Here is a persuasive advertisement for ghostwriting services for the theme of “Countable Unions of Countable Sets”:
**Get Ahead in Math with Our Expert Ghostwriting Services**
Struggling to prove countable unions of countable sets? Our team of math experts can help! With our ghostwriting services, you’ll get a well-structured, error-free proof that showcases your understanding of the concept.
**Our Services:**
* Customized proof writing for countable unions of countable sets
* Expert guidance on systematic counting through listed elements
* Timely delivery to meet your deadlines
**Our Advantage:**
* Our writers hold advanced degrees in mathematics and have years of experience in proof writing
* We guarantee 100% originality and accuracy in our work
* Our services are confidential and tailored to your specific needs
**Why Choose Us?**
* Save time and reduce stress with our expert assistance
* Improve your understanding of the concept and achieve better grades
* Get a competitive edge in your math course with our high-quality proof writing
Don’t let math struggles hold you back. Order now and get a well-crafted proof that showcases your expertise!
countable. The fact that it is a countable union means we are allowed to write out sets in a list
\[X_{1},X_{1},X_{3},\ldots\]
and then the fact that each \(X_{n}\) is countable allow us to list its elements as \(x_{n1},x_{n2},x_{n3},\ldots\). We then finish the proof showing some systematic way of counting through \(x_{nm}\). In this proof we made an infinite number of unspecified choices; we “choose” a list of elements of \(X_{n}\) without specifying the choice we had made. Moreover since we know nothing about the sets \(X_{n}\), it is impossible to say how we list them. For a long time mathematicians discussed among themselves about the use of the axiom of choice and seemingly contradictory results it produced. Why do people make a fuss about the axiom of choice? The main reason is that if it is used in a proof, then the part of that proof is nonconstructive. Now most seem to agree that the advantages of the axiom of choice outweigh the disadvantages.
**Definition 5**.: Suppose \(X\) is a set, \(I\) is an index set (not assumed to be countable), and \(\{X_{i}\}_{i\in I}\) is a collection of nonempty subsets of \(X\). We call \(\psi\) a _choice function_ if \(\psi:I\to\bigcup_{i\in I}X_{i}\), defined as \(i\mapsto x_{i}\) such that \(\psi(i)\in X_{i}\) for all \(i\). The axiom of choice can be stated as follows:
**The Axiom of Choice:** For every collection of nonempty sets, there exists a choice function.
Note that the axiom of choice is a nonconstructive assertion of existence. It postulates the existence of certain objects without giving any indication of how to find these objects. There are various logically equivalent statements to the axiom of choice, such as the Hausdorff Principle, Zorn’s lemma, and Well-Ordering Principle which we state below. Among these are two forms of the axiom of choice that are more often used in mathematics than the basic form we stated above. One is the _well-ordering principle_ which states every set can be well-ordered. The other is _Zorn’s lemma_ which states under certain circumstances “maximal” element exists. This is hugely important; for example, a basis for a vector space is precisely a maximal linearly independent set, and it turns out that Zorn’s lemma applies to collections of linearly independent sets in a vector space, which shows that every vector space has a basis.
To define these equivalent concepts, we first need the definitions of _partially ordered set_, _totally ordered set_, and _well-ordered set_.
**Definition 6**.: A _partially ordered set_ is a set \(X\) with a relation \(\leq\) that is reflexive, transitive, and antisymmetric. By antisymmetric we mean if \(x\leq y\) and \(y\leq x\), then \(x=y\). In a partially ordered set, an element \(y\) is _maximal_ if \(x\geq y\) implies \(x=y\). Also, in a partially ordered set \(X\) with \(Y\subset X\), an _upper bound_ for \(Y\) is an element \(x\in X\) such that \(y\leq x\) for all \(y\in Y\).
A _totally ordered set_ is a partially ordered set with an additional property that for any two elements, say \(x,y\in X\) (\(x\) and \(y\) distinct), either \(x\leq y\) or \(y\leq x\) (but not both). Finally, a _well-ordered set_\(X\) is a totally ordered set in which any nonempty subset \(E\subset X\) has a least element. That is an element \(x^{*}\in E\) such that \(x^{*}\leq x\) for any other \(x\in E\).
**Example 14**.: Simple examples of totally ordered sets are \((\mathbb{N},\leq)\), \((\mathbb{Q},\leq)\), and \((\mathbb{R},\leq)\). Let \(X\) be a set, and let \(\mathcal{P}(X)\) denote the collection of all subsets of \(X\). Then it is not hard to see that \((\mathcal{P}(X),\subseteq)\) is a partially ordered set.
**Hausdorff Maximality Principle:** Every partially ordered set \(X\) contains a totally ordered subset \(Y\) that is maximal with respect to the ordering on \(\mathcal{P}(X)\).
