## June 13th, 2021

### On the Baire category theorem and the open mapping theorem

Originally published at 狗和留美者不得入内. You can comment here or there.

Theorem 1 (Baire Category Theorem) Let be a complete metric space.

1. The intersection of a countable collection of open dense sets is dense.
2. 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 [2]) 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 [3] 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.

References

### Using the Minkowski functional to prove separation of sets via a hyperplane

Originally published at 狗和留美者不得入内. You can comment here or there.

We shall use n.v.s to refer to normed vector space.

Definition 1 Let be a n.v.s. A set of the form is called a hyperplane.

Definition 2 We call a set of the form a half-space determined by . Replacing with gives the other half-space.

Proposition 1 Any half-space is a convex set.

Proof: Trivial.

Proposition 2 Let be a n.v.s. such that . For any , the hyperplane is closed iff is continuous.

Proof: open iff is closed. Suppose is continuous and that for , . Take some open neighborhood of not containing . Then, is open and disjoint with .

To prove that is continuous, it suffices to prove that is bounded, by Proposition 2 of [1]. To show that is bounded, showing that suffices, where is the unit ball centered at zero. To do so, it suffices to show that every such that has an open neighborhood of the form such that for any , , since if this holds, then for ,

Thus, .

Assume that is open. Then, every such that is contained by a . Suppose that satisfies . Then, there must exist some such that , a contradiction. This completes our proof.

Definition 3 Let be two subsets of and with . If is such that of its two half-planes, one contains and the other contains , then we say that separates and . We say that strictly separates and if there exists some such that

We now wish to prove that for any open convex set containing and any , there exists a hyperplane that separates and . Let denote the linear functional associated with this separation. We prescribe for some . If we can find some sublinear function that is strictly bounded above on by such that on the subspace satisfies , then we can apply the Hahn-Banach theorem to extend to all in order separate from by a hyperplane.

For this desired sublinear function , we try the following.

Definition 4 Let be a subset of n.v.s . Then the gauge or Minkowski functional with respect to is defined by .

Proposition 3 If contains an open ball centered at , the Minkowski functional satisfies the condition that for all , .

Proof: Trivial and left to the reader.

We want that on , , which is satisfied when we let . We let . That is open means there is some open ball centered at the origin . For any , obviously holds. This is a simple stupid way to uniformly upper bound on .

Proposition 4 For any subset , is bounded above by for . If is open, then for all , .

Proof: Trivially proven.

Proposition 5 If is an open convex set containing , then .

Proof: Since we can scale down arbitrarily by Proposition 3, it suffices to prove this triangle inequality on an open ball centered at that is contained by . Moreover it suffices to prove that for any ,

is satisfied. We notice that

with

Let . Obviously . With this in mind, that is convex implies that for any for any , . We use to set the value of noticing that we can multiply by the denominator to derive the desired inequality. We calculate

We notice that if we replace with , we would obtain the desired inequality. Because is convex must be connected. Assuming that , we have . The connectedness hypothesis implies in fact that all such that must be in . We have that

which then implies for . Thus, we are able to make the aforementioned replacement, which then completes our proof.

Proposition 6 If is an open convex set containing , then its associated Minkowski functional is a subadditive function.

Proof: The two requirement of subadditive were proven in and respectively.

Theorem 1 Let be an open convex subset of a n.v.s . For arbitrary , there exists such that for all . In particular, the hyperplane separates and .

Proof: After a translation, we may always assume that . The Minkowski functional by Proposition 6 is a subadditive function. We define on to be linear. We can extend to on all of by the Hahn-Banach theorem (see [2] for details on it). Since by Proposition 4, for any , , we have for all , . Thus, setting gives us what we want to prove.

In [3], propositions of more general separation of sets by hyperplanes were proven. I shall not write them up here because I believe the foundational ideas behind their proofs have already been explained in the propositions above. I had learned of the Minkowski functional this week but initially did not feel like I grasped the motivation behind it. Such difficulty was resolved after reading 1.2 in [3], the title of which is “The Geometric Forms of the Hahn-Banach Theorem: Separation of Convex Sets”. Interestingly, I read about the Minkowski functional on Wikipedia before writing up [2], in the process of which I gained non-trivial understanding of the Hahn-Banach Theorem, which I had learned in 2017 or 2018 but forgotten in 2021 due to more or less superficial understanding. Certainly, the Hahn-Banach theorem can come across as very formal and abstract at first encounter, and it might not be immediate why it’s so significant. However, the geometric interpretation of it via separation of convex sets gives it more concrete meaning.

References