# Why is the All but Finitely Many Theorem Important?

• Rasalhague
In summary: So |P\cup Q| is finite.In summary, the "all but finitely many" theorem states that if a set has a countable cardinality and two of its subsets have finite complements, then the intersection of the two subsets is not empty. This can be shown through a counterexample using the set of integers and the concepts of open balls and finite sets.
Rasalhague
"All but finitely many" theorem

Let $S$ be a set with cardinality $|S|=\aleph_0$. Let $A,B \subseteq S$. Let $S\backslash A$ and $S\backslash B$ be finite. Then $A \cap B \neq \varnothing$.

How can this be shown? I came across it as an assumption in a proof that a sequence in a metric space can converge to at most one limit.

There's something wrong with that assumption. I'll show you a counterexample.

S = the set of integers
A = all integers except 0
B = all integers except 1

S-A = {0}
S-B = {1}

##A\cap B## = all integers except 0 and 1
##A\cap B## ≠ ∅.

Also, you don't need any assumptions that look anything like that to prove that limits with respect to a metric are unique, but you probably know that. ##x_n\to x## means that every open ball around x contains all but a finite number of terms of the sequence. So if ##x_n\to x##, ##x_n\to y## and x≠y, just define r=(1/2)d(x,y) and consider the open balls B(x,r) and B(y,r). Since all but a finite number of terms are in B(x,r), and since B(y,r) is disjoint from B(x,r), B(y,r) contains at most a finite number of terms. This contradicts the assumption that ##x_n\to y##.

Edit: Apparently I can't even read today. thought your post said =∅, not ≠∅.

Last edited:

Hint:

$$S\setminus (A\cap B)=?$$

micromass said:
Hint:

$$S\setminus (A\cap B)=?$$

Ah, I see, thanks!

$S\setminus (A\cap B) = (S\setminus A)\cup (S\setminus B).$

If $A\cap B = \varnothing$, then $S\setminus (A\cap B) = S\setminus \varnothing = S$. So $(S\setminus A)\cup (S\setminus B) = S$. But $|S|=\aleph_0$, whereas $S\setminus A$ and $S\setminus B$ are each finite, and the union of two finite sets is finite. Contradiction. Therefore $A\cap B \neq \varnothing$.

To see that the union of two finite sets is finite, let $P,Q$ be finite. $P\cup Q = P\cup (Q\setminus P)$. As a subset of a finite set, $Q\setminus P$ is finite. As a http://planetmath.org/encyclopedia/CardinalityOfDisjointUnionOfFiniteSets.html , $|P\cup (Q\setminus P)| = |P| + |Q\setminus P|$. So $|P\cup Q|=|P\cup (Q\setminus P)|=|P| + |Q\setminus P| \in \left \{ 0 \right \}\cup \mathbb{N}$.

Last edited by a moderator:

The "all but finitely many" theorem is a fundamental result in mathematics that states that if a set has an infinite cardinality and two of its subsets are finite, then their intersection must be non-empty. This theorem is often used in various proofs, such as the one you mentioned regarding the convergence of a sequence in a metric space.

To understand why this theorem holds, let's consider an example. Suppose we have a set S with infinitely many elements, and two subsets A and B with finite elements. This means that there are only a finite number of elements that are not in A and B, and the rest of the elements must be in both A and B. In other words, the intersection of A and B must contain at least one element.

Now, let's generalize this to any set with infinite cardinality. Since there are infinitely many elements in S, there must be infinitely many elements that are not in either A or B. However, since both S\backslash A and S\backslash B are finite, there must be a finite number of elements that are not in both A and B. Therefore, the intersection of A and B must contain at least one element, proving the "all but finitely many" theorem.

This theorem has many applications in mathematics, such as in topology, number theory, and analysis. It also has implications in real-world problems, such as in computer science and economics. Overall, the "all but finitely many" theorem is a powerful tool that helps us understand the behavior of infinite sets and their subsets.

## What is the "All but finitely many theorem"?

The "All but finitely many theorem" is a mathematical theorem that states that for any infinite set of numbers, there exists a finite subset of that set that has the same properties as the infinite set.

## What is an example of the "All but finitely many theorem"?

An example of the "All but finitely many theorem" is the statement that for any infinite set of even numbers, there exists a finite subset of even numbers that has the same properties as the infinite set.

## How is the "All but finitely many theorem" used in mathematics?

The "All but finitely many theorem" is used in various areas of mathematics, including number theory and set theory. It is often used to prove the existence of certain objects or to show that certain properties hold for infinite sets.

## Is the "All but finitely many theorem" a generalization of other theorems?

Yes, the "All but finitely many theorem" is a generalization of other theorems such as the Pigeonhole principle and the Dirichlet principle. It extends these principles to infinite sets.

## Are there any limitations to the "All but finitely many theorem"?

Yes, the "All but finitely many theorem" does have limitations. It only applies to infinite sets and does not hold for finite sets. Additionally, it does not specify exactly how many elements are in the finite subset with the same properties as the infinite set.

• Calculus and Beyond Homework Help
Replies
1
Views
774
• Set Theory, Logic, Probability, Statistics
Replies
2
Views
2K
• Calculus and Beyond Homework Help
Replies
1
Views
777
• Calculus
Replies
5
Views
443
• Topology and Analysis
Replies
2
Views
213
• Set Theory, Logic, Probability, Statistics
Replies
1
Views
849
• Calculus and Beyond Homework Help
Replies
3
Views
774
• Set Theory, Logic, Probability, Statistics
Replies
27
Views
3K
• Differential Equations
Replies
4
Views
2K
• Set Theory, Logic, Probability, Statistics
Replies
6
Views
2K