Proving that given topological sets forms a partition.

Question

Let $A \subset \mathbb{R}$ be an open set.

1. Show that for a given $x \in A$ the set $$R_x = \{y \in A \ | \ (x,y] \subset A\}$$ is not-empty, open and path-connected. Conclude that $R_x$ is an open interval of the form $(x,r)$.

2. Show that the set $$C_x = \{y \in A \ | [\min\{x,y\},max\{x,y\}] \subset A\}$$ is an open interval and that for any $a,b \in A$ holds $C_a = C_b$ or $C_a \cap C_b = \emptyset$.

3. By considering the rational numbers in $\{C_x\}_{x \in A}$ conclude that one can write $$ A = \cup_{i \in \mathbb{N}}(a_i,b_i) $$ where $a_i < b_i < a_{i+1}$.

This is what I got:

1. Non-emptiness: Since $A$ is open we can just take an (suffieciently small) interval $[x,y]$ that lies completely in $A$. (The point $y \ne x$ exists because $A$ is open and thus not a singleton.)

Openness: We observe that for all $y \in R_x$ holds $(x,y] \subseteq R_x$. Since $A$ is open we can find an open ball $B_\epsilon (y) \subset A$. Finally we observe $(x,y+\epsilon /2] \subset (x,y] \cup B_\epsilon (y) \subset A$ and so $B_{\epsilon /2}(y) \subset R_x$.

Path connected: Since $(x,x] = \emptyset \subset A$ we see that for all $x,y \in A$ holds $[x,y] \subset A$. So the path connectedness is clear.

All in all we got that $R_x$ is open and connected and since it is in $\mathbb{R}$ it is thus an open interval.

I have done $1)$ and $3)$ (under assumption that $2)$ holds) but I do not know how to prove $2)$. In particular I do not understand why it should hold that $C_a = C_b$ or $C_a \cap C_b = \emptyset$. Could you help me?

Answer 1

For 2.Obviously $a\in C_a\subset A. $ Show that $C_a$ is an interval. Show that if $J$ is an interval such that $a\in J\subset A $ then $J\subset C_a.$ (That is, $C_a$ is the $\subseteq$-largest interval that contains $a$ and is a subset of $A.$)

Now if $P$ and $Q$ are intervals with non-empty intersection then $P\cup Q$ is an interval.

So suppose $C_a$ and $ C_b$ intersect. Let $J=C_a\cup C_b.$ Then $J$ is an interval, and $a\in J\subset A,$ so $J\subset C_a .$ That is, $$C_a\cup C_b=J\subset C_a$$ which implies $C_b\subset C_a.$

Interchanging $a,b$ in the preceding paragraph we also obtain $C_b\subset C_a.$ So $C_a=C_b$ if $C_a\cap C_b\ne \emptyset.$

Remark: Let $In[a,y]=[\min(a,y),\max(a,y)]=[a,y]\cup [y,a].$ If $J$ is an interval such that $a\in J\subset A$ then for every $y\in J$ we have $y\in A$ and $ In[a,y]\subset J\subset A,$ so $y\in C_a.$ Hence $J\subset C_a$.