Skip to main content
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

# 6.8: Baire Categories. More on Linear Maps

$$\newcommand{\vecs}{\overset { \rightharpoonup} {\mathbf{#1}} }$$ $$\newcommand{\vecd}{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}{\| #1 \|}$$ $$\newcommand{\inner}{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$

We pause to outline the theory of so-called sets of Category I or Category II, as introduced by Baire. It is one of the most powerful tools in higher analysis. Below, $$(S, \rho)$$ is a metric space.

Definition 1

A set $$A \subseteq(S, \rho)$$ is said to be nowhere dense (in $$S)$$ iff its closure $$\overline{A}$$ has no interior points (i.e., contains no globes): $$(\overline{A})^{0}=\emptyset$$.

Equivalently, the set $$A$$ is nowhere dense iff every open set $$G^{*} \neq \emptyset$$ in $$S$$ contains a globe $$\overline{G}$$ disjoint from $$A.$$ (Why?)

Definition 2

A set $$A \subseteq(S, \rho)$$ is meagre, or of Category I (in $$S),$$ iff

$A=\bigcup_{n=1}^{\infty} A_{n},$

for some sequence of nowhere dense sets $$A_{n}$$.

Otherwise, $$A$$ is said to be nonmeagre or of Category II.

$$A$$ is residual iff $$-A$$ is meagre, but $$A$$ is not.

Examples

(a) $$\emptyset$$ is nowhere dense.

(b) Any finite set in a normed space $$E$$ is nowhere dense.

(c) The set $$N$$ of all naturals in $$E^{1}$$ is nowhere dense.

(d) So also is Cantor's set $$P$$ (Problem 17 in Chapter 3, §14); indeed, $$P$$ is closed $$(P=\overline{P})$$ and has no interior points (verify!), so $$(\overline{P})^{0}=P^{0}=\emptyset$$.

(e) The set $$R$$ of all rationals in $$E^{1}$$ is meagre; for it is countable (see Chapter 1, §9), hence a countable union of nowhere dense singletons {$$r_{n}$$}, $$r_{n} \in R.$$ But $$R$$ is not nowhere dense; it is even dense in $$E^{1},$$ since $$\overline{R}=E^{1}$$ (see Definition 2, in Chapter 3, §14). Thus a meagre set need not be nowhere dense. (But all nowhere dense sets are meagre why?)

Examples (c) and (d) show that a nowhere dense set may be infinite (even uncountable). Yet, sometimes nowhere dense sets are treated as "small" or "negligible," in comparison with other sets. Most important is the following theorem.

Theorem $$\PageIndex{1}$$ (Baire)

In a complete metric space $$(S, \rho),$$ every open set $$G^{*} \neq \emptyset$$ is nonmeagre. Hence the entire space $$S$$ is residual.

Proof

Seeking a contradiction, suppose $$G^{*}$$ is meagre, i.e.,

$G^{*}=\bigcup_{n=1}^{\infty} A_{n}$

for some nowhere dense sets $$A_{n}.$$ Now, as $$A_{1}$$ is nowhere dense, $$G^{*}$$ contains a closed globe

$\overline{G}_{1}=\overline{G_{x_{1}}\left(\delta_{1}\right)} \subseteq-A_{1}.$

Again, as $$A_{2}$$ is nowhere dense, $$G_{1}$$ contains a globe

$\overline{G}_{2}=\overline{G_{x_{2}}\left(\delta_{2}\right)} \subseteq-A_{2}, \quad \text { with } 0<\delta_{2} \leq \frac{1}{2} \delta_{1}.$

By induction, we obtain a contracting sequence of closed globes

$\overline{G}_{n}=\overline{G_{x_{n}}\left(\delta_{n}\right)}, \quad \text { with } 0<\delta_{n} \leq \frac{1}{2^{n}} \delta_{1} \rightarrow 0.$

As $$S$$ is complete, so are the $$\overline{G}_{n}$$ (Theorem 5 in Chapter 3, §17). Thus, by Cantor's theorem (Theorem 5 of Chapter 4, §6), there is

$p \in \bigcap_{n=1}^{\infty} \overline{G}_{n}.$

As $$G^{*} \supseteq \overline{G}_{n},$$ we have $$p \in G^{*}.$$ But, as $$\overline{G}_{n} \subseteq-A_{n},$$ we also have $$(\forall n) p \notin A_{n}$$ ; hence

$p \notin \bigcup_{n=1}^{\infty} A_{n}=G^{*}$

(the desired contradiction!).$$\quad \square$$

We shall need a lemma based on Problems 15 and 19 in §7. (Review them!)

lemma

Let $$f \in L\left(E^{\prime}, E\right), E^{\prime}$$ complete. Let $$G=G_{0}(1)$$ be the unit globe in $$E^{\prime}.$$ If $$\overline{f[G]}$$ (closure of $$f[G]$$ in $$E$$ ) contains a globe $$G_{0}=G_{0}(r) \subset E,$$ then $$G_{0} \subseteq f[G].$$

Note. Recall that we "arrow" only vectors from $$E^{\prime}$$ (e.g., $$\overrightarrow{0}),$$ but not those from $$E$$ (e.g., 0).

Proof

Let $$A=f[G] \cap G_{0} \subseteq G_{0}.$$ We claim that $$A$$ is dense in $$G_{0};$$ i.e., $$G_{0} \subseteq \overline{A}.$$ Indeed, by assumption, any $$q \in G_{0}$$ is in $$f[G].$$ Thus by Theorem 3 in Chapter 3, §16, any $$G_{q}$$ meets $$f[G] \cap G_{0}=A$$ if $$q \in G_{0}.$$ Hence

$\left(\forall q \in G_{0}\right) \quad q \in \overline{A},$

i.e., $$G_{0} \subseteq \overline{A},$$ as claimed.

Now fix any $$q_{0} \in G_{0}=G_{0}(r)$$ and a real $$c(0<c<1).$$ As $$A$$ is dense in $$G_{0}$$,

$A \cap G_{q_{0}}(c r) \neq \emptyset;$

so let $$q_{1} \in A \cap G_{q_{0}}(c r) \subseteq f[G].$$ Then

$\left|q_{1}-q_{0}\right|<c r, \quad q_{0} \in G_{q_{1}}(c r).$

As $$q_{1} \in f[G],$$ we can fix some $$\vec{p}_{1} \in G=G_{0}(1),$$ with $$f\left(\vec{p}_{1}\right)=q_{1}.$$ Also, by Problems 19(iv) and 15(iii) in §7, $$c A+q_{1}$$ is dense in $$c G_{0}+q_{1}=G_{q_{1}}(c r)$$. But $$q_{0} \in G_{q_{1}}(cr)$$. Thus

$G_{q_{0}}\left(c^{2} r\right) \cap\left(c A+q_{1}\right) \neq \emptyset;$

so let $$q_{2} \in G_{q_{0}}\left(c^{2} r\right) \cap\left(c A+q_{1}\right),$$ so $$q_{0} \in G_{q_{2}}\left(c^{2} r\right),$$ etc.

Inductively, we fix for each $$n>1$$ some $$q_{n} \in G_{q_{0}}\left(c^{n} r\right),$$ with

$q_{n} \in c^{n-1} A+q_{n-1},$

i.e.,

$q_{n}-q_{n-1} \in c^{n-1} A.$

As $$A \subseteq f\left[G_{0}(1)\right],$$ linearity yields

$q_{n}-q_{n-1} \in f\left[c^{n-1} G_{0}(1)\right]=f\left[G_{0}\left(c^{n-1}\right)\right], \quad n>1.$

Thus for each $$n>1,$$ there is $$\vec{p}_{n} \in G_{0}(c^{n-1})$$, (i.e., $$|\vec{p}_{n}|<c^{n-1})$$ such that $$f(\vec{p}_{n})=q_{n}-q_{n-1}.$$ Now, as $$|\vec{p}_{n}|<c^{n-1}$$ and $$0<c<1$$,

