Ref: “Introductory Real Analysis” by Kolmogorov, Fomin.

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Def [Nowhere dense sets]
Let ${ (X, d) }$ be a metric space. Let ${ A \subseteq X . }$
We say ${ A }$ is nowhere dense if for every open ball ${ B }$ the set ${ A \cap B }$ is not dense in ${ B . }$

Let ${ (X, d) }$ be a complete metric space. How large can (countable) unions of nowhere dense sets be?

Thm [Baire]
Let ${ (X, d) }$ be a complete metric space. Then ${ X }$ cannot be represented as a countable union of nowhere dense sets.

Pf: Suppose to the contrary

\[{ X = \bigcup _{n=1} ^{\infty} A _n }\]

where each ${ A _n }$ is nowhere dense.

Consider a closed ball ${ \mathbf{B} _0 }$ of radius ${ 1 . }$

Since ${ A _1 }$ is nowhere dense, there is a closed ball ${ \mathbf{B} _1 }$ of radius ${ < \frac{1}{2} }$ such that

\[{ \mathbf{B} _1 \subseteq \mathbf{B} _0, \quad A _1 \cap \mathbf{B} _1 = \emptyset . }\]

Since ${ A _2 }$ is nowhere dense, there is a closed ball ${ \mathbf{B} _2 }$ of radius ${ < \frac{1}{3} }$ such that

\[{ \mathbf{B} _2 \subseteq \mathbf{B} _1, \quad A _2 \cap \mathbf{B} _2 = \emptyset. }\]

And so on.

Note that the sequence of centers of

\[{ \mathbf{B} _0 \supseteq \mathbf{B} _1 \supseteq \mathbf{B} _2 \supseteq \ldots }\]

is a Cauchy sequence. Hence the sequence converges to a point ${ x . }$ Note that

\[{ x \in \bigcap _{n=1} ^{\infty} \mathbf{B} _n . }\]

Hence ${ x }$ is not in any ${ A _n , }$ a contradiction.

Hence ${ X }$ cannot be represented as a countable union of nowhere dense sets, as needed. ${ \blacksquare }$

Can we characterize nowhere dense sets ${ A }$ in terms of operations on ${ A }$?

Obs [Nowhere dense sets]
Let ${ (X, d) }$ be a metric space. Let ${ A \subseteq X . }$
Then ${ A }$ is nowhere dense if and only if ${ \text{int}(\overline{A}) = \emptyset . }$

Pf: Note that ${ A }$ is nowhere dense if and only if for every open ball ${ B }$ there is a point ${ x \in B , }$ ${ x \not \in \overline{A} . }$ Hence ${ A }$ is nowhere dense if and only if ${ \overline{A} ^c }$ is dense. Note that ${ \overline{A} ^c }$ is dense if and only if every open ball intersects ${ \overline{A} ^c , }$ if and only if there is no open ball contained in ${ \overline{A}, }$ if and only if ${ \text{int}(\overline{A}) = \emptyset. }$
Hence ${ A }$ is nowhere dense if and only if ${ \text{int}(\overline{A}) = \emptyset , }$ as needed. ${ \blacksquare }$