# intersection of infinite collection of finite sets?

I know that there are questions asking like "intersection of a infinite collection of sets" and I can understand that the answer for that one is a null set, but I got a question here, in which all sets are finite and nonempty. Please take a look at the pic below.

It's a True/False question.

What I'm confused about is that the "infinity" sign in the interception. Since all sets are finite, does it mean that there are many sets in the chain are the same? (because the symbol used is for subsets not for proper subsets). Please give me some hints how I should think about this question.

Thank you

• Maybe to understand the question, realize that if each is non-empty and finite, they have a cardinality, which is a function from a set to R. Since each is non-empty, each A has a positive cardinality... where should you go from here? – Alfred Yerger Jan 28 '15 at 22:14
• @Alfred: Since when cardinals are real numbers? – Asaf Karagila Jan 28 '15 at 22:25
• Let $c_n$ be the cardinality of $A_n$. Then the set of all $c_n$ is a non-empty set of positive integers, so has a smallest element. – André Nicolas Jan 28 '15 at 22:29
• @AsafKaragila I just meant to give some rationale, by no means an attempt at a proof. – Alfred Yerger Jan 28 '15 at 22:37
• thanks. Since An are finite sets, can I say that the set of cn is a finite sets? Because the number of subsets of A1 is finite. In this way, I can always "rank" a finite series and thus the "final intersection" asked in the question is finite and nonempty. Does this sound right? @AndréNicolas – JasonHu Jan 28 '15 at 22:43

Yes; it actually means that, there is some $N$ such that, for all $n>N$ we have $A_{N}=A_{n}$ - that is, all the sets are the same. This is basically what you're thought that $A_1$ has only finitely many subsets amounts to. The proof of this is basically the same as saying that the sequence of cardinalities $$|A_1|\geq |A_2|\geq |A_3|\geq \ldots$$ will be constant for large enough $N$ - that is, you cannot have an infinite descending sequence of natural numbers. (This is provable by induction, and essentially the idea induction captures in the first place)
Then, you can safely say that, for that $N$, we have $A_{i}\supseteq A_N$ for all $i$. Thus, the intersection of all of the terms of the sequence (which is what the $\infty$ in the intersection is trying to say) must contain $A_N$, since all the terms in the intersection do. However, it is also contained in $A_N$, since $A_N$ is a member of the set over which the intersection is being taken. Thus, it equals $A_N$, which is a finite, non-empty set.