# Lecture 10: induction

• proof by induction
• example: size of the power set

## Definitions

Definition: The power set of a set $$X$$, written $$Pow(X)$$ or $$2^X$$ is the set of all subsets of $$X$$.

For example,

• $$Pow(\{1,2\}) = \{\emptyset, \{1\}, \{2\}, \{1,2\}\}$$.
• $$Pow(\emptyset) = \{\emptyset\}$$
• $$Pow(\{\emptyset\}) = \{\emptyset, \{\emptyset\}\}$$

Notation: If $$X$$ and $$Y$$ are sets, then $$X \setminus Y$$ (pronounced "$$X$$ minus $$Y$$") is the set of all elements of $$X$$ that are not in $$Y$$.

## Induction

To demonstrate how induction works, we will prove that $$|Pow(X)| = 2^{|X|}$$ for finite $$X$$.

Big Claim: For all $$n \in \mathbb{N}$$, if $$|X| = n$$ then $$|Pow(X)| = 2^n$$.

Claim 0: If $$|X| = 0$$ then $$|Pow(X)| = 2^0$$.

Proof: If $$|X| = 0$$, then $$X$$ must be the empty set. $$Pow(\emptyset) = \{\emptyset\}$$, which has $$1$$ element. We are done since $$1 = 2^0$$.

We'll prove the next claim in a more roundabout way for reasons that will become clear soon:

Claim 1: If $$|X| = 1$$ then $$|Pow(X)| = 2^1$$.

Proof: Since $$|X| = 1$$, we can write $$X = \{x_1\}$$. We can split $$Pow(X)$$ into two groups of subsets: let $$A$$ be the set of those that contain $$x_1$$ and $$B$$ be the set of those that don't. Formally, $$A = \{S \subseteq X \mid x_1 \in S\}$$ and $$B = \{S \subseteq X \mid x_1 \notin S\}$$.

Now, $$B$$ is just the power set of $$X \setminus \{x_1\}$$ (convince yourself with a small example). Since $$|X| = 1$$, $$X \setminus \{x_1\}$$ has cardinality 0. Therefore we can apply claim 0 to conclude that $$|B| = |Pow(X \setminus \{x_1\})| = 2^0$$.

Moreover, $$|B| = |A|$$, because the function $$f : B → A$$ given by $$f(S) = S \cup \{x_1\}$$ is a bijection (check this as an exercise).

So we have $$|Pow(X)| = |A \cup B| = |A| + |B| = 2^0 + 2^0 = 2 \cdot 2^0 = 2^1$$ as required.

2 steps down, an infinite number to go. On to claim 2:

Claim 2: If $$|X| = 2$$ then $$|Pow(X)| = 2^2$$.

Proof: Since $$|X| = 2$$, we can write $$X = \{x_1,\dots,x_2\}$$. We can split $$Pow(X)$$ into two groups of subsets: let $$A$$ be the set of those that contain $$x_2$$ and $$B$$ be the set of those that don't. Formally, $$A = \{S \subseteq X \mid x_2 \in S\}$$ and $$B = \{S \subseteq X \mid x_2 \notin S\}$$.

Now, $$B$$ is just the power set of $$X \setminus \{x_2\}$$. Since $$|X| = 2$$, $$X \setminus \{x_2\}$$ has cardinality 1. Therefore we can apply claim 1 to conclude that $$|B| = |Pow(X \setminus \{x_2\})| = 2^1$$.

Moreover, $$|B| = |A|$$, because the function $$f : B → A$$ given by $$f(S) = S \cup \{x_2\}$$ is a bijection.

So we have $$|Pow(X)| = |A \cup B| = |A| + |B| = 2^1 + 2^1 = 2 \cdot 2^1 = 2^2$$ as required.

There seems to be a pattern. Here's claim 3:

Claim 3: If $$|X| = 3$$ then $$|Pow(X)| = 2^3$$.

Proof: Since $$|X| = 3$$, we can write $$X = \{x_1,\dots,x_3\}$$. We can split $$Pow(X)$$ into two groups of subsets: let $$A$$ be the set of those that contain $$x_3$$ and $$B$$ be the set of those that don't. Formally, $$A = \{S \subseteq X \mid x_3 \in S\}$$ and $$B = \{S \subseteq X \mid x_3 \notin S\}$$.

Now, $$B$$ is just the power set of $$X \setminus \{x_3\}$$. Since $$|X| = 3$$, $$X \setminus \{x_3\}$$ has cardinality 2. Therefore we can apply claim 2 to conclude that $$|B| = |Pow(X \setminus \{x_3\})| = 2^2$$.

Moreover, $$|B| = |A|$$, because the function $$f : B → A$$ given by $$f(S) = S \cup \{x_3\}$$ is a bijection.

So we have $$|Pow(X)| = |A \cup B| = |A| + |B| = 2^2 + 2^2 = 2 \cdot 2^2 = 2^3$$ as required.

At this point, you are probably convinced that we could use claim 3 to prove claim 4 (with almost exactly the same proof as the proof of claim 3), and we could use claim 4 to prove claim 5, and claim 5 to prove claim 6, and so on. In fact, for any $$N$$, you could generate proofs of the claims all the way up to claim $$N$$ using the following template:

Claim n: If $$|X| = n$$ then $$|Pow(X)| = 2^n$$.

