Answer:
If the intersection is finite the statement is true, but if the intersection is infinite the statement is false.
Step-by-step explanation:
From the statement of the problem I am not sure if the intersection is finite or infinite. Then, I will study both cases.
Let us consider first the finite case: [tex]A = \cap_{i=1}^{n}A_i[/tex]. Because the condition A1 ⊇ A2 ⊇ A3 ⊇ A4 ... we can deduce that the set [tex]A_n[/tex] is a subset of each set [tex]A_i[/tex] with [tex] i\leq n[/tex]. Thus,
[tex]\cap_{i=1}^{n}A_i = A_n[/tex].
Therefore, as [tex]A_n[/tex] is infinite, the intersection is infinite.
Now, if we consider the infinite intersection, i.e. [tex]A = \cap_{k=1}^{\infty}A_k[/tex] the reasoning is slightly different. Take the sets
[tex]A_k = (0,1/k)[/tex] (this is, the open interval between 0 and [tex]1/k[/tex].)
Notice that (0,1) ⊇ (0,1/2) ⊇ (0, 1/3) ⊇(0,1/4) ⊇...So, the hypothesis of the problem are fulfilled. But,
[tex]\cap_{k=1}^{\infty}(0,1/k) = \empyset[/tex]
In order to prove the above statement, choose a real number [tex]x[/tex] between 0 and 1. Notice that, no matter how small [tex]x[/tex] is, there is a natural number [tex]K[/tex] such that [tex]1/K<x[/tex]. Then, the number [tex]x[/tex] is not in any interval [tex](0,1/k)[/tex] with [tex]k>K[/tex]. Therefore, [tex]x[/tex] is not in the set [tex]\cap_{k=1}^{\infty}(0,1/k)[\tex].
