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I have always wanted to ask this question and finally encountered an exercise that incentivized the question further.

In Tao's Analysis I, the reader is asked to prove the following proposition:

Let $X$ be a set with some cardinality $n$. Then $X$ cannot have any other cardinality, i.e. $X$ cannot have cardinality $m$ for any $m \neq n$.

I used induction (and proof by contradiction within the induction) to prove this statement. Now, importantly, baked into this proposition is the idea that $X$ is finite. Specifically, in assuming that "$X$ be a set with some cardinality $n$", we have intrinsically asserted that "$X$ is finite".

Here is the definition that Tao provides for finite sets, confirming the above statement:

A set is finite iff it has cardinality $n$ for some natural number $n$; otherwise, the set is call infinite.

So my question is as follows:

Is induction (in this context) proving a statement about a finite entity (i.e. the set $X$) infinitely many times? What role does "infinity" actually play when talking about induction?

There was a previous question posed several years ago (Why doesn't induction extend to infinity? (re: Fourier series)), but I am not sure if this really gets at my question (or perhaps I have misunderstood the answers provided).

Cheers~

S.C.
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1 Answers1

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Induction proves a statement P(x) for each integer x.
It assumes P(0) and for all n in N, (P(n) implies P(n++)).
That way, step by step, each integer is reached.
Thus one can conclude for all n in N, P(n).

One cannot conclude P($\infty$) because
there is no integer n for which n + 1 = $\infty$.
Yes, n + 1 is not infinity, it is just another finite integer.

For example, let P(x) be x is finite number.
Clearly P(0) and P(n) implies P(n+1).
Thus by induction for all n in N, P(n).
As infinity is not finite, to claim induction proves P($\infty$) is absurd.

William Elliot
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