Proof: Since $$|X| = n$$, we can write $$X = \{x_1,\dots,x_n\}$$. We can split $$Pow(X)$$ into two groups of subsets: let $$A$$ be the set of those that contain $$x_n$$ and $$B$$ be the set of those that don't. Formally, $$A = \{S \subseteq X \mid x_n \in S\}$$ and $$B = \{S \subseteq X \mid x_n \notin S\}$$.

Now, $$B$$ is just the power set of $$X \setminus \{x_n\}$$. Since $$|X| = n$$, $$X \setminus \{x_n\}$$ has cardinality n-1. Therefore we can apply claim (n-1) to conclude that $$|B| = |Pow(X \setminus \{x_n\})| = 2^{n-1}$$.

Moreover, $$|B| = |A|$$, because the function $$f : B → A$$ given by $$f(S) = S \cup \{x_n\}$$ is a bijection.

So we have $$|Pow(X)| = |A \cup B| = |A| + |B| = 2^{n-1} + 2^{n-1} = 2 \cdot 2^{n-1} = 2^n$$ as required.

This is how a proof by induction works. To do a proof by induction:

• You first clearly describe what "claim $$n$$" says (this is often written $$P(n)$$ and is called the inductive hypothesis)

• You then prove the first claim directly (claim 0 in our example above, whose proof was different from the others). This is called the base case.

• You then give a "template" proof of claim $$n$$, but in that proof you may use claim $$n-1$$. This is called the inductive step.

• You can then conclude that for all $$n$$, claim $$n$$ holds, by induction.

Here is how I might put together the above pieces to prove the "big claim" if I were submitting it for a homework assignment, for example:

Big Claim: For all $$n \in \mathbb{N}$$, if $$|X| = n$$ then $$|Pow(X)| = 2^n$$.

Proof: by induction. Let $$P(n)$$ be the statement "if $$|X| = n$$ then $$|Pow(X)| = 2^n$$.

In the base case, we want to show $$P(0)$$, i.e. if $$|X| = 0$$ then $$|Pow(X)| = 2^0$$. Well, if $$|X| = 0$$, then $$X$$ must be the empty set. $$Pow(\emptyset) = \{\emptyset\}$$, which has $$1$$ element. We are done since $$1 = 2^0$$.

For the inductive step, fix an arbitrary $$n$$ and assume $$P(n-1)$$ holds. We want to show that if $$|X| = n$$ then $$|Pow(X)| = 2^n$$.

Since $$|X| = n$$, we can write $$X = \{x_1,\dots,x_n\}$$. We can split $$Pow(X)$$ into two groups of subsets: let $$A$$ be the set of those that contain $$x_n$$ and $$B$$ be the set of those that don't. Formally, $$A = \{S \subseteq X \mid x_n \in S\}$$ and $$B = \{S \subseteq X \mid x_n \notin S\}$$.

Now, $$B$$ is just the power set of $$X \setminus \{x_n\}$$. Since $$|X| = n$$, $$X \setminus \{x_n\}$$ has cardinality n-1. Therefore we can apply P(n-1) to conclude that $$|B| = |Pow(X \setminus \{x_n\})| = 2^{n-1}$$.

Moreover, $$|B| = |A|$$, because the function $$f : B → A$$ given by $$f(S) = S \cup \{x_n\}$$ is a bijection.

So we have $$|Pow(X)| = |A \cup B| = |A| + |B| = 2^{n-1} + 2^{n-1} = 2 \cdot 2^{n-1} = 2^n$$, which completes the inductive step.

Therefore, by induction, for all $$n$$, if $$|X| = n$$ then $$|Pow(X)| = 2^n$$.

Here is a second example of a proof by induction. Note that here we are assuming $$P(n)$$ and proving $$P(n+1)$$ in the inductive step. Either style (proving $$P(n) ⇒ P(n+1)$$ or proving $$P(n-1) ⇒ P(n)$$) is fine: in each case you are showing how to reduce a complex case (either one of size $$n+1$$ or one of size $$n$$) to a simpler case (either one of size $$n$$ or one of size $$n-1$$).

Claim: for all $$n$$, $$\sum_{i=0}^n i = \frac{n(n+1)}{2}$$

Proof: by induction. Let $$P(n)$$ be the statement "$$\sum_{i=0}^n i = \frac{n(n+1)}{2}$$".

For the base case, we want to show $$P(0)$$, i.e. that $$\sum_{i=0}^0 i = \frac{0(0+1)}{2}$$. By inspection, both the left and right hand sides are 0, so $$P(0)$$ holds.

For the inductive step, assume $$P(n)$$; we wish to show $$P(n+1)$$; that is, we want to show $$\sum_{i=0}^{n+1} i = \frac{(n+1)(n+2)}{2}$$.

We simplify the left-hand side: \begin{aligned} \sum_{i=0}^{n+1} i &= (n+1) + \sum_{i=0}^n i && \text{by definition} \\ &= (n+1) + \frac{n(n+1)}{2} && \text{by P(n)} \\ &= \frac{2(n+1) + n(n+1)}{2} = \frac{(n+1)(n+2)}{2} && \text{by arithmetic} \\ \end{aligned} as required.

Therefore, by induction, $$\sum_{i=0}^n i = \frac{n(n+1)}{2}$$ for all $$n$$.