Baire's Theorem and Applications

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Dense and Nowhere Dense Sets

Dense Sets in a Topological Space

Definition: Let be a topological space. The set is said to be Dense in if the intersection of every nonempty open set with is nonempty, that is, for all .

Given any topological space it is important to note that is dense in because every is such that , and so for all .

For another example, consider the topological space where is the usual topology of open intervals. Then the set of rational numbers is dense in . If not, then there exists an such that .

Since we have that for some open interval with and . Suppose that . Then we must also have that:

The intersection above implies that there exists no rational numbers in the interval , i.e., there exists no such that . But this is a contradiction since for all with there ALWAYS exists a rational number such that , i.e., . So for all . Thus, is dense in .

We will now look at a very important theorem which will give us a way to determine whether a set is dense in or not.

Theorem 1: Let be a topological space and let . Then is dense in if and only if .

  • Proof: Suppose that is dense in . Then for all we have that . Clearly so we only need to show that .

Nowhere Dense Sets in a Topological Space

Definition: Let be a topological space. A set is said to be Nowhere Dense in if the interior of the closure of is empty, that is, .

For example, consider the topological space where is the usually topology of open intervals on , and consider the set of integers . The closure of , is the smallest closed set containing . The smallest closed set containing is since is open as is an arbitrary union of open sets:

So what is the interior of ? It is the largest open set contained in . All open sets of with respect to this topology are either the empty set, an open interval, a union of open intervals, or the whole set (the union of all open intervals). But no open intervals are contained in and so:

Therefore is a nowhere dense set in with respect to the usual topology on .


The Baire Category Theorem

Lemma 1: Let be a topological space and let . If is a nowhere dense set then for every there exists a such that .

Theorem 1 (The Baire Category Theorem): Every complete metric space is of the second category.

  • Proof: Let be a complete metric space. Then every Cauchy sequence of elements from converges in . Suppose that is of the first category. Then there exists a countable collection of nowhere dense sets such that:
  • Let . For each nowhere dense set , there exists a set such that .
  • Let be a ball contained in such that . Let be a ball contained in whose radius is and such that . Repeat this process. For each let be a ball contained in whose radius is and such that and such that .
  • The sequence is Cauchy since as gets large, the elements are very close. Since is a complete metric space, we must have that this Cauchy sequence therefore converges to some , i.e., .
  • Now notice that for all because if not, then there exists an such that for all . Hence is open and so there exists an open ball such that but then because for all .
  • Since for all then since we must have that then

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