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\begin{center} \textbf{Bounded Variation and Nowhere Dense Sets.} \end{center}

Definition.  The subset
$M$ of $\mathbb{R}$ is said to be \textit{nowhere dense} means that if
$U$ is a non-empty open set then there is a non-empty open
subset $V$ of $U$ that does not intersect $M$.

[Note: the word ``segment'' can replace the word ``open set'' in the above and the definition is equivalent.]

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Exercises:  Determine which of these sets are nowhere dense:

\qquad (a) The integers.

\qquad (b) A finite set.

\qquad (c) The rational numbers.

\qquad (d) $\{ \frac 1n | n \in \mathbb{N} \}$.

\qquad (e) An open set.

\qquad (f) The boundary of an open set.

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Theorem 4.9.  If $M$ is a nowhere dense  then $\overline{M}$ is nowhere dense.

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Some lemmas:

(i) The set $M$ is nowhere dense if and only if ${\overline{M}}^c$ is a dense open set.

(ii) If the set $M$ is not nowhere dense then there exists an open set $U$ so that if $V \subset U$ is open and non-empty, then $V\cap M \ne \emptyset$.

(iii) If the set $M$ is not nowhere dense then there exists an open set $U$ so that every point of $U$ is a limit point of $M$.

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Theorem 4.10.  If $M$ is is an interval $[a,b]$, then $M$ is not the union of
countable many nowhere dense sets.
[Hint: recall that the monotonic common part of non-empty compact sets is not empty.]

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Definition.  The subset $M$ of the space $X$ is said to be
\textit{perfect} if and only if every point of $M$ is a limit point
of $M$.

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Theorem 4.11.  There exists a closed perfect nowhere dense subset of
the reals.

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Lemma 4.12.  Suppose $f: [a,b] \rightarrow \mathbb{R}$ is an increasing function and $M$ is the set of numbers in the domain of $f$ at which $f$ is discontinuous.  Then $\{f(x) | x \in M \}$ is nowhere dense in $\mathbb{R}$.

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Lemma 4'. Same as lemma 4 except replace ``increasing'' with ``non-decreasing''.

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Theorem 4.13.  Suppose $f: [a,b] \rightarrow \mathbb{R}$ is an increasing function.  Then $f$ is continuous at some point.  Furthermore, if $M = \{ x | f \mbox{ is continuous at } (x, f(x)) \}$ then $M$ is dense in $[a,b]$

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Theorem 4.14.  Suppose $f: [a,b] \rightarrow \mathbb{R}$ is a bounded variation function.  Then $f$ is continuous at some point.  Furthermore, if $M = \{ x | f \mbox{ is continuous at } (x, f(x)) \}$ then $M$ is dense in $[a,b]$.

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