Loading [MathJax]/jax/output/HTML-CSS/jax.js

Lecture 8: Combinatorics

• we'll start by finishing our discussion of well-defined functions; review the lecture 7 notes. I have updated those notes with the new material we covered.

• we'll start talking about combinatorics

• reading: MCS 15.1–15.3

Finite cardinality

Definition: We say the cardinality (or size) of a set $X$ is $n$ (written $|X| = n$) if $|X| = |\{1,2,\dots,n\}|$. Here we are using the definition of same-cardinality defined in lecture 5; that is, there must exists a bijection between $X$ and $\{1,2,3,\dots,n\}$.

If this is the case, we say $X$ is finite.

Sum rule

Claim: if $A$ and $B$ are disjoint (that is, if $A \cap B = \emptyset$) then $|A| + |B| = |A \cup B|$.

Proof: Suppose $|A| = i$ and $|B| = j$. Then there exists bijections $f_A : \{1,\dots,i\} → A$ and $f_B : \{1,\dots,j\} → B$. We can define a bijection $f : \{1, \dots, i+j\} → A \cup B$ as follows: on input $n$, if $1 \leq n \leq i$ then let $f(n) ::= f_A(n)$, and if $i+1 \leq n \leq i+j$ then let $f(n) ::= f_B(n-i)$.

We can easily check that $f$ is a bijection:

• it is a function because every $n$ is either $\leq i$ or $\geq i+1$, and not both.

• it is injective; if $f(n_1) = f(n_2)$ then either $f(n_1) \in A$, in which case both $n_1$ and $n_2$ are between $1$ and $i$, and we have $f_A(n_1) = f(n_1) = f(n_2) = f_A(n_2)$. This implies that $n_1 = n_2$ since $f_A$ is injective. Similarly, if $f(n_1) \in B$, we use the injectivity of $f_B$.

• it is also surjective; given $y \in A \cup B$, either $y \in A$ or $y \in B$. if $y \in A$, then (since $f_A$ is surjective), there exists an $n$ between $1$ and $i$ with $f_A(n) = y$. But then $f(n) = y$ by definition, so in this case there exists an $n$ with $f(n) = y$. On the other hand, if $y \in B$, then there exists some $n \in \{1,\dots,j\}$ with $f_B(n) = y$. But then $f(i+n) = f_B(n) = y$, so again, $y$ is in the image of $f$. Since $y$ is in the image in either case, we see that $f$ is surjective.

Basic inclusion/exclusion

Claim: in general, $|A \cup B| = |A| + |B| - |A \cap B|$.

Note: several students pointed out that we are am using the fact that $A \cup B = (A \setminus B) \cup (A \cap B) \cup (B \setminus A)$, and also that these three sets are disjoint. I did not prove these facts, but they are certainly provable, using the definitions of intersection, union, and difference. For example, to show that $A \cap B$ and $A \setminus B$ are disjoint, we could do a proof by contradiction. Assume that there is some $x$ in $A \cap B$ and $A \setminus B$. By the definition of $\cap$, this means $x \in B$, but by the definition of $\setminus$, we see that $x \notin B$. This is a contradiction.

Proof: We could construct a bijection as we did above, but it's easier to reuse what we've already proven. Let $|A \setminus B| = i$, $|A \cap B| = j$ and $|B \setminus A| = k$.

Clearly (see note above) $A = (A \setminus B) \cup (A \cap B)$. Note these are disjoint sets, so by above, $|A| = i + j$. Similarly, $|B| = j + k$. Moreover, $A \cup B = (A \setminus B) \cup (A \cap B) \cup (B \setminus A)$ so $|A \cup B| = i + j + k = (i + j) + (j + k) - j = |A| + |B| - |A \cap B|$ as required.