**Zorn’s Lemma:** If a nonempty partially ordered set has the property that every nonempty totally ordered subset has an upper bound, then the partially ordered set has a maximal element.
**Well-Ordering Principle:** Any set \(X\) can be well-ordered.
Perhaps it is obvious that the Well-Ordering Principle implies the Axiom of Choice, because if we well-order \(X\), we can choose \(x_{i}\) to be the smallest element of \(X_{i}\), and in this way we have constructed the required choice function. However, it is not so easy to show that the Axiom of Choice implies the Well-Ordering Principle. The proofs of the equivalence of the Axiom of Choice, the Hausdorff Maximality Principle, Zorn’s lemma, and the Well-Ordering Principle are all explained in the book by Paul J. Sally [43], p. 40.
We will need the axiom of choice in Sections 2.7 and 2.8 of Chapter 2, when we are discussing non-measurable sets and the proof of the Banach-Tarski paradox. The Banach-Tarski paradox states that there is a way of dividing a solid unit sphere into a finite number of subsets and then reassembling these subsets using rotations, reflections, and translations to form two solid unit spheres. The proof does not provide an explicit way of defining subsets. Why does it seem strange and paradoxical? It is because we feel volume has not been preserved. But how do we know we can give a sensible definition of volume for all subsets of the sphere? This leads us to measure theory and non-measurable sets.
### Exercises
1. For two sets \(A\) and \(B\), show that the following statements are equivalent:
1. \(A\subseteq B\).
2. \(A\cup B=B\).
3. \(A\cap B=A\).
2. Establish the following set theoretic relations:1. \(A\cup B=B\cup A\) and \(A\cap B=B\cap A\) (Commutativity).
2. \(A\cup(B\cup C)=(A\cup B)\cup C\) and \(A\cap(B\cap C)=(A\cap B)\cap C\) (Associativity).
3. \(A\cup(B\cap C)=(A\cup B)\cap(B\cup C)\) and \(A\cap(B\cup C)=(A\cap B)\cup(B\cap C)\) (Distributivity).
4. \(A\subseteq B\iff B^{c}\subseteq A^{c}\).
5. \(A\setminus B=A\cap B^{c}\).
6. \((A\cup B)^{c}=A^{c}\cap B^{c}\) and \((A\cap B)^{c}=A^{c}\cup B^{c}\). (De Morgan’s laws).
7. If \(A,B\), and \(C\) are sets, show that 1. \(A\times B=\emptyset\iff A=\emptyset\) or \(B=\emptyset\).
8. \((A\cup B)\times C=(A\times C)\cup(B\times C)\).
9. \((A\cap B)\times C=(A\times C)\cap(B\times C)\).
10. Suppose \(f:A\to B\) and \(g:B\to C\) are functions, show that 1. If both \(f\) and \(g\) are one-to-one, then \(g\circ f\) is one-to-one.
11. If both \(f\) and \(g\) are onto, then \(g\circ f\) is onto.
12. If both \(f\) and \(g\) are bijection, then \(g\circ f\) is a bijection.
13. For an arbitrary function \(f:X\to Y\), prove that the following relations hold: 1. \(f(\bigcup_{i\in I}A_{i})=\bigcup_{i\in I}f(A_{i})\). 2. \(f(\bigcap_{i\in I}A_{i})\subseteq\bigcap_{i\in I}f(A_{i})\). 3. Give a counterexample to show that \(f(\bigcap_{i\in I}A_{i})=\bigcap_{i\in I}f(A_{i})\) is not always true.
14. Given \(f:A\to B\), suppose there exists two functions \(g:B\to A\) and \(h:B\to A\) such that \(f\circ g=I_{B}\) and \(h\circ f=I_{A}\). Show that \(f\) is a bijection and that \(f^{-1}=g=h\).
15. For a function \(f:X\longrightarrow Y\), show that the following statements are equivalent: 1. \(f\) is one-to-one. 2. \(f\left(A\cap B\right)=f\left(A\right)\cap f\left(B\right)\) holds for all \(A,B\in\mathcal{P}\left(X\right)\).
8. Let \(A\) be a set, and let \(\mathcal{P}(A)\) denote the set of all subsets of \(A\) (i.e., the power set of \(A\)). Prove that \(A\) and \(\mathcal{P}(A)\) do not have the same cardinality.
9. If \(A\) and \(B\) are sets, then show that
1. \(\mathcal{P}(A)\cup\mathcal{P}(B)\subseteq\mathcal{P}(A\cup B)\).
2. \(\mathcal{P}(A)\cap\mathcal{P}(B)=\mathcal{P}(A\cap B)\).
