## Finite Sets

June 2020

**Problem 4: Let \(A\) be a nonempty finite simply ordered set.**

- (a) Show that \(A\) has a largest element.
- (b) Show that \(A\) has the order type of a section of the positive integers.

Since \(A\) is finite and simply ordered, we can define the order-preserving bijection \(f: A \rightarrow \{1, \ldots, n\}\). Since \(n\) is the largest element in the image and \(f\) is a bijection, it follows that \(f^{-1}(n)\) is the largest element in \(A\). The answer to (b) is fairly obvious since we're able to construct an order-preserving bijection.

\(\tag*{$\blacksquare$}\)

**Problem 6:**

- (a) Let \(A = \{1, \ldots, n\}\). Show there is a bijection of \(\mathcal{P}(A)\) with the cartesian product \(X^n\) where \(X = \{0, 1\}\).
- (b) Show that if \(A\) is finite, then \(\mathcal{P}(A)\) is finite.

(a) We know that, given \(A= \{1, \ldots, n\}\), the cardinality of \(\mathcal{P}(A)\) is \(2^{n}\). So we must prove that the cardinality of \(X^n\) is also \(2^{n}\). To be thorough, we will also prove the cardinality of the power set.

Starting with \(\mathcal{P}(A)\), given \(A = \{1\}\), \(\mathcal{P}(A) = \{\varnothing, \{1\}\}\), which validates our claim (\(2^{1} = 2\)). Having shown this for \(n\), we show that this is also true for \(n+1\).

Let \(B = \{1, \ldots, n, n+1\}\). Then \(\mathcal{P}(B)\) contains
every \(s \in \mathcal{P}(A)\), and, since \(B = A \cup \{n+1\}\),
\(\mathcal{P}(B)\) also contains every \(s \cup \{n+1\}\) (for every
\(s \in \mathcal{P}(A)\), we can define a \(s \cup \{n+1\} \in
\mathcal{P}(B)\) ).
Thus, since there are the same number of elements \(s\) as there
are \(s \cup\{n+1\}\), \(B\)'s cardinality is precisely double the
cardinality of \(A\).

Now onto \(X^{n}\). For \(n=1\), the cardinality of \(X^{n}\) is 2, since \(X^{1} = X\).Having shown this is true for \(n\), we show it is true for\(n+1\).

Any tuple in \(X^{n+1}\) takes on the form \((x_1, x_2, \ldots, x_n, x_{n+1})\) where the elements \((x_1, \ldots, x_{n})\) correspond to a tuple in \(X^{n}\). Thus, half the tuples in \(X^{n+1}\) take on the form \((x_1, x_2, \ldots, 0)\), and half take on the form \((x_1, x_2, \ldots, 1)\), meaning that \(X^{n+1}\) has double the cardinality of \(X^{n}\). If \(|X^{n}|=2^{n}\), and \(|X^{n+1}|=2|X^{n}|\), then the cardinality of \(X^{n+1}\) also follows the \(2^{n}\) rule.

Thus, since both \(\mathcal{P}(A)\) and \(X^{n}\) have the same cardinality, we can find a bijection between the two sets.

(b) If \(A\) is finite, then \(|A| = n, n \in \mathbb{Z}_+ \implies |\mathcal{P}(A)| = 2^{n}\).

\[\tag*{$\blacksquare$}\]