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**Example 5**.: Prove that
\[1+2+3+\cdots+n=\frac{1}{2}n(n+1)\]
for every natural number \(n\).
Proof
: Let \(P(n)\) be the statement \(1+2+3+\cdots+n=\frac{1}{2}n(n+1)\). Then \(P(1)=1=\frac{1}{2}(1)(1+1)\) is true, and this establishes the basis for induction. Next we suppose that \(P(k)\) is true, where \(k\in\mathbb{N}\) (induction hypothesis). That is, we assume
\[1+2+3+\cdots+k=\frac{1}{2}k(k+1).\]
Since we wish to show \(P(k+1)\) is true, we add \(k+1\) to both sides to obtain
\[1+2+3+\cdots+k+(k+1) =\frac{1}{2}k(k+1)+(k+1)\] \[=\frac{1}{2}\left[k(k+1)+2(k+1)\right]\] \[=\frac{1}{2}(k+1)(k+2)\] \[=\frac{1}{2}(k+1)\left[(k+1)+1\right].\]
Thus \(P(k+1)\) is true whenever \(P(k)\) is true, and by induction we conclude that \(P(n)\) is true for all \(n\).
**Remark 1**.: There is a generalization of the principle of mathematical induction that enables us to conclude that a given statement is true for all natural numbers sufficiently large. In this case, let \(P(n)\) be a proposition about \(n\) and \(a\in\mathbb{N}\). Suppose that:
* \(P(a)\) is true.
* For each \(k\geq a\), if \(P(k)\) is true, then \(P(k+1)\) is true.
Then \(P(n)\) is true for all \(n\geq a\).
**Example 6**.: Suppose we want to show for all \(n\geq 5\), \(4n<2^{n}\). We again use induction, but base case is now \(k=5\) since \(4k=20<32=2^{k}\). By the inductive hypothesis we assume for some \(k\geq 5\) we have \(4k<2^{k}\). Next we consider \(P(k+1)\) and write
\[4(k+1)=4k+4<2^{k}+4=2^{k+1}.\]
Thus by induction we have that \(4n<2^{n}\) for all \(n\geq 5\).
### Cardinality
We start by defining _finite sets_. A set \(X\) is finite if \(X\) is empty or there exists \(n\in\mathbb{N}\) such that there is a bijection \(f:X\to\{1,2,\ldots,n\}\), where \(\{1,2,\ldots,n\}\) is the set of all natural numbers less than or equal to \(n\). If \(X\) is a finite set, then the _cardinality_ of \(X\), denoted \(|X|\) or \(card(X)\), is the number of elements in \(X\). For example,
\[|\{a,b,c,d\}|=4.\]
Note that the cardinality of a finite set is a natural number. Indeed, the set of natural numbers is precisely the set of cardinalities of finite sets, provided we include \(0=|\emptyset|\). We can start with some useful observations:
**Observation 3**.: There is an injective map \(f:X\to Y\) if and only if \(|X|\leq|Y|\). Why? Well, define a map \(f:X\to Y\) by choosing an element of \(Y\) for each element of \(X\), and since we are trying to make \(f\) injective, we want to avoid reusing elements. If \(|Y|<|X|\), then we will run out of elements of \(Y\) before we have a complete function, so we will have to reuse some elements. On the other hand, if \(|Y|\geq|X|\), then we will be able to avoid reusing elements of \(Y\) and possibly still have elements of \(|Y|\) to spare. Thus, there exists an injective map \(f:X\to Y\) if and only if \(|X|\leq|Y|\).
**Observation 4**.: There is a surjective map \(f:X\to Y\) if and only if \(|X|\geq|Y|\). Again imagine going through the elements of \(X\), assigning an element of \(Y\) to each. We will have a complete function when all elements of \(X\) have an \(f(x)\) assigned, and if \(|X|<|Y|\), then we will run out elements of \(X\) before all of the elements of \(Y\) are used. On the other hand, if \(|X|\geq|Y|\), then there are enough elements \(x\in X\) to assign every \(y\in Y\) to some \(x\). Thus, there exists a surjective function \(f:X\to Y\) if and only if \(|X|\geq|Y|\).
Putting these two together, we have the following definition:
**Definition 1**.: Two sets \(X\) and \(Y\) have the same cardinality if and only if there exists a bijection \(f:X\to Y\).
