Final answer:
Using Cantor's intersection theorem for non-empty closed intervals in a complete metric space, it is shown that the intersection of an infinite sequence of nested intervals is not empty, due to increasing lower bounds and decreasing upper bounds converging to a non-empty limit interval.
Step-by-step explanation:
The question asks to show that the intersection of an infinite sequence of non-empty closed intervals K_i, such that each interval is a subset of or equal to the one before it (K_i ⊃ K_{i+1}), is not empty. This can be proven using the Cantor's intersection theorem, which states that for a sequence of non-empty closed nested intervals in a complete metric space (the real numbers, R, in this case), the intersection of these intervals is not empty.
To prove this, we assume that K_1 is a closed interval [a_1, b_1], with a_1 ≤ b_1. For each i ≥ 1, because K_i is a closed interval itself and K_i ⊃ K_{i+1}, it follows that K_{i+1} will also be a closed interval of the form [a_{i+1}, b_{i+1}], with a_1 ≤ a_{i+1} ≤ b_{i+1} ≤ b_1.
Considering that we have a sequence of increasing lower bounds a_i and a sequence of decreasing upper bounds b_i, and since R is a complete metric space, both sequences have limits, say a and b, respectively. Since a_i ≤ b_i for all i, it follows by the properties of limits that a ≤ b. Therefore, the interval [a, b] is non-empty and lies within the intersection of all K_i. Hence, the intersection ∩i=1∞K_i is not empty, as required.