Show that $(0,1)$ is connected and not compact.

Answer

In Rudin's book, the definition of "connected" is

A set $E \subset X$ is said to be connected if $E$ is not a union of two nonempty separated sets.

Two subsets $A$ and $B$ of a metric space $X$ are said to be separated if both $A \cap \overline{B}$ and $\overline{A} \cap B$ are empty

(1) Connected

Suppose $(0,1)$ is not connected. Then there are separated sets $A$ and $B$ such that $(0,1)=A \cup B$. Let $a \in A, b \in B$ and assume $a<b$ without loss of generality. Since $C:=[a,b] \cap \overline{A}$ is closed, $x:=\sup C \in C \subset \overline{A}$. Note that $x \neq b$ because $\overline{A} \cap B = \emptyset$. Then, $(x,b] \subset B$  $\Rightarrow$  $[x,b] \subset \overline{B}$  $\Rightarrow$  $x \in \overline{B}$. This implies $x \notin A,B$, but $a \leq x \leq b$  $\Rightarrow$  $x \in (0,1)=A \cup B$, which is contradiction.

(2) Not compact

Consider the open cover $\{(0,1-1/n)\}_{n=2}^\infty$. For any finite subcover $(0,1-1/n_1),\cdots,(0,1-1/n_N)$, $1-1/n_\ast \notin \bigcup_{k=1}^N(0,1-1/n_k)$ for $n_\ast>\max(n_1,\cdots,n_N)$.

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