This definition provides a nice example of how intuition can go wrong. For example, if asked to provide a definition in terms of set theory for when one set is larger than another, many of us might suggest that \(|X|<|Y|\) if \(X\subsetneq Y\). Certainly it works this way for finite sets; however, infinite sets have the counter-intuitive property that they can be put in bijective correspondence with proper subsets of themselves. That is, for infinite sets, a proper subset can be the same size as the whole set. Indeed, this is the very definition of what it means for a set to be infinite. In particular, we have the following definition:
**Definition 2**.: A set \(X\) is _infinite_ if there is a proper subset \(S\subsetneq X\) and a bijection \(f:X\to S\).
**Example 7**.: The set of natural numbers
\[\mathbb{N}=\{1,2,3,4,\ldots,n,\ldots\}\]is infinite since there is a bijection \(f:\mathbb{N}\to 2\mathbb{N}\) between the set of all natural numbers \(\mathbb{N}\) and the set of even natural numbers \(2\mathbb{N}\) given by \(f(n)=2n\), where \(n\in\mathbb{N}\). The function \(f\) is injective since \(2n=2n^{\prime}\) implies \(n=n^{\prime}\) and \(f\) is surjective since every even natural number is \(2n\) for some natural number \(n\). More visually, we can see the correspondence with a table:
\[\begin{array}{c|cccccccc}\mathbb{N}&1&2&3&4&5&6&\dots\\ \hline 2\mathbb{N}&2&4&6&8&10&12&\dots\end{array}.\]
Thus, there are exactly as many even natural numbers as natural numbers, and the set of natural numbers is infinite. Similarly, there are exactly as many natural numbers as there are integers, which we can see most easily by listing a correspondence:
\[\begin{array}{c|cccccccccccc}\mathbb{N}&1&2&3&4&5&6&7&8&9&\dots\\ \hline\mathbb{Z}&0&-1&1&-2&2&-3&3&-4&4&\dots\end{array}\]
or, for those preferring a formula, define
\[f(n)=\left\{\begin{array}{cc}-\dfrac{n}{2}&n\text{ even}\\ \\ \dfrac{n-1}{2}&n\text{ odd}.\end{array}\right.\]
While this property is very counter-intuitive, we should not expect to have much natural intuition about infinity since it is something we have never (and can never) directly experience. Nonetheless, with careful mathematical reasoning, we can learn many interesting facts about infinity.
The following theorem is very useful because it allows us to conclude two sets have the same cardinality without explicitly constructing a bijection between the sets. This theorem is sometimes called the Schroder-Bernstein theorem. E. Schroder and F. Bernstein gave independent proofs of this theorem.
**Theorem 1** (Schroder-Bernstein Theorem).: _Let \(X\) and \(Y\) be nonempty sets. If there exist a one-to-one map \(f:X\to Y\), from \(X\) into \(Y\), and a one-to-one map \(g:Y\to X\), from \(Y\) into \(X\), then there is a map \(h:X\to Y\) that is both one-to-one and onto._
Proof
: Recall that the image of \(f\) is defined as
\[f(X)=\{y\in Y:\ y=f(x)\ \text{ for some }x\in X\}.\]
Because \(f\) is not necessarily onto, the image \(f(X)\) may not be all of \(Y\). Let \(x\in X\) be arbitrary, and let \(C_{x}\) be the list of all elements of the form
\[C_{x}=\{x,\,g^{-1}(x),\,f^{-1}\circ g^{-1}(x),\,g^{-1}\circ f^{-1}\circ g^{-1} (x),\dots\}.\]
The elements of this sequence are called _predecessors_ of \(x\). Notice that since we started with \(x\in X\), then if \(g^{-1}(x)\) exists, in other words if \(x\) is in the image of \(g\), then \(g^{-1}(x)\in Y\). For each \(x\in X\), one of the three following possibilities happens:* The list \(C_{x}\) is infinite. In this case \(x\) has infinitely many predecessors and the corresponding subset of \(X\) will be denoted by \(X_{1}\).
* The last term in the list is an element of \(X\). The corresponding subset of \(X\) will be denoted by \(X_{2}\).
* The last term in the list is an element of \(Y\). Let \(X_{3}\) denote the corresponding subset of \(X\).
