# Infinite intersection

1. Apr 5, 2014

### Atomised

1. The problem statement, all variables and given/known data

Let $\Lambda$ = N and set A$_{j}$ = [j, $\infty$) for j$\in$ N Then

j=1 to $\infty$ $\bigcap$ A$_{j}$ = empty set

Explanation: x$\in$ j=1 to $\infty$ $\bigcap$ provided that x belongs to every A$_{j}$.

This means that x satisfies j <= x <= j+1, $\forall$ j$\in$N. But clearly this fails whenever j is a natural number strictly greater than x. In other words there are no real numbers which satisfy this criterion.

2. Relevant equations

I understand the importance of demonstrating that x belongs to Aj for all j

3. The attempt at a solution

Why not just choose x = j+1, thus it will belong to Aj

1. The problem statement, all variables and given/known data

I know this contradicts the x <= j+1 condition but I do not understand this condition, why can't x exceed j+1?

Apologies if formatting unclear.

Last edited: Apr 5, 2014
2. Apr 5, 2014

### Zondrina

I think I can make out what you wrote there. Write a few terms of your intersection out...

$A_1 \cap A_2 \cap A_3 \cap ...$
$= [1,∞) \cap [2,∞) \cap [3,∞) \cap ...$

Notice as your intersection proceeds, the $(j+1)^{th}$ interval does not contain the previous $j$. So when you intersect $[1,∞) \cap [2,∞)$, the $1$ will not be included in the intersection and then the $2$ and so on...

So eventually you should wind up with nothing at all.

EDIT: Choosing $x > j$ ensures $x \in A_j$.

Last edited: Apr 5, 2014
3. Apr 5, 2014

### Atomised

But if you choose x to be j+1 it will always belong to Aj. ??

4. Apr 5, 2014

### statdad

You need to choose a fixed x, not one that changes with j. Your comment simply means that you can find, for any set, a number in THAT set - that doesn't imply that it is in all sets.

Even more specific. If you choose x to be 51 then yes, it is in A50 and A51, but it won't be in A52.

5. Apr 5, 2014

### HallsofIvy

Yes, that x will belong to Aj but it won't belong to A(j+1) or any A with a larger subscript. The point is that j can take on any integer value- there is no x that is in all Aj.

6. Apr 6, 2014

### Atomised

Here is my naïve response:

Is it possible that statdad's answer (which I'm sure is technically correct) simply exposes a flaw in this approach?

It seems to me that the delta epsilon approach to limits is of the nature of a convergent sequence, but this is of the nature of a divergent sequence and not susceptible to this approach. Of course j will eventually exceed any given x, but it is not as if any Aj will ever run out of elements.

Another garbled idea: R & N have different orders of infinitude, but all Aj are countably infinite.

Is the given answer not just a convention rather than a truth?

-----------------------------------------------------------------------------------------

If I do accept the conventional argument would it not be more correct to say

j <= x < j + k, k some arbitrary k in Z?

I mean the j+1 is arbitrary is it not?

This example was drawn from an excellent set of notes for an introductory course in real analysis.

Last edited: Apr 6, 2014
7. Apr 6, 2014

### statdad

Again, the idea is: in order for the intersection of these (or any collection) of sets to be non-empty, there has to be at least one fixed number in each one. By setting x = j but then continuing to discuss sets Aj you are not giving a fixed value: every time you select a new set you select a new x.

To attempt to save that line: suppose you fix a value j, and let x = j. Then this x is not in any Ak for k > j, so this x is not in the intersection.

8. Apr 6, 2014

### Atomised

Statdad: thanks your answer helps me to understand what is required. I still have a doubt and I will revisit the topic once I have more fully studied sequences.

I have proven that the problem I am currently working on is non-understandable at n, for all n.

9. Apr 7, 2014

### Atomised

Can somebody tell me which law of mathematics is contradicted by defining x = j+1 and rather than assuming that one must always choose x and test it first, assume instead that the mathematician must choose a (high) value for j, in order to 'escape' from x. Of course with this set up x will always belong to Aj. Is it not just a game where whoever goes first loses?

10. Apr 7, 2014

### PeroK

$$Let \ A_1 = [1, 1.9] \ and \ A_2 = [2, 2.9]$$
$$And \ consider \ A_1 \cap A_2$$
$$Let \ x = j + 0.5$$
$$If \ j = 1, \ then \ x = 1.5 \in A_1 \ and \ if \ j = 2, \ then \ x = 2.5 \in A_2.$$
$$Hence \ x \in A_1 \cap A_2$$

Hence the intersection is not empty.

11. Apr 7, 2014

### Atomised

Thanks Perok, point taken. How about this scheme:

Choose $x_0$ and when $j<=x_0<=j+1$
Choose $x_1$ and when $j<=x_1<=j+1$
Choose $x_2$ and when $j<=x_2<=j+1$
etc ad infinitum

Then $x_i$$\in$$A_j$,$\forall$j

Is this not legitimate mathematically?

12. Apr 7, 2014

### PeroK

Mathematically, the intersection is non-empty if you can find an x that is in every set. You can define x however you like, but it must be a single x. You can't choose a different x for different sets. All you're showing then is that every A_j is non-empty.

This is an important point of logic and the definition of intersection.

You mentioned a game where who goes first loses. This is quite a good analogy.

Intersection game: you specify x and I see if I can find a set that does not contain x. If the intersection is non-empty, you win; if it's empty I win.

Every set is non-empty game: you specify a set and I see whether it contains an element. If every set is non-empty I win; if at least one of the sets is empty you win.

If you're playing the intersection game, you have to stick by the rules of that game!

13. Apr 7, 2014

### Atomised

Although perhaps only relevant for this particular example with the interval [j,$\infty$) my point is precisely that $x_i$ belongs to all $A_j$ if j<i, and that i can be set arbitrarily large - so that it is not just a case of $A_j$ being non-empty, but $A_j$ containing the same element.

How about an Aussie Rules Intersection game:

You specify subset $A_j$ and I give you an element that is in this and all prior $A_j$?

Is that illogical?

14. Apr 7, 2014

### PeroK

That proves that every finite sub-family of sets A_j has a non-empty intersection.

15. Apr 7, 2014

### az_lender

It's not illogical, but it loses track of the original proposition, which was that the intersection of ALL $A_j$ is empty. If the intersection of all $A_j$ were not empty, it would be possible to specify at least one real number that belongs to all of them.

BTW, I agree with your reluctance (in the original post) to accept the condition x <= j+1 as a condition of x's membership in $A_j$. The only condition of x's membership in a particular $A_j$ is j <= x.

16. Apr 7, 2014

### Atomised

Perok, az_lender,

Is there a mathematical double standard here, in the following theorem

a = b iff | a - b | < ε, $\forall$ ε > 0

it is accepted that the difference between a and b being arbitrarily small stands for equality, why then is $x$'s ability to be arbitrarily large insufficient to secure membership in any A_j?

17. Apr 7, 2014

### SammyS

Staff Emeritus
"Is there a mathematical double standard here" ?

No.

According to the Law of Trichotomy, exactly one of the following must be true.
(i) a-b > 0
(ii) a-b = 0
(iii) a-b < 0​
If either (i) or (iii) is true, then you can find ε > 0 such that | a-b | > ε .

Otherwise, it must be true that a-b = 0, in which case a = b .

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted