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Lecture 10: induction

• reading: MCS 5.1

• 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 $|Pow(X)| = 2^n$.

We start with a baby version:

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 $n$.