Why doesn't induction extend to infinity? (re: Fourier series)

Solution 1:

A trivial case where $P(n)$ is true for all $n\in\mathbf N$ but $P(\infty)$ is false is the statement "$n$ is finite".

Solution 2:

Here is a quote from B. Russell's Introduction to mathematical philosophy, pages 27-28, that I think describes well this limitation of induction:

Mathematical induction affords, more than anything else, the essential characteristic by which the finite is distinguished from the infinite. The principle of mathematical induction might be stated popularly in some such form as "what can be inferred from next to next can be inferred from first to last." This is true when the number of intermediate steps between first and last is finite, not otherwise. Anyone who has ever watched a goods train beginning to move will have noticed how the impulse is communicated with a jerk from each truck to the next, until as last even the hindmost truck is in motion. When the train is very long, it is a very long time before the last truck moves. If the train were infinitely long, there would be an infinite succession of jerks, and the time would never come when the whole train would be in motion. Nevertheless, if there were a series of trucks no longer than the series of inductive numbers..., every truck would begin to move sooner or later if the engine persevered, though there would always be other trucks further back which had not yet begun to move.

There are contexts in which a statement $P(n)$ can be proved for all $n\in\mathbb N$ by induction, and has a counterpart $P(\infty)$ that is false. In other contexts, $P(\infty)$ may be true. But even then, induction on $\mathbb N$ does not prove the $P(\infty)$ case.

Getting back to limits of functions, note for example that:

• A finite sum of continuous functions is continuous.

• A pointwise convergent series of continuous functions need not be continuous.

• But, a uniformly convergent series of continuous functions is continuous.

So in this case, going from finite sums to infinite series requires new tools, different types of convergence, to obtain the desired properties. As for real analytic functions, I don't know what can be said along these lines. For complex analytic functions there are nicer results, such as the fact that a locally uniformly convergent sequence of complex analytic functions is complex analytic. In the real case, to give a stark contrast, every continuous function on a bounded interval is a uniform limit of polynomials (as analytic as you can get), but there are continuous functions that are differentiable nowhere. Similarly, a continuously differentiable function of period $2\pi$ is the uniform limit of its Fourier series, but continuously differentiable functions need not even be twice differentiable, let alone analytic.

Solution 3:

An example where "$P(n)$ for all $n$" does not imply $P(\infty)$, not from the realm of analysis:

Let $P(n)$ be "the set $\cup_{k=1}^n [\frac1k, 1]$ is closed." Clearly it is true for every positive integer $n$, since the union is $[\frac1n,1]$.

But $\cup_{k=1}^\infty [\frac1k, 1] = (0, 1]$ is not closed, so $P(\infty)$ is false.

Solution 4:

With all of these counter-examples, perhaps it would be important to mention that there is a version of induction which "extends to infinity". It is called Transfinite induction.

Transfinite induction can be used to prove some pretty surprising things: for instance, it can be used to show that there exists a subset of $\mathbf R^2$ which intersects every line in exactly 2 points. Or that $\mathbf R^3-\{0\}$ can be partitioned into disjoint lines.