in A$and$b\in B$such ... View answer 4 answers 7 votes 2k views 9 votes To show that$A$has no greatest element, you have to show that there is no$q\in A$which is greater than any$p\in A$, the opposite of your first line. If$p^2<2$, we take the number$q=p+\frac{...

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Take two elements, $0$ and $1$, and define the poset $X$ with $0\leq 1$ (and the reflexivity conditions), and $Y$ the poset only with $0\leq 0$ and $1\leq 1$ (i.e., we can't compare $0$ and $1$ in $Y$....

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One of the main tools on infinite matrices and Hilbert spaces operators is the so-called Schur's test. This is Exercise 45 in Halmos' A Hilbert Space Problem Book. Schur's test. Let $A=[a_{ij}]_{i,... View answer 2 answers 4 votes 85 views 7 votes You proof is correct and looks like the easiest one. Here's one without using the fundamental theorem of fg abelian groups, but on the same spirit: By Bézout's theorem, take$k,l\in\mathbb{Z}$with$...

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If $a=0$ then this is trivial, so assume $a\neq 0$. $\mathbb{Z}_5=\mathbb{Z}/5\mathbb{Z}$ is a field, so $\mathbb{Z}_5^\times$, the group of invertible elements, is a group of $4$ elements. In ...

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The last argument where you say that $f(x)$ would be unbounded does not seem valid since you only have $f(x)\leq \Vert x\Vert_\infty\sum_{k=1}^\infty|f(e_k)|$. If $\sum_{k=1}^\infty |f(e_k)|=\infty$, ...

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No. Let $A=\left\{0,1,2\right\}$ with the operation $+:A\times A\rightarrow A$ given by $0+0=0$, $0+1=1$, $0+2=2$, $1+0=1$, $1+1=0$, $1+2=1$, $2+0=2$, $2+1=2$, $2+2=0$. Then $0$ is the (bilateral) ...

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It's true in general that $rank(A^tA)=rank(A)$. To prove that, it clearly suffices to show that $kerA^tA=ker A$, since $rank(T)=n-dim(ker T)$ for every $n\times n$ matrix $T$ (where we denote $ker T=\... View answer 3 answers 3 votes 768 views 6 votes Remember from one-variable calculus that, given a (continuous) function$f:\mathbb{R}\to\mathbb{R}$, the indefinite integral of$f$is the integral of$f$(with respect to Lebesgue measure) without a ... View answer 4 answers 4 votes 85 views Accepted answer 6 votes Well, you have to look at several cases: Since$\varnothing$is a member of the left-hand side, then it is also a member of the right-hand side. Since the members of the right-hand side are$y$and$\...

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Take any set of odd numbers. $\hspace{0pt}$

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These are called magmas, not groupoids. The midpoint'' operation $s\ast t=\frac{s+t}{2}$ on $\mathbb{R}$ makes it a magma which is not a semigroup.

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Denote by $d_G$ and $d_{G'}$ the distances in $G$ and $G'$, respectively. Take vertices $u,v$. There are a few cases of interest. If $d_G(u,v)\geq 2$, then the edge $uv$ is not in $G$, so $uv\in G'$ ...

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Let's recap some things: For general $f=(f_1,\ldots,f_m):U\subseteq\mathbb{R}^n\to\mathbb{R}^m$ (where $U$, the domain of $f$, is open in $\mathbb{R}^n$) and $x\in U$, $Df(x)$ is the Jacobian (or ...
Since $X\in\tau\subseteq\tau^*$, $X$ is open in $X^*$. For density, the only point in $X^*\setminus X$ is $\infty$. Let $V$ be an open neighbourhood (in $X^*$) of $\infty$. Then $X\setminus V$ is ...