0

Let $(R,d)$ be the space of real numbers with the usual metric. Let $I$ be an interval such that $I \subseteq R$. We need to show that $I$ is connected.

Let us say that $I$ is not connected , thus for non-empty disjoint sets $A$ and $B$ , we have $I = A \cup B$ with $A \cap \bar{B} = \phi$ and $\bar{A} \cap {B} = \phi$.

Let $ x \in A$ , $y \in B$ and assume $x<y$. We notice $[x,y] \subseteq I$.

Now , let $z = sup(A\cap[x,y])$. I couldn't understand the following statements :

(1) If $z \in A$ , then $z \not \in \bar{B}$ (Got this). Now it says , thus there exists a $\delta > 0$ such that $( z - \delta , z+ \delta) \cap B = \phi$. How's that ?

Also ,

(2) This implies that there exists a $z_1 \not \in B$ such that $ z < z_1 < y$. Can anyone explain ? Thanks in advance.

User9523
  • 2,094

1 Answers1

1

(1) $\overline{B}$ is closed, so its complement $\overline{B}^c$ is open.

$z \in \overline{B}^c$ implies that $(z- \delta , z + \delta) \subseteq \overline{B}^c$ just because $\overline{B}^c$ is open. But this is equivalent to $$(z- \delta , z + \delta) \cap \overline{B} = \emptyset$$ in particular, $$\emptyset \subseteq (z- \delta , z + \delta) \cap B \subseteq (z- \delta , z + \delta) \cap \overline{B} = \emptyset$$ so that $(z- \delta , z + \delta) \cap B$ is empty.

(2) Pick $z_1 = z+ \frac{\delta}{2}$.

Crostul
  • 36,738
  • 4
  • 36
  • 72