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: {{begin-eqn}} {{eqn | n = 1       | l = \map a x       | r = \inf \set {z: \openint z x \subseteq I} }} {{eqn | n = 2       | l = \map b x       | r = \sup \set {y: \openint x y \subseteq I} }} {{end-eqn}} 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 {{iff}} $I \setminus \map {K_\pm} a$ 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 $x \sim y$ {{iff}} $\map J x = \map J y$ (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 are Everywhere Dense in Topological Space of 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. {{qed}} Category:Open Sets Category:Real Number Line with Euclidean Metric biss2fk2311lbh0sxdynpd1iji8vold