Final Answer:
The intersection of an arbitrary collection of closed sets is closed.
Step-by-step explanation:
When considering the intersection of closed sets, each set's complement is open by definition. For any set, its complement being open implies that the set itself is closed. Now, when multiple closed sets are intersected, the resulting set consists of elements present in all of these closed sets. Consequently, the complement of this intersection set comprises elements that are not in any of the original sets.
Since each original set's complement is open, the absence of these elements in the intersection set's complement implies that the intersection set is closed. Therefore, irrespective of the number of sets being intersected, the resulting intersection remains closed.
In simpler terms, if you have several rooms (closed sets) and you take the intersection of all these rooms, whatever remains in this common space must be closed off from the outside, preserving the characteristic of closedness.
This holds true even when dealing with an arbitrary number of rooms (sets), validating that their shared intersection remains a closed set. The definition of closed sets, which states that their complements are open, is the key here. Consequently, the intersection of closed sets retains closure due to the nature of open complements and the properties of set intersections.
This explanation demonstrates that the intersection of closed sets preserves the closed property due to the complement's openness, ensuring that the resulting set maintains closure.