We can use mathematical induction which is deduced from Peano axioms and illustrated on Terence Tao's Real Analysis(here it is)

  • Axiom 2.1 $0$ is a natural number.
  • Axiom 2.2 If $n$ is a natural number, then $n++$ is also a natural number.
  • Axiom 2.3 $0$ is not the successor of any natural number; i.e., we have $n++\neq 0$ for every natural number $n$.
  • Axiom 2.4 Different natural numbers must have different successors; i.e., if $n,m$ are natural numbers and $n\neq m$, then $n++\neq m++$. Equivalently, if $n++=m++$, then we must have $n=m$.
  • Axiom 2.5 (Principle of mathematical induction). Let $P(n)$ be any property pertaining to a natural numbers $n$. Suppose that $P(0)$ is true, and suppose that whenever $P(n)$ is true, $P(n++)$ is also true. Then $P(n)$ is true for every natural number $n$.

to prove a proposition such as $$1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$$ Here is an exercise about limitation $$\lim_{n\to\infty}\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+\cdots+\frac{1}{\sqrt{n^2+n}}\right)=1\neq 0.$$ so that we couldn't use mathematical induction here. Notice that I don't mean you can exchange the order of the limits operation and addition operation.
An other example from Royden and Fitzpatrick's Real Analysis:

Proposition 5 The union of a finite collection of measurable sets is measurable.

Proposition 7 The union of a countable collection of measurable sets is measurable.

The authors use the definition to prove Proposition 7. However, why don't we just use the technology above with Proposition 5 to obtain Proposition 7?

  • $\begingroup$ Let $a_1,a_2,\dots$ be a sequence. Sometimes one uses induction to prove that $a_n$ satisfies certain inequalities. These may be useful in finding $\lim_{n\to\infty}a_n$. $\endgroup$ Aug 12 '13 at 4:21
  • $\begingroup$ Another "induction limit proof": Proof of 1=0 by mathematical induction? $\endgroup$ Aug 12 '13 at 8:18

I'm not fully clear on what you mean, but if I'm interpreting your question correctly, the answer is

Just because $P(n)$ is true for every natural number $n$ doesn't mean that $P(\infty)$ is true.

Consider this math.SE thread and this math.SE thread.


If $P$ is a proposition, then mathematical induction can show that $P(n)$ is true for every $n$ in the natural numbers. That is, it shows that a statement is true for every finite case - but a countable union doesn't have to be a finite union, so the usual form of induction won't give us a strong enough result. That is, we can prove the result for infinitely many finite cases, but no infinite ones.


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