What's wrong with my logic ?

1. Aug 23, 2013

jetplan

Hi,

Consider the following two logics:

Logic 1:
Given an infinite countable set of real number A = {a1, a2, a3 ...... }
If there exist a real number M such that: a1 + a2 + a3 + ... + an < M for all integer n,
then the infinite sum a1 + a2 + a3 ... < M

Logic 2:
Given an infinite collection of open set U = {A1, A2, A3 ...... }
If A1 $\cap$ A2 $\cap$ ... $\cap$ An is an open set for all integers n, then the infinite intersection A1 $\cap$ A2 $\cap$ ... is also an open set

I am pretty sure logic 1 is legit while logic 2 is fallacious
My question is, what is wrong with the logic of the form:

If Predicate P(n) is true for all integer n, then $\\lim_{n\rightarrow +\infty}P(n)$ is also true.

Is there a general rule as to when such logic holds, and when not ?
Or, it is totally irrelevant and different problem yields different result ?

Thanks and you folks are great !

2. Aug 23, 2013

Staff: Mentor

Correct, if you change "<M" to "≤M" in the first example.

That is not true for all P.

As another example, consider the set A = {a1, a2, a3 ...... } with $a_i=\frac{1}{i^2}$. The sum of the first n elements is always a rational number, but the limit $\frac{\pi^2}{6}$ is not.

Right.

3. Aug 23, 2013

jetplan

Hi mfb,

Thanks for the cool explanation. So, do you think it makes sense if I say:

+++++

In general, "Predicate P(n) is true for all integer n" has nothing to do with the truth value of $\\lim_{n\rightarrow +\infty}P(n)$.

We need to apply different tricks specific to the problem itself to determine if $\\lim_{n\rightarrow +\infty}P(n)$ is true, and even if it is, it probably has nothing to do with "Predicate P(n) is true for all integer n"

Am I right ?

+++++

Sorry for the mumble mumble but my teacher grill me real hard on this.

Thanks

4. Aug 23, 2013

rubi

The general case of this is transfinite induction. If you want to prove something for a bigger class of ordinal numbers than just the natural numbers, it's not enough to just prove $P(\alpha)\rightarrow P(\alpha+1)$, but you also need to prove $P(\alpha)$ for all limit ordinals. In your case, you'd have to prove the case $\alpha=\omega$ (the ordinal corresponding to the case $n\rightarrow\infty$) separately.

One doesn't need to use induction for such proofs, but i thought, this might enlight you.

Last edited: Aug 23, 2013
5. Aug 23, 2013

D H

Staff Emeritus
Your "logic 1" is also fallacious. Consider 1/2+1/4+1/8+...+1/2n. This is less than 1 for all n, yet the infinite sum is exactly 1. As mfb already noted, you have to replace < with ≤.

What's wrong is that (ordinary) mathematical induction is a statement about finite numbers. A limit to infinity (or to negative infinity) is a horse of a different color.

6. Aug 23, 2013

economicsnerd

Given a sequence $\{P(n)\}_{n=1}^\infty$ of predicates, it's not in general clear what one means by $\lim_{n\to\infty}P(n).$ Without a general meaning for it, we don't have a general meaning for the statement $$\text{}P(n)\text{ true for all }n\in\mathbb N\implies \lim_{n\to\infty}P(n) \text{ true."}$$ So if the above statement doesn't have a precise meaning, it's hard to make precise statements about its general validity.

7. Aug 23, 2013

jetplan

Let's say I have a countable set of real number X = {a1, a2, a3 ...... }
Define
A1 = {a1}
A2 = {a1,a2 }
.....
An = {a1, a2, ... , an}

Is it legit to write that

$\displaystyle\lim_{n\rightarrow +\infty} {A_n} = X$ ?

Is there an ε-δ notation for set operation, equivalent to what we have for sequence and function ?
Thx!!

8. Aug 23, 2013

Staff: Mentor

I think that is (at best) a very strange notation. You can write X as union of all those A_n.

9. Aug 23, 2013

eigenperson

There is such a thing as the set-theoretic limit. And yes, in your example, X is the limit.

Just like ordinary limits, the set-theoretic limit does not always exist.