# 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

## 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.