# Let $(\Bbb R, \tau)$ be the usual topological space. Show that any interval $C \subseteq \Bbb R$ is connected.

Let $$(\Bbb R, \tau)$$ be the usual topological space. Show that any interval $$C \subseteq \Bbb R$$ is connected.

Attempt: Let $$C \subseteq \Bbb R$$ be arbitrary interval. If $$C = \Bbb R$$, then $$C$$ is connected since the only closed and open set in $$C$$ are $$\emptyset$$ and $$C$$. So, assume that $$C \ne \Bbb R$$, that is, $$C \subset \Bbb R$$. Suppose for the sake of contradiction that $$C$$ is disconnected. Then $$C$$ has a separation, say, $$A$$ and $$B$$ with $$A$$ and $$B$$ are open in $$C$$, $$A \ne \emptyset, B \ne \emptyset, A \cap B = \emptyset$$, and $$A \cup B = C$$. Let $$a \in A$$ and $$b \in B$$. Without of loss generality, let $$a. Since $$a,b \in C$$ and $$C$$ is an interval, then $$[a,b] \subseteq C$$. Define $$A_1=A \cap [a,b]$$ and $$B_1=B \cap [a,b]$$. Since $$A$$ and $$B$$ are closed in $$C$$ and $$[a,b] \subseteq C$$, then $$A_1$$ and $$B_1$$ are closed in $$[a,b]$$. Now, since $$[a,b]$$ is closed in $$\Bbb R$$, then $$A_1$$ and $$B_1$$ are closed in $$\Bbb R$$. Let $$c:=\sup A_1$$. Since $$A_1$$ is closed, then $$c \in A_1$$. Since $$b \in B_1$$ and $$A_1 \cap B_1 = \emptyset$$, then $$c. But, $$A_1$$ is open in $$[a,b]$$, so there is $$r>0$$ such that $$(c-r,c+r) \cap [a,b] \subseteq A_1$$. Next, since $$c, there exists $$d \in (c,c+r) \cap [a,b]$$ such that $$d \in A_1$$. Hence, $$c. But then, $$d \le c$$ by definition of $$c$$, which is a contradiction. Therefore, $$C$$ is connected. Since $$C$$ was arbitrarily taken, we can conