Sheng Li (gmachine1729) wrote,
Sheng Li
gmachine1729

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 [公式] be a complete metric space.

  1. The intersection of a countable collection of open dense sets is dense.
  2. [公式] is not a countable union of nowhere dense sets.

Proof: Let [公式] be our intersection of a (countable) sequence of open dense sets. It suffices to prove that [公式] does not contain any nonempty open sets, or equivalently, that any open ball [公式] intersects with [公式] . By closure under finite intersection, we have that for all [公式] , [公式] is open. That [公式] is an open dense set implies that [公式] is open, which means it contains some closed ball [公式] of radius [公式] . With [公式] in mind, we also have that [公式] for some closed ball [公式] of radius less than [公式] . In this fashion we construct a sequence of closed balls [公式] for which [公式] , which implies that [公式] . Cantor’s intersection theorem (Theorem 1 of [2]) tells us that [公式] is non-empty, which means [公式] is non-empty, which completes our proof.

A set is nowhere dense iff its complement is dense in [公式] . Suppose by contradiction that [公式] is a collection of nowhere dense sets that such that [公式] . The application of de Morgan’s law tells us that [公式] , which the intersection of countably many dense open sets. This is impossible since [公式] is obviously not dense in [公式] . [公式]

Definition 1 If [公式] and [公式] are topological spaces, then a map [公式] is called open if [公式] is open in [公式] whenever [公式] is open in [公式] .

Proposition 1 For metric spaces [公式] and [公式] , [公式] is open iff for all any ball [公式] centered at [公式] , [公式] contains a ball centered at [公式] .

Proof: Suppose by contradiction that [公式] is not open yet any ball [公式] centered at [公式] is such that [公式] contains a ball centered at [公式] . In this case, there exists a ball [公式] centered at [公式] such that [公式] contains an element on its boundary, which we shall denote with [公式] . Every point [公式] has a open ball neighborhood [公式] contained in [公式]. By our hypothesis, [公式] must contain an open ball centered at [公式] that is also contained in [公式] , which is impossible since [公式] .

For the other direction, [公式] open implies that for any ball [公式] centered at [公式] , [公式] is open, which means it must contained a balled centered at [公式] . [公式]

Proposition 2 If [公式] and [公式] are normed vector spaces and [公式] is linear, then [公式] is open iff [公式] contains a ball centered at [公式] when [公式] is a ball of radius [公式] about [公式] .

Proof: A normed vector space is a metric space. Thus by Proposition 1, [公式] is open iff for all any ball [公式] over radius [公式] centered at [公式] , [公式] contains a ball centered at [公式] . We can openly and bijectively map [公式] to [公式] via [公式] . It is easy to see that

[公式]

contains an open balled centered at [公式] iff [公式] is open. [公式]

Lemma 1 In a normed vector space [公式] , for any [公式] , [公式] .

Proof: Trivial from triangle inequality. [公式]

Lemma 2 Let [公式] be normed vector spaces. Let [公式] be a surjective linear map such that for all [公式] , [公式] contains a neighborhood of [公式] , or equivalently that every ball [公式] is contained by [公式] for some [公式] . If [公式] converges to [公式] and [公式] converges to [公式] , then [公式] .

Proof: By the argument in the proof of Proposition 2, we can replace in the hypothesis neighborhood of [公式] with neighborhood of [公式] and closed ball of radius [公式] centered at [公式] with the one centered at [公式] . Suppose by contradiction that [公式] . Because every neighborhood of [公式] has infinitely many points [公式] such that [公式] is arbitrarily close to [公式] , there exists some open ball [公式] centered at [公式] such that [公式] does not contain any neighborhood of [公式] . This open ball necessarily contains for any [公式] , [公式] . Thus, the hypothesis of the Lemma we are trying to prove has been violated. [公式]

Theorem 2 (Open mapping theorem) A surjective linear map [公式] between two Banach spaces is necessarily an open map.

Proof: Let [公式] denote the ball of radius [公式] centered at [公式] . By Proposition 2, it suffices to prove that [公式] for some [公式] , or equivalently, that for some [公式] , any [公式] such that [公式] implies that some [公式] satisfies [公式] .

That [公式] is surjective means that [公式] . By the Baire Category Theorem, some [公式] is not nowhere dense and thus its closure must contain an open ball. This tells us that [公式] must contain an open ball. Every non-zero [公式] corresponds to a one-dimensional subspace. The restriction of [公式] to that subspace is of course continuous. Thus, [公式] . For any [公式] , there is a neighborhood [公式] of [公式] such that [公式] . Thus, [公式] can not contain any such [公式] , which means that [公式] . This means that for some [公式] , [公式] .

We now attempt to show a slightly weaker statement than desired, namely that for some [公式] , [公式] . Any [公式] satisfies [公式] . That [公式] implies that [公式] has a larger diameter than [公式] . We already know that [公式] . We also observe that [公式] is equivalent to [公式] by the linearity of [公式] .

We want to upper bound a ball centered at [公式] . To do so, we note that [公式] is equivalent to

[公式]

We also note that [公式] , which means by linearity of [公式] , [公式] . Thus, that [公式] implies that [公式] . This tells us that

[公式]

Combining [公式] and [公式] gives us

[公式]

which equates to

[公式]

which tells us that [公式] suffices.

With Theorem 1 of [3] in mind, we try to construct a sequence [公式] such that [公式] for all [公式] and strict inequality for at least one [公式] , which means that with [公式] , [公式] converges to some [公式] such that [公式] . This is because if for any [公式] , we can construct such a sequence such that [公式] , our proof is complete. By linearity of [公式] , [公式] gives us for all [公式] ,

[公式]

We notice that for any [公式] , there exists [公式] such that [公式] . Then, by [公式] , there exists [公式] such that [公式] . If [公式],

Let [公式] , the distance between [公式] and [公式] . If [公式] , then [公式] for [公式] such that [公式] is in [公式] . In this case, we can perturb [公式] and [公式] accordingly so that [公式] . Inductively, we derive that for any [公式] for all [公式] , there exists [公式] such that

[公式]

Moreover, for each [公式] , the finite set of [公式] s associated with it contains the finite set of [公式] s associated with all [公式] such that [公式] . In this way, we have a sequence [公式] and a sequence of partial sums [公式] that converges to some [公式] . Via [公式] , the sequence [公式] , which converges to [公式] is induced. By Lemma 2, we have that [公式] . This completes our proof. [公式]

References

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