If the last term in the list is an element of \(X\), then the last term is of the form \(y=f^{-1}\circ g^{-1}\circ\cdots\circ g^{-1}(x)\) or \(y=x\), and \(g^{-1}(y)\) does not exist. In this case \(C_{x}\) stops in \(X\). This explains case 2 above. If the last term in the list is an element of \(Y\), that is, the last term is of the form \(w=g^{-1}(x)\) or \(w=g^{-1}\circ f^{-1}\circ\cdots\circ g^{-1}(x)\) and \(w\) is not in the image of \(f\), then we are describing case 3 above. Just like the subsets \(X_{1},X_{2}\), and \(X_{3}\) described above, we define the corresponding subsets \(Y_{1},Y_{2}\), and \(Y_{3}\) as follows:
* \(Y_{1}=\{y\in Y:\ y\ \text{has infinitely many predecessors}\}\).
* \(Y_{2}=\{y\in Y:\ \text{the predecessors of}\ y\ \text{stop in}\,X\}\).
* \(Y_{3}=\{y\in Y:\ \text{the predecessors of}\ y\ \text{stop in}\,Y\}\).
Now observe that \(X_{1},X_{2},X_{3}\) and \(Y_{1},Y_{2},Y_{3}\) partition \(X\) and \(Y\), respectively, and the mappings \(f:X_{1}\to Y_{1}\), \(g:Y_{1}\to X_{1}\) are both bijections. The same is true for \(f:X_{2}\to Y_{2}\), \(g:Y_{2}\to X_{2}\) and \(f:X_{3}\to Y_{3}\), \(g:Y_{3}\to X_{3}\).
**Definition 3**.: A set \(X\) has cardinality \(\aleph_{0}\) (pronounced as “aleph null”) if there is a bijection between the sets \(X\) and \(\mathbb{N}\). Naturally the set \(\mathbb{N}\) has cardinality \(\aleph_{0}\).
**Definition 4**.: A set is called _countably infinite_ if it has cardinality \(\aleph_{0}\), that is, if it is in one-to-one correspondence with natural numbers. The term “countable” or “denumerable” is also used by some to refer to infinite sets that are in one-to-one correspondence with \(\mathbb{N}\), while others include finite sets when they say countable. We will say _countable_ if it is finite or countably infinite.
Let \(\mathbb{Q}_{+}\) denote the positive rational numbers. Then we claim that \(\mathbb{Q}_{+}\) has cardinality \(\aleph_{0}\). Even though in the following we do not write down explicitly what the bijection between \(\mathbb{Q}_{+}\) and \(\mathbb{N}\) looks like, we understand the idea of bijection by the following _diagonalization_ argument. We write all the positive fractions in a grid with all fractions with denominator 1 in the first row, all fractions with denominator 2 in the second row, all the fractions with denominator 3 in the third row, etc. Next go through row by row, and remove all fractions that are not written in the lowest terms. Then start with the upper left-hand corner, and draw a path through all the remaining numbers as shown in the following diagram below:
[MISSING_PAGE_EMPTY:348]
numbers. As we defined before sets that correspond bijectively with the natural numbers \(\mathbb{N}\) (or the integers \(\mathbb{Z}\) or the rational numbers \(\mathbb{Q}\)) are called _countably infinite_; larger sets are called _uncountable_.
One of the most important sets of numbers is the set \(\mathbb{R}\) of real numbers. In the coming sections we study the properties of \(\mathbb{R}\) in detail. There are two things to emphasize about \(\mathbb{R}\) at this point. The first one is the fact that the reals are uncountable, and the second one is that the cardinality of the real numbers is \(c=2^{\aleph_{0}}\). Note that some real numbers, precisely all those numbers with finite decimal expression, have two different expansions, one ending in an infinite string of zeros and the other ending with infinite string of nines. For example, \(0.125=0.1249999\dots\). For our discussion in Theorem 2 below, we consider the real numbers to be the set of all terminating or infinite decimals with the convention that no decimal expansion can terminate in all nines. Then the decimal expression of a real number is unique, since it does not end all in nines.
**Theorem 2**.: _The set of real numbers between \(0\) and \(1\) is not countable._
Proof
: This proof is done by contradiction. Assume that there is a bijection between \(\mathbb{N}\) and \((0,1)\). Thus for each \(n\in\mathbb{N}\), \(f(n)\) is a real number in \((0,1)\), and we can represent it using decimal notation as
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