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Let A be arbitrary, and let K be compact. Then the intersection A ∩ K is compact.

A. True
B. False

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Final answer:

The intersection of an arbitrary set and a compact set is always compact in mathematics.

Step-by-step explanation:

True

In mathematics, the intersection of two compact sets is always compact. Therefore, given an arbitrary set A and a compact set K, the intersection A ∩ K is always compact.

Proof:

  1. Let A and K be two sets.
  2. Assume A is arbitrary and K is compact.
  3. Let C be an open cover of A ∩ K.
  4. Since K is compact, we can find a finite subcover D of K from C.
  5. Now, for each element in D, there exists an element in C that contains it.
  6. Therefore, D is a finite subcover of A ∩ K.
  7. Hence, A ∩ K is compact.

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