# Help understanding every open set in $\mathbb{R}$ is the union of a countable collection of disjoint open intervals

I am sorry if this is a duplicate question (although I couldn't find an explanation in other posts)

While I follow the common proof of the above which goes like:

Let $\mathit{E} \subset \mathbb{R}$ be open. By definition, for each $\mathit{x} \in \mathit{E}$, $\exists$ a small interval around $\mathit{x}$ contained in $\mathit{E}$.
Now let $\mathit{a} = \mathit{inf}$ {$\mathit{y}$ | $\mathit{(y,x)} \subset \mathit{E}$ } and $\mathit{b} = \mathit{sup}$ {$\mathit{z}$ | $\mathit{(x,z)} \subset \mathit{E}$ } So, $\mathit{a \lt x \lt b}$ and $\mathit{I_x} = (a,b)$ is an open interval containing $\mathit{x}$ and $\mathit{I_x} \subset E$ with $\mathit{a,b} \notin E$.
Varying $\mathit{x}$ over $\mathit{E}$, $\mathit{E} = \bigcup\limits_x I_x$
Now, for $I_x = (p, q)$ and $I_y = (r,s)$ $p,q,r,s \notin E$ and it can be shown that $p \leq r \lt q$ cannot be true and hence disjoint.

I have trouble understanding/visualizing the above proof for an open interval (open set) say $(0,1)$ in $\mathbb{R}$. So, how can it be a union of open intervals? We surely cannot have $I_x$ and $I_y$ with $0 \leq p \lt q \lt r \lt s \leq 1$? Or is it that all $I_x$ are identical?

• $(0, 1)$ is the union of one disjoint open interval in $\mathbb{R}$, namely itself. Commented Aug 25, 2014 at 19:51
• Nobody said there was more than one in the collection of intervals. The confusion is thinking of "union" as a binary operator, but the definition of arbitrary unions of sets is not defined in terms of a binary union operator. Commented Aug 25, 2014 at 19:52

If $A$ and $I$ are sets, and $\mathcal P(A)$ is the power set of $A$. Then a function:$$U:I\to \mathcal P(A)$$ is sometimes seen as a indexed set of subsets of $A$, also written as $\{U_i\}_{i\in I}$, where $U_i$ is just shorthand for $U(i)$. Then the "union" of this indexed set:

$$\bigcup_{i\in I} U(i)$$

is defined as the set: $\{x\in A\mid \exists i\in I\, x\in U(i)\}$.

When $I$ is a singleton, such as $I=\{1\}$, then the "union" of one subset of $A$ is just that subset of $A$.

Now, the above theorem says:

If $V$ is open in $\mathbb R$, then there is a set $I$ and a map $U:I\to\mathcal P(\mathbb R)$ so that for all $i\in I$, $U(i)$ is an open interval, and for all $i,j\in I$, $i\neq j\implies U(i)\cap U(j)=\emptyset$, and $V=\bigcup_{i\in I} U(i)$.

When $V=(0,1)$, $I=\{1\}$ and $U(1)=V$. Then $\forall i\in I: U(i)$ is an open interval, and $\forall i,j\in I:i=j$, so $i\neq j$ is false, and thus $i\neq j\implies U(i)\cap U(j)=\emptyset$ is true, because $p\implies q$ is true whenever $p$ is false.

(Perhaps even stranger is the case $I=\emptyset$. At first, it often seems odd to define a function on the empty set.)

• Thanks for your response. So, the union contains only the maximal interval (i.e, (0,1) itself in this case, or any (a,b) generally). But isn't this sort of trivial? Further, even for R^2, R^3,...R^n the union would still contain the interval itself right? (an open rectangle is the union of only that disjoint open rectangle and so on...). Wherein lies the utility of this theorem apart from seemingly strong result? Commented Aug 26, 2014 at 5:58