Baire's Theorem and Applications

Dense and Nowhere Dense Sets

Dense Sets in a Topological Space

Definition: Let $(X,\tau )$ be a topological space. The set $A\subseteq X$ is said to be Dense in $X$ if the intersection of every nonempty open set with $A$ is nonempty, that is, $A\cap U\neq \emptyset$ for all $U\in \tau \setminus \{\emptyset \}$.

Given any topological space $(X,\tau )$ it is important to note that $X$ is dense in $X$ because every $U\in \tau$ is such that $U\subseteq X$, and so $X\cap U=U\neq \emptyset$ for all $U\in \tau \setminus \{\emptyset \}$.

For another example, consider the topological space $(\mathbb {R} ,\tau )$ where $\tau$ is the usual topology of open intervals. Then the set of rational numbers $\mathbb {Q} \subset \mathbb {R}$ is dense in $\mathbb {R}$. If not, then there exists an $U\in \tau \setminus \{\emptyset \}$ such that $\mathbb {Q} \cap U=\emptyset$.

Since $U\in \tau$ we have that $(a,b)\subseteq U$ for some open interval $(a,b)$ with $a,b\in \mathbb {R}$ and $a<b$. Suppose that $\mathbb {Q} \setminus U=\emptyset$. Then we must also have that:

${\begin{aligned}\quad \mathbb {Q} \cap U=\mathbb {Q} \cap (a,b)=\emptyset \end{aligned}}$

The intersection above implies that there exists no rational numbers in the interval $(a,b)$, i.e., there exists no $q\in \mathbb {Q}$ such that $a<q<b$. But this is a contradiction since for all $a,b\in \mathbb {R}$ with $a<b$ there ALWAYS exists a rational number $q\in \mathbb {Q}$ such that $a<q<b$, i.e., $q\in (a,b)$. So $\mathbb {Q} \cap (a,b)\neq \emptyset$ for all $U\in \tau \setminus \{\emptyset \}$. Thus, $\mathbb {Q}$ is dense in $\mathbb {R}$.

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

Theorem 1: Let $(X,\tau )$ be a topological space and let $A\subseteq X$. Then $A$ is dense in $X$ if and only if ${\bar {A}}=X$.

• Proof: $\Rightarrow$ Suppose that $A$ is dense in $X$. Then for all $U\in \tau \setminus \{\emptyset \}$ we have that $A\cap U=\emptyset$. Clearly ${\bar {A}}\subseteq X$ so we only need to show that $X\subseteq {\bar {A}}$.

Nowhere Dense Sets in a Topological Space

Definition: Let $(X,\tau )$ be a topological space. A set $A\subseteq X$ is said to be Nowhere Dense in $X$ if the interior of the closure of $A$ is empty, that is, $\mathrm {int} ({\bar {A}})=\emptyset$.

For example, consider the topological space $(\mathbb {R} ,\tau )$ where $\tau$ is the usually topology of open intervals on $\mathbb {R}$, and consider the set of integers $\mathbb {Z}$. The closure of $\mathbb {Z}$, ${\bar {\mathbb {Z} }}$ is the smallest closed set containing $\mathbb {Z}$. The smallest closed set containing $\mathbb {Z}$ is $\mathbb {Z}$ since $\mathbb {Z} ^{c}$ is open as $\mathbb {Z} ^{c}$ is an arbitrary union of open sets:

${\begin{aligned}\quad \mathbb {Z} ^{c}=...(-2,-1)\cup (-1,0)\cup (0,1)\cup (1,2)\cup ...\end{aligned}}$

So what is the interior of ${\bar {\mathbb {Z} }}=\mathbb {Z}$? It is the largest open set contained in ${\bar {\mathbb {Z} }}=\mathbb {Z}$. All open sets of $\mathbb {R}$ with respect to this topology $\tau$ 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 $\mathbb {Z}$ and so:

${\begin{aligned}\quad \mathrm {int} ({\bar {\mathbb {Z} }})=\emptyset \end{aligned}}$

Therefore $\mathbb {Z}$ is a nowhere dense set in $\mathbb {R}$ with respect to the usual topology $\tau$ on $\mathbb {R}$.

The Baire Category Theorem

Lemma 1: Let $(X,\tau )$ be a topological space and let $A\subseteq X$. If $A$ is a nowhere dense set then for every $U\in \tau$ there exists a $B\subseteq U$ such that $A\cap {\bar {B}}=\emptyset$.

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

• Proof: Let $X$ be a complete metric space. Then every Cauchy sequence $(x_{n})_{n=1}^{\infty }$ of elements from $X$ converges in $X$. Suppose that $X$ is of the first category. Then there exists a countable collection of nowhere dense sets $A_{1},A_{2},...\subset X$ such that:

${\begin{aligned}\quad X=\bigcup _{i=1}^{\infty }A_{i}\end{aligned}}$

• Let $U\subset X$. For each nowhere dense set $A_{i}$, $i\in \{1,2,...\}$ there exists a set $B_{i}\subset U$ such that $A_{i}\cap {\bar {B_{i}}}=\emptyset$.

• Let $B_{1}(x_{1},r)\subset U$ be a ball contained in $U$ such that $A_{1}\cap {\bar {B_{1}}}=\emptyset$. Let $B_{2}\left(x_{2},{\frac {r}{2}}\right)\subseteq B_{1}$ be a ball contained in $B_{1}$ whose radius is ${\frac {r}{2}}$ and such that $A_{2}\cap {\bar {B_{2}}}=\emptyset$. Repeat this process. For each $n\in \{2,3,...\}$ let $B_{n}\left(x_{n},{\frac {r}{n}}\right)$ be a ball contained in $B_{n-1}$ whose radius is ${\frac {r}{n}}$ and such that $A_{n}\cap {\bar {B_{n}}}=\emptyset$ and such that $A_{n}\cap {\bar {B_{n}}}=\emptyset$.

• The sequence $(x_{n})_{n=1}^{\infty }$ is Cauchy since as $n\in \mathbb {N}$ gets large, the elements $x_{n}$ are very close. Since $X$ is a complete metric space, we must have that this Cauchy sequence therefore converges to some $p\in X$, i.e., $\lim _{n\to \infty }x_{n}=p$.

• Now notice that $x\in {\bar {B_{n}}}$ for all $n\in \mathbb {N}$ because if not, then there exists an $m\in \mathbb {N}$ such that $x\not \in {\bar {B_{n}}}$ for all $n\geq m$. Hence $({\bar {B_{n}}})^{c}$ is open and so there exists an open ball $X$ such that $x\in B\subset ({\bar {B_{n}}})^{c}$ but then $\lim _{x\to \infty }x_{n}\neq x$ because $x_{m}\not \in B$ for all $m\geq n$.

• Since $x\in {\bar {B_{n}}}$ for all $n\in \mathbb {N}$ then since $A_{n}\cap {\bar {B_{n}}}=\emptyset$ we must have that then $x\not \in A_{n}$

Licensing

Content obtained and/or adapted from:

• The Baire Category Theorem, mathonline.wikidot.com under a CC BY-SA license