Theorem 1 (Baire Category Theorem) Let be a complete metric space.
- The intersection of a countable collection of open dense sets is dense.
- is not a countable union of nowhere dense sets.
Proof: Let be our intersection of a (countable) sequence of open dense sets. It suffices to prove that does not contain any nonempty open sets, or equivalently, that any open ball intersects with . By closure under finite intersection, we have that for all , is open. That is an open dense set implies that is open, which means it contains some closed ball of radius . With in mind, we also have that for some closed ball of radius less than . In this fashion we construct a sequence of closed balls for which , which implies that . Cantor’s intersection theorem (Theorem 1 of ) tells us that is non-empty, which means is non-empty, which completes our proof.
A set is nowhere dense iff its complement is dense in . Suppose by contradiction that is a collection of nowhere dense sets that such that . The application of de Morgan’s law tells us that , which the intersection of countably many dense open sets. This is impossible since is obviously not dense in .
Definition 1 If and are topological spaces, then a map is called open if is open in whenever is open in .
Proposition 1 For metric spaces and , is open iff for all any ball centered at , contains a ball centered at .
Proof: Suppose by contradiction that is not open yet any ball centered at is such that contains a ball centered at . In this case, there exists a ball centered at such that contains an element on its boundary, which we shall denote with . Every point has a open ball neighborhood contained in . By our hypothesis, must contain an open ball centered at that is also contained in , which is impossible since .
For the other direction, open implies that for any ball centered at , is open, which means it must contained a balled centered at .
Proposition 2 If and are normed vector spaces and is linear, then is open iff contains a ball centered at when is a ball of radius about .
Proof: A normed vector space is a metric space. Thus by Proposition 1, is open iff for all any ball over radius centered at , contains a ball centered at . We can openly and bijectively map to via . It is easy to see that
contains an open balled centered at iff is open.
Lemma 1 In a normed vector space , for any , .
Proof: Trivial from triangle inequality.
Lemma 2 Let be normed vector spaces. Let be a surjective linear map such that for all , contains a neighborhood of , or equivalently that every ball is contained by for some . If converges to and converges to , then .
Proof: By the argument in the proof of Proposition 2, we can replace in the hypothesis neighborhood of with neighborhood of and closed ball of radius centered at with the one centered at . Suppose by contradiction that . Because every neighborhood of has infinitely many points such that is arbitrarily close to , there exists some open ball centered at such that does not contain any neighborhood of . This open ball necessarily contains for any , . Thus, the hypothesis of the Lemma we are trying to prove has been violated.
Theorem 2 (Open mapping theorem) A surjective linear map between two Banach spaces is necessarily an open map.
Proof: Let denote the ball of radius centered at . By Proposition 2, it suffices to prove that for some , or equivalently, that for some , any such that implies that some satisfies .
That is surjective means that . By the Baire Category Theorem, some is not nowhere dense and thus its closure must contain an open ball. This tells us that must contain an open ball. Every non-zero corresponds to a one-dimensional subspace. The restriction of to that subspace is of course continuous. Thus, . For any , there is a neighborhood of such that . Thus, can not contain any such , which means that . This means that for some , .
We now attempt to show a slightly weaker statement than desired, namely that for some , . Any satisfies . That implies that has a larger diameter than . We already know that . We also observe that is equivalent to by the linearity of .
We want to upper bound a ball centered at . To do so, we note that is equivalent to
We also note that , which means by linearity of , . Thus, that implies that . This tells us that
Combining and gives us
which equates to
which tells us that suffices.
With Theorem 1 of  in mind, we try to construct a sequence such that for all and strict inequality for at least one , which means that with , converges to some such that . This is because if for any , we can construct such a sequence such that , our proof is complete. By linearity of , gives us for all ,
We notice that for any , there exists such that . Then, by , there exists such that . If ,
Let , the distance between and . If , then for such that is in . In this case, we can perturb and accordingly so that . Inductively, we derive that for any for all , there exists such that
Moreover, for each , the finite set of s associated with it contains the finite set of s associated with all such that . In this way, we have a sequence and a sequence of partial sums that converges to some . Via , the sequence , which converges to is induced. By Lemma 2, we have that . This completes our proof.
-  Gerald B. Folland.Real Analysis – Modern Techniques and their Applications. John Wiley & Sons, Inc., 1999.
-  gmachine1729：More basic point-set topology definitions and properties
-  gmachine1729：On normed vector spaces