$\sum_{1}^{\infty}\left|\vec{p}_{n}\right|<+\infty;$

so by the completeness of $$E^{\prime}, \sum \vec{p}_{n}$$ converges in $$E^{\prime}$$ (Theorem 1 in Chapter 4, §13). Let $$\vec{p}=\sum_{k=1}^{\infty} \vec{p}_{k};$$ then

\begin{aligned} f(\vec{p}) &=f\left(\lim _{n \rightarrow \infty} \sum_{k=1}^{n} \vec{p}_{k}\right)=\lim _{n \rightarrow \infty} f\left(\sum_{k=1}^{n} \vec{p}_{k}\right) \\ &=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} f\left(\vec{p}_{k}\right) \text { for } f \in L\left(E^{\prime}, E\right). \end{aligned}

But $$f(\vec{p}_{k})=q_{k}-q_{k-1}(k>1),$$ and $$f(\vec{p}_{1})=q_{1};$$ so

$\sum_{k=1}^{n} f\left(\vec{p}_{k}\right)=q_{1}+\sum_{k=2}^{n}\left(q_{k}-q_{k-1}\right)=q_{n}.$

Thus

$f(\vec{p})=\lim _{n \rightarrow \infty} \sum_{k=1}^{n} f\left(\vec{p}_{k}\right)=\lim _{n \rightarrow \infty} q_{n}=q_{0}.$

Moreover, $$\left|\vec{p}_{k}\right|<c^{k-1}(k \geq 1).$$ Thus

$|\vec{p}| \leq \sum_{k=1}^{\infty}\left|\vec{p}_{k}\right|<\sum_{k=1}^{\infty} c^{k-1}=\frac{1}{1-c};$

i.e.,

$\vec{p} \in G_{\overrightarrow{0}}\left(\frac{1}{1-c}\right).$

But $$q_{0}=f(\vec{p});$$ so

$q_{0} \in f\left[G_{\overrightarrow{0}}\left(\frac{1}{1-c}\right)\right].$

As $$q_{0} \in G_{0}(r)$$ was arbitrary, we have

$G_{0}(r) \subseteq f\left[G_{0}\left(\frac{1}{1-c}\right)\right],$

or by linearity,

$G_{0}(r(1-c)) \subseteq f\left[G_{0}(1)\right]=f[G].$

This holds for any $$c \in(0,1).$$ Hence

$f[G] \supseteq \bigcup_{0<c<1} G_{0}(r(1-c))=G_{0}(r). \quad \text {(Verify!)}$

Thus all is proved.$$\quad \square$$

We can now establish an important result due to S. Banach.

Theorem $$\PageIndex{2}$$ (Banach)

Let $$f \in L\left(E^{\prime}, E\right),$$ with $$E^{\prime}$$ complete. Then $$f\left[E^{\prime}\right]$$ is meagre in $$E$$ or $$f\left[E^{\prime}\right]=E,$$ according to whether $$f\left[G_{0}(1)\right]$$ is or is not nowhere dense.

Proof

If $$f\left[G_{0}(1)\right]$$ is nowhere dense in $$E,$$ so also is $$f\left[G_{0}(n)\right], n>0.$$ (Verify by Problems 15 and 19 in §7.) But then

$f\left[E^{\prime}\right]=f\left[\bigcup_{n=1}^{\infty} G_{0}(n)\right]=\bigcup_{n=1}^{\infty} f\left[G_{\overrightarrow{0}}(n)\right]$

is a countable union of nowhere dense sets, hence meagre, by definition.

