# Open Sets in Real Number Line

## Theorem

Every non-empty open set $I \subseteq \R$ can be expressed as a countable union of pairwise disjoint open intervals.

If:

- $\displaystyle I = \bigcup_{n \mathop \in \N} J_n$
- $\displaystyle I = \bigcup_{n \mathop \in \N} K_n$

are two such expressions, then there exists a permutation $\sigma$ of $\N$ such that:

- $\forall n \in \N: J_n = K_{\map \sigma n}$

## Proof

We know that $\R$ is a complete metric space.

Let $x \in I$.

By the definition of open sets in a metric space, there is an open interval $I_x$, included in $I$, that contains $x$.

Define the following:

\(\text {(1)}: \quad\) | \(\displaystyle \map a x\) | \(=\) | \(\displaystyle \inf \set {z: \openint z x \subseteq I}\) | ||||||||||

\(\text {(2)}: \quad\) | \(\displaystyle \map b x\) | \(=\) | \(\displaystyle \sup \set {y: \openint x y \subseteq I}\) |

The existence of $I_x$ assures us that we have $\map a x < x < \map b x$.

In this way we associate a non-empty open interval:

- $\map J x = \openint {\map a x} {\map b x}$

to $x \in I$.

Let $\epsilon > 0$ be arbitrary.

As $\map a x$ is an infimum, there exists $z < \map a x + \epsilon$ such that:

- $\openint z x \subseteq I$

Thus certainly:

- $\openint {\map a x + \epsilon} x \subseteq I$

It follows that we have:

- $\displaystyle \openint {\map a x} x = \bigcup_{n \mathop = 1}^\infty \openint {\map a x + \frac 1 n} x \subseteq I$

By a similar argument:

- $\openint x {\map b x} \subseteq I$

Therefore, as we have:

- $\map J x = \openint {\map a x} x \cup \set x \cup \openint x {\map b x}$

we conclude that:

- $\map J x \subseteq I$

Suppose now that we have $\map a x = -\infty$ or $\map b x = +\infty$.

If both are the case, it must be that $I = \R$, and hence $I$ is open.

It can then also be written uniquely as the pairwise disjoint union of the single open interval $\R$.

In other cases, we observe that intervals of the type:

- $\map {K_-} a = \openint {-\infty} a$

or:

- $\map {K_+} a = \openint a {+\infty}$

are open.

Assume $I$ contains such an interval $\map {K_\pm} a$ with $a \notin I$.

Then we have:

- $\displaystyle I = \paren {I \cap \map {K_-} a} \cup \paren {I \cap \map {K_+} a}$

Thus $I$ is open if and only if $I \setminus \map {K_\pm} a$ is open, since the $\map {K_\pm} a$ are disjoint, open sets.

It is also clear that a unique decomposition of $I \setminus \map {K_\pm} a$ into disjoint open intervals will give rise to such a decomposition for $I$.

Therefore, without loss of generality, we assume $I$ contains no such interval.

Define a relation $\sim$ on $I$ by $x \sim y$ if and only if $\map J x = \map J y$ (the induced equivalence of $J$).

Then $\sim$ is an equivalence relation on $I$ by Relation Induced by Mapping is Equivalence Relation.

Therefore, by the Fundamental Theorem on Equivalence Relations, $\sim$ partitions $I$.

In fact, the equivalence classes are open intervals.

To prove this, let $x < y$ with $x, y\in I$. We see that when $x \sim y$, we have $x \in \map J x = \map J y \ni y$.

Also, when $x \in \map J y$, it follows that $\openint x y \subseteq I$.

Therefore, we must have:

- $\map a x = \map a y$

and:

- $\map b x = \map b y$

hence:

- $\map J x = \map J y$

This implies that the equivalence class of $x$ is precisely $\map J x$.

Finally, as:

- $\openint {\map a x} {\map b x} \ne \O$

by Rationals Dense in Reals there exists:

- $q \in \openint {\map a x} {\map b x} \cap \Q$

Therefore each set in the partition of $I$ can be labelled with a rational number.

Since the Rational Numbers are Countably Infinite, the partition is countable.

Then enumerating the disjoint intervals of $I$, we have an expression:

- $\displaystyle I = \bigcup_{n \mathop \in \N} J_n $

Now let:

- $\displaystyle I = \bigcup_{n \mathop \in \N} J_n,\qquad I = \bigcup_{n \mathop \in \N} K_n$

be two such expression for $I$.

Suppose that there exists $m_0 \in \N$ such that for all $n \in \N$, $K_{m_0} \ne J_n$.

Write $K_{m_0} = \openint {a_0} {b_0}$.

Since $I$ contains $K_{m_0}$ there exists $J_{m_0'} = \openint {a_0'} {b_0'}$ such that $J_{m_0'} \cap K_{m_0} \ne \O$.

This means that either $a_0' \in \openint {a_0} {b_0}$ or $b_0' \in \openint {a_0} {b_0}$.

Therefore either:

- $J_{m_0'} \setminus K_{m_0} = \hointr {b_0'} {b_0}$

or:

- $J_{m_0'} \setminus K_{m_0} = \hointl {a_0} {a_0'}$

Then we have one of the following two expressions:

- $\displaystyle I \setminus K_{m_0} = \bigcup_{n \mathop \ne m_0} \paren {J_n \setminus K_{m_0} } \cup \hointr {b_0'} {b_0}$

- $\displaystyle I \setminus K_{m_0} = \bigcup_{n \mathop \ne m_0} \paren {J_n \setminus K_{m_0} } \cup \hointl {a_0} {a_0'}$

Now by Half-Open Real Interval is neither Open nor Closed, and since this union is disjoint, we have that $I\setminus K_{m_0}$ is not open.

On the other hand, we have:

- $\displaystyle I \setminus K_{m_0} = \bigcup_{n \mathop \ne m_0} K_n$

which is a union of open sets, hence open.

This contradiction shows that for every $m \in \N$ there exists $\map \sigma m \in \N$ such that $K_m = J_{\map \sigma m}$.

Moreover since the $J_n$, $n \in \N$ are disjoint, there can be only one such $\map \sigma m$, so $\sigma$ is a permutation.

$\blacksquare$