10. Prove \((1+x)^{n}\geq 1+nx\) for all real \(x>-1\) and all positive integers \(n\) (_Bernoulli’s inequality_).
11. Prove \(2^{n}\geq n^{2}\) for all \(n\geq 4\).
12. An _algebraic number_ is a root of a polynomial whose coefficients are rational. Show that the set of all algebraic numbers is countable.
13. Show that a countable union of finite or countable sets is countable.
14. Show that \((0,1)\) is uncountable if and only if \(\mathbb{R}\) is uncountable.
15. Let \(X=\mathbb{R}^{2}\), and define \(\vec{u}\sim\vec{v}\) if \(\vec{u}-\vec{v}\in\mathbb{Z}^{2}\), i.e., two vectors in the plane are equivalent if their components differ by integers. Then show that \(\sim\) is an equivalence relation.
### 1.2 Number Systems
For most of us, our first exposure to mathematical reasoning involves numbers and computation, and sadly too many of us never encounter any mathematics past this early stage. Even worse, the interesting aspects of numbers are too often ignored in favor of mechanical algorithms for computation. In this section we will explore the various number systems that are used throughout mathematics. We will focus on the interesting side too often ignored in favor of the mechanics of computation, which we assume the reader is familiar with from high school algebra.
### The Natural Numbers
Recalling the definition given in Example 7, the _natural numbers_
\[\mathbb{N}=\{1,2,3,\dots,n,\dots\}\]
are the kinds of numbers, which can be cardinalities of finite sets. Indeed, this is how we first conceptualize natural numbers – to teach a child the concept of the number two, we might show him her many examples of pairs of things.
Thinking of the natural numbers as cardinalities of finite sets gives us two operations on \(\mathbb{N}\): _addition_ comes from taking unions of disjoint sets, and _multiplication_ comes from taking Cartesian products:
\[|A|+|B|=|A\cup B|\quad(\text{if }A\cap B=\emptyset)\quad\text{and}\quad|A|\times|B|=|A \times B|.\]
In order to minimize writing, we generally omit the “\(\times\)” symbol when multiplying, writing “\(xy\)” instead of “\(x\times y\).”
These operations of addition and multiplication have some important properties:
* **Commutativity.** If we are adding or multiplying two natural numbers, the order of the two numbers does not matter: \[x+y=y+x\quad\text{and}\quad xy=yx.\]
* **Associativity.** If we are adding or multiplying three numbers, it does not matter what order we resolve the operations: \[x+(y+z)=(x+y)+z\quad\text{and}\quad x(yz)=(xy)z.\]
* **Identity Elements.** There are natural numbers \(0\) and \(1\) which are _identity elements_ with respect to addition and multiplication, respectively: \[0+x=x\quad\text{and}\quad 1x=x.\]
Not every operation on a set has the properties above; they are useful enough that mathematicians have a name for a set with an operation satisfying all the properties above. A set with an associative operation is called a _semigroup_; a semigroup with an identity element is called a _monoid_, and if the operation is commutative, we have a _commutative semigroup_ or _commutative monoid_.
### The Integers and Rational Numbers
As someone playing with sets quickly observes, some additions and multiplications can be undone, while others cannot – we can undo multiplication by two if our input is even, but not if it is odd.
One approach is to define new operations that undo the old ones, e.g., subtraction and division. However, there are some problems (or at least, inelegancies) in this approach – the “undoing operations” may not share the nice properties of the operation they undo, and worse, they may not even make sense at all in some cases.
For example, subtraction and division are both non-commutative and non-associative:
\[3-2\neq 2-3\quad\text{and}\quad 2\div 3\neq 3\div 2,\]
while
\[1-(2-3)=1-(-1)=2\neq-4=-1-3=(1-2)-3\]and
\[1\div(2\div 3)=\frac{3}{2}\neq\frac{1}{6}=(1\div 2)\div 3.\]
More problematic is the fact that the operation may not even be defined for every pair of natural numbers. For example, division by zero does not make any sense – if \(x\div 0=y\), then \(0y=x\), which is not possible if \(x\) and \(y\) are both nonzero natural numbers.
A better solution is to expand the set of numbers we are using so that we can undo addition with another addition. That is, for every natural number \(x\neq 0\), we would like there to be another number \(-x\) such that \(-x+x=0\) and a number \(x^{-1}\) such that \(xx^{-1}=1\). Including these gives us the _negative integers_ and _rational numbers_, respectively. In particular, subtraction \(x-y\) is simply addition to \(x\) of \(-y\) and division \(x\div y\) is simply multiplication of \(x\) by \(y^{-1}\). The numbers \(-x\) and \(x^{-1}\) are the _additive inverse_ and _multiplicative inverse_, respectively, of \(x\).
发表回复