Solve Math Puzzle: Proving Nonemptiness of Intersection

[tauon]

Here's a math problem that's giving me a head ache (though to some of you it might seem to be quite trivial)

r,s\in N^\ast , r+1\leq s ; |A_i|=r, \forall i\in \{1,2,...,s\}
the intersection of any r+1 of sets A_i is nonempty [1]

prove that
\bigcap_{i=1,s} A_i \neq \emptyset [C]
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At first I thought about it like this:

by reductio ad absurdum, suppose
\bigcap_{i=1,s} A_i = \emptyset
we can arbitrarily chose an A_k =\{x_1 , ..., x_r\} then \exists i_1, ... , i_r \in \{1, ... ,s\} so that x_1 \not\in A_{i_1}, ... , x_r \not\in A_{i_r}
\Rightarrow A_k\cap A_{i_1} \cap ... \cap A_{i_r} = \emptyset (dem: if the contrary is true and the intersection is nonempty then at least one element of A_k is in all the sets- contradiction with how the A_{i_j} family of sets was defined)

So there is a family of r+1 sets with an empty interesection- contradiction with [1] therefore [C] is true.

But then I thought: wait a minute. That doesn't prove [C], it only "verifies" [1], since the last A_s might contain no element from \bigcap_{i=1,r+1} A_i so in the end it is possible \bigcap_{i=1,s} A_i = \emptyset for some arbitrary A_k thus constructed.

And now I'm confused and I don't know if I solved the problem right or not... what part am I doing wrong here?

ps: sorry for the messy post, I'm a beginner with the math bbcode.

 

[bpet]

No your original proof looks right - only note that the i1,...,ir are not necessarily distinct but this is easy to fix - shows why r+1<=s is necessary.

 

[tauon]

No your original proof looks right - only note that the i1,...,ir are not necessarily distinct but this is easy to fix

oh, yes. I think you're right.

- shows why r+1<=s is necessary.

what do you mean?

edit: never mind, I understand what you meant now. thanks.