# Tag Info

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Sets are collections of mathematical objects. Months are not mathematical objects, they are an arbitrary period of time relevant to the physical universe. If you model the universe as something with a timeline, and define some partition of the timeline to intervals which mimics the idea of months and days, then the answer is positive. But once you get down ...

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In order to answer your question, you need to preliminary answer this one: What is the length in days of each month? It seems to me that the function length on the set of months is not defined (it could serve as an example of something that taking two different values on a point in the domain is not a function). Let $\cal M$ be the set of months. In order ...

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Given a $\mathbf{Pos}$-enriched category $\mathcal{C}$ and an object $Y$ in $\mathcal{C}$, a downward-closed subobject of $Y$ is a monomorphism $m : X \to Y$ in $\mathcal{C}$ such that, for every object $T$ in $\mathcal{C}$, the induced map $$\mathcal{C} (T, m) : \mathcal{C} (T, X) \to \mathcal{C} (T, Y)$$ is the embedding of a downward-closed subset. In ...

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It is the Cartesian product defined by $X \times Y = \{ (x,y) | x \in X \quad y \in Y \}$ So for example $\{ 1 , 2 \} \times \{3,4\} = \{(1,3),(1,4),(2,3),(2,4)\}$

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First, the definition that you have written is not correct. The sign '$\le$' should be '$<$'. Now, to the answer. The set $U=\{z\in\Bbb C:|z|<1\}$ is open. To prove it, take any point $z_0\in U$. Since $|z_0|<1$, the number $r=\frac12(1-|z_0|)$ is positive, so we can define $$V=\{z\in\Bbb C: |z-z_0|<r\}$$ Then, for any $z\in V$ we have ...

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As @Hoot says, the splitting should be a map $s_n : C_n \to C_{n+1}$ (and it is indeed what's written in my version of the book). Note that in the formula you have written, the domain of the RHS is not equal to the domain of the LHS... If you want to remember how the splitting map works, look at the definition of a chain homotopy below (in the book). A ...

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Of course it's needed. The codomain is how we designate our "ending set." If we want to talk about a function from R to R that is not surjective, then defining R as the codomain is how we specify that we are talking about the function from R to R. If we abandoned the idea of a codomain, we would have no notion of non-surjective functions. I couldn't even ...

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No, in order for an integral to converge both $\int_0^{\infty} f(x) dx$ and $\int_{-\infty}^{0} f(x) dx$ must converge. $\int_{-\infty}^{\infty}\sin(x)dx$ converges in principal value however. $\int_{-\infty}^{\infty}\sin(x)dx$ is DEFINED as the sum $\int_{-\infty}^{0} f(x) dx+\int_0^{\infty} f(x) dx$ where both converge, so no $\int_{-\infty}^{\infty} ... 2 First bullet: This important theorem is about polynomials, not general functions, so it only involves polynomials. A polynomial is by definition a sum of constant coefficients times an integer power of the variable. So the expression of$q(x)$does not depend on$x$just because the theorem is looking for a polynomial, it is a "design" choice. Polynomials ... 2 Given that you say that the "Set of eleven best cricketers of the world [...] is not a set because the criteria for best cricketer changes from person to person," then wouldn't the "Set of all months having more than 28 days" also be not a set because the number of days of the month changes from year to year? 2 If$f\colon A\to A$is surjective and not injective, then it has at least one infinite "orbit". Namely, there is at least one$a\in A$such that$\{x\in A\mid\exists n\in\Bbb Z, f^n(x)=a\text{ or }f^n(a)=x\}$is infinite. Assume otherwise, then each "orbit" is finite and closed under$f$. But for finite sets, surjective implies injective. Therefore on every ... 1$(1, 2, 0)$is not the zero vector in$\mathbb{Z}^3_5$. A vector$(v_1, v_2, v_3) \in \mathbb{Z}^3_5$is zero if and only if$v_1 \equiv v_2 \equiv v_3 \equiv 0 \mod{5}$. When you multiply a vector in$\mathbb{Z}^3_5$by a scalar, you should realize that sinze$\mathbb{Z}_5$is the base field of this vector space, any scalar should be thought of as an ... 1 Since the integrand is non-negative, it suffices to verify finiteness: $$\text{For which p is }\int_{\mathbb R^+}x^p\tan^{-1} x\ dx<\infty\text{?}$$ Since$\tan^{-1}x=x+o(x)$, we have the bound$\tan^{-1}x\geq cx$for$x\in [0,\epsilon)$. Thus$p$must satisfy $$\int_{0}^{\epsilon}x^{p+1}<\infty,$$ which means$p>-2$. On the other hand, ... 1 Lets denote with $$a_n := area(E_n), \, b_n :=area(F_n), \, c_n=area(U_n), \, d_n=area(V_n)$$ and assume$\lim a_n = \lim b_n=:x$and$\lim c_n = \lim d_n=:y$. Your question is, why is$x = y $? Because of $$E_n \subset A \subset V_n$$ we have$a_n \leq d_n$and thus$x \leq y.$Also we have $$U_n \subset A \subset F_n$$ which shows$c_n \leq b_n$and ... 1 While your reasoning is correct that "every function is an operation" under that extremely general definition of "operation", I would say that a more common definition of an "operation" on a set$S$would be a function $$\alpha: S^n\to S\quad\text{ for some }n\geq 0$$ or, to allow "partial" operations, $$\alpha: X\to S\quad\text{ where }X\subset S^n\text{ ... 1 You have two mistakes, as pointed by Andres in the comments. The claim that for an infinite set A, the set B=\{X\subseteq A\mid X\text{ finite}\} has the same cardinality as A requires the axiom of choice. For a Dedekind-finite set this is most certainly false in any case (there's an injection from A into B, and this injection is not a ... 1 Let's forget the word "division" for a moment. The claim is that if f and g are polynomials in one variable (i.e., in one indeterminate) with coefficients in a field (e.g., with rational coefficients), and if g is not the zero polynomial, then there exist unique polynomials q and r, with \deg r < \deg g, such that$$ f = qg + r. $$The ... 1 I think this comment of yours is the heart of your problem. I can solve any question on long division, be it for numbers or polynomials. But I do not understand the logic behind long division. I know how the algorithm works but I don't know why it works. Once you understand that, you should be able to answer the rest of your questions. I could ... 1 Lets say a function \zeta :\Bbb{R} \to \Bbb{R} where \zeta(x)=x^2+1 . Here the range of of this will be [1,\infty) and the codomain is \Bbb{R} . S range and codomain are not the same and hence this function is not a surjection. That is that the codomain equals the range. So we see that codomain is important when you want to think about special ... 1 Look at the section on that page regarding "Normal Operators" (Normal operators include Self-Adjoint operators. There it is stated that "A is normal if and only if there exists a unitary matrix U such that$$A=U D U^*$$So the spectral theorem directly addresses your question in that it is saying: You can diagonalize with a Unitary matrix precisely ... 1 For a normal matrix$A$(in particular, for a positive definite one), there exist a unitary matrix$U$and diagonal matrix$D$such that$A = U D U^H$. The diagonal elements of$D$are the eigenvalues, and the columns of$U$are eigenvectors of$A\$ corresponding to those eigenvalues.

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