# 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.