Induction extended to ordered pairs: Application of the First Principle of Finite Induction to proving Bertrand’s Ballot Theorem.

Question

In this question, the comments stated the existence of Bertrand’s Ballot Theorem. In reading the article, I found a peculiar application of proof by induction. The text is as follows:

• We loosen the condition $p>q$ to ${\displaystyle p\geq q}$. Clearly, the theorem is correct when $p=q$, since in this case the first candidate will not be strictly ahead after all the votes have been counted (so the probability is 0). (Comment: Base Case)
• Clearly the theorem is true if $p > 0$ and $q = 0$ when the probability is 1, given that the first candidate receives all the votes; it is also true when $p = q > 0$ as we have just seen.(Comment: Base Case)
• (Comment: Induction Hypothesis) Assume it is true both when $p = a − 1$ and $q = b$, and when $p = a$ and $q = b − 1$, with $a > b > 0$. (We don't need to consider the case $a=b$ here, since we have already disposed of it before.)(Comment: Induction Step) Then considering the case with $p = a$ and $q = b$, the last vote counted is either for the first candidate with probability $\dfrac{a}{(a + b)}$, or for the second with probability $\dfrac{b}{(a + b)}$. So the probability of the first being ahead throughout the count to the penultimate vote counted (and also after the final vote) is:$$\dfrac{a}{a+b}\frac{(a-1)-b}{a+b-1}+ \dfrac{b}{a+b}\frac{a-(b-1)}{a+b-1} =\frac{a-b}{a+b}.$$ And so it is true for all $p$ and $q$ with $p > q > 0$.

I want to understand how the First Principle of Finite Induction can be applied rigorously here.

The First Principle of Finite Induction states that

Let S be a set of positive integers with the following properties:
(i) The integer $1$ belongs to S,
(ii) Whenever the integer $k$ is in S, the next integer $k+1$ must also be in S.
Then S is the set of all positive integers.

But here the statement seems to be as such (at least this is what I formulated/interpreted):

Define set S such that if

• $(a,0)\in S$ and $(0,a)\in S$
• Whenever $(a,b-1)\in S$ and $(a-1,b)\in S$ it so happens that $(a,b)\in S$ $\forall (a,b)\in \mathbb N\times \mathbb N$

Then $S= \mathbb N\times \mathbb N$.

In order to prove this, I tried to formulate an analogous version of the Well-Ordering Principle, but failed, because it (almost obviously) doesn’t seem to exist for sets of ordered pairs.

How to apply the First Principle of Finite Induction on the above mentioned proof? Otherwise it is not exactly intuitive to me to apply induction to ordered pairs like this.

Answer 1

Assuming you know that $S$ is a subset of $\mathbb{N}\times\mathbb{N}$, then you can indeed prove that. Going for a contradiction, take the least pair under the lexographical ordering belonging to $\mathbb{N}\times\mathbb{N}$ but not in $S$, both the coordinates $a,b$ of such a pair must be non-zero but then both $(a-1,b)$ and $(a,b-1)$ must be in $S$, a contradiction, hence no least pair exists which means that the intersection between $S$ and $\mathbb{N}\times\mathbb{N}$ is $\mathbb{N}\times\mathbb{N}$.

Answer 2

I’ll write a proof on the lines of the standard proof of the First Principle of Finite Induction:

The Lexicographical ordering of a Cartesian product $A\times B$ of two sets $A$ and $B$ is defined as :

If $a, a’\in A$ and $b, b’\in B$, then $$(a,b)\leq (a’,b’)\quad\text{if and only if}\quad a<a’ \quad\text{or}\quad (a=a’ \ \text{and}\ b\leq b’)$$

Consider the set $S$ such that it satisfies the given criteria. Consider the set $T$ such that it is the set of all the elements of $\mathbb N^2$ which are not in $S$, i.e., $T=\mathbb N^2\setminus S$. Assume $T$ to be non-empty.

Under the lexicographical ordering of $T$, consider $(a,b)$ to be the least element of $T$. Then, $a,b≠0$ because $(a,0)\in S$ and $(0,a)\in S$ $\forall a\in \mathbb N$ as per the hypothesis.

Now, by the minimal nature of $(a,b)$, along with the fact that $(a-1, b)<(a,b)$ and $(a,b-1)<(a,b)$ by the definition of the lexicographic order, $(a,b-1)$ and $(a-1,b)$ do not belong to $T$; consequently, they belong to the set $S$.

But by the hypothesis, $(a,b)$ must belong to $S$, which is a contradiction. Thus, our assumption is wrong and $T$ is empty. Thus, $S=\mathbb N\times \mathbb N$.

Hence proved. Q.E.D.$\qquad\blacksquare$

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P.S. What I learnt from this was the lexicographical ordering for a Cartesian Product.