Now suppose $$f\left[G_{0}(1)\right]$$ is not nowhere dense; so $$\overline{f\left[G_{0}(1)\right]}$$ contains some $$G_{q}(r) \subseteq E.$$ We may assume $$q \in f\left[G_{\overrightarrow{0}}(1)\right]$$ (if not, replace $$q$$ by a close point from $$f\left[G_{0}(1)\right]).$$ Then $$q=f(\vec{p})$$ for some $$\vec{p} \in G_{0}(1).$$ The latter implies

$|-\vec{p}|=|\vec{p}|=\rho(\vec{p}, \overrightarrow{0})<1;$

so

$G_{-\vec{p}}(1) \subseteq G_{\overrightarrow{0}}(2).$

Also, as $$\overline{f\left[G_{\overline{0}}(1)\right]} \supseteq G_{q}(r),$$ translation by $$-q=f(-\vec{p})$$ yields

$\overline{f\left[G_{\overline{0}}(1)\right]}+f(-\vec{p}) \supseteq G_{q}(r)-q=G_{0}(r),$

i.e.,

$G_{0}(r) \subseteq \overline{f\left[G_{-\vec{p}}(1)\right]} \subseteq \overline{f\left[G_{\overrightarrow{0}}(2)\right]}.$

Hence $$\overline{f\left[G_{\overrightarrow{0}}(1)\right]} \supseteq G_{0}\left(\frac{1}{2} r\right)$$ (why?); so, by the Lemma

$f\left[G_{\overrightarrow{0}}(1)\right] \supseteq G_{0}\left(\frac{1}{2} r\right) \text { in } E.$

This implies $$f\left[G_{\overrightarrow{0}}(2 n)\right] \supseteq G_{0}(n r),$$ and so

$f\left[E^{\prime}\right] \supseteq \bigcup_{n=1}^{\infty} G_{0}(n r)=E,$

i.e., $$f\left[E^{\prime}\right]=E,$$ as required. Thus the theorem is proved.$$\quad \square$$

Theorem $$\PageIndex{3}$$ (Open map principle)

Let $$f \in L\left(E^{\prime}, E\right),$$ with $$E^{\prime}$$ and $$E$$ complete. Then the map $$f$$ is open on $$E^{\prime}$$ iff $$f\left[E^{\prime}\right]=E,$$ i.e., iff $$f$$ is onto $$E$$.

Proof

If $$f\left[E^{\prime}\right]=E,$$ then by Theorem 1, $$f\left[E^{\prime}\right]$$ is nonmeagre in $$E,$$ as is $$E$$ itself. Thus by Theorem 2, $$f\left[G_{\overrightarrow{0}}(1)\right]$$ is not nowhere dense, and (1) follows as before. Hence by Problems 15(iii) and 19 in §7, $$f\left[G_{\vec{p}}\right] \supseteq$$ some $$G_{q}$$ whenever $$q=f(\vec{p})$$. (Why?) Therefore, $$G_{\vec{p}} \subseteq A \subseteq E^{\prime}$$ implies

$G_{f(\vec{p})} \subseteq f\left[G_{\vec{p}}\right] \subseteq f[A];$

i.e., $$f$$ maps any interior point $$\vec{p} \in A$$ into such a point of $$f[A].$$ By Problem 8 in §7, $$f$$ is open on $$E^{\prime}.$$

Conversely, if so, then $$f\left[E^{\prime}\right]$$ is an open set $$\neq \emptyset$$ in $$E,$$ a complete space; so by Theorems 1 and 2, $$f\left[E^{\prime}\right]$$ is nonmeagre and equals $$E.$$ (See also Problem 16(ii) in §7.)$$\quad \square$$

Note 1. Theorem 3 holds even if $$f$$ is not one-to-one.

Note 2. If in Theorem 3, however, $$f$$ is bijective, it is open on $$E^{\prime},$$ and so $$f^{-1} \in L\left(E, E^{\prime}\right)$$ by Note 1 in §7. (This is the promised general proof of Corollary 2 in §6.)

Theorem $$\PageIndex{4}$$ (Banach-Steinhaus uniform boundedness principle)

Let $$E^{\prime} b e$$ complete. Let $$\mathcal{N}$$ be a family of maps $$f \in L\left(E^{\prime}, E\right)$$ such that

$\left(\forall x \in E^{\prime}\right)\left(\exists k \in E^{1}\right)(\forall f \in \mathcal{N}) \quad|f(\vec{x})|<k.$

("$$\mathcal{N}$$ is bounded at each $$\vec{x}$$.")

Then $$\mathcal{N}$$ is "norm-bounded," i.e.,

$\left(\exists K \in E^{1}\right)(\forall f \in \mathcal{N}) \quad\|f\|<K,$

with $$\|\quad\|$$ as in §2.

Proof

It suffices to show that $$\mathcal{N}$$ is "uniformly" bounded on some globe,

$\left(\exists c \in E^{1}\right)\left(\exists G=G_{\vec{p}}(r)\right)(\forall f \in \mathcal{N})(\forall \vec{x} \in G) \quad|f(\vec{x})| \leq c.$

For then $$|\vec{x}-\vec{p}| \leq r$$ implies

$2 c>|f(\vec{x})-f(\vec{p})|=|f(\vec{x}-\vec{p})|,$

or (setting $$\vec{x}-\vec{p}=r \vec{y} )|\vec{y}|<1$$ implies

$(\forall f \in \mathcal{N}) \quad|f(\vec{y})|<\frac{2 c}{r} \quad\text {(why?);}$

so

$(\forall f \in \mathcal{N}) \quad\|f\|=\sup _{|\vec{y}| \leq 1}|f(\vec{y})|<\frac{2 c}{r}.$

Thus, seeking a contradiction, suppose (3) fails and assume its negation:

$\left(\forall c \in E^{1}\right)\left(\forall G=G_{\vec{p}}(r)\right)(\exists f \in \mathcal{N})\left(\exists \vec{x} \in G=G_{\vec{p}}(r)\right) \quad|f(\vec{x})|>c.$

Then for $$c=1,$$ we can fix some $$f_{1} \in \mathcal{N}$$ and $$G_{\vec{x}_{1}}\left(r_{1}\right)$$ such that $$0<r_{1}<1$$ and

$\left|f_{1}\left(\vec{x}_{1}\right)\right|>1.$

By the continuity of the norm $$| |$$ , we can choose $$r_{1}$$ so small that

$\left(\forall \vec{x} \in \overline{G_{\vec{x}_{1}}\left(r_{1}\right)}\right) \quad|f(\vec{x})|>1.$

Again by (4), we fix $$f_{2} \in \mathcal{N}$$ and $$\vec{x}_{2} \in G_{\vec{x}_{1}}\left(r_{1}\right)$$ such that $$\left|f_{2}\right|>2$$ on some globe

$\overline{G_{\vec{x}_{2}}\left(r_{2}\right)} \subseteq G_{\vec{x}_{1}}\left(r_{1}\right),$

with $$0<r_{2}<1 / 2.$$ Inductively, we thus form a contracting sequence of closed globes

$\overline{G_{\vec{x}_{n}}\left(r_{n}\right)}, \quad 0<r_{n}<\frac{1}{n},$

and a sequence $$\left\{f_{n}\right\} \subseteq \mathcal{N},$$ such that

$(\forall n) \quad\left|f_{n}\right|>n \text { on } \overline{G_{\vec{x}_{n}}\left(r_{n}\right)} \subseteq E^{\prime}.$

As $$E^{\prime}$$ is complete, so are the closed globes $$\overline{G_{\vec{x}_{n}}\left(r_{n}\right)} \subseteq E^{\prime}.$$ Also, $$0<r_{n}<$$ 1$$/ n \rightarrow 0.$$ Thus by Cantor's theorem (Theorem 5 of Chapter 4, §6), there is

$\vec{x}_{0} \in \bigcap_{n=1}^{\infty} \overline{G_{\vec{x}_{n}}\left(r_{n}\right)}.$

As $$\vec{x}_{0}$$ is in each $$\overline{G_{\vec{x}_{n}}\left(r_{n}\right)},$$ we have

$(\forall n) \quad\left|f_{n}\left(\vec{x}_{0}\right)\right|>n;$

so $$\mathcal{N}$$ is not bounded at $$\vec{x}_{0},$$ contrary to (2). This contradiction completes the proof.$$\quad \square$$

Note 3. Complete normed spaces are also called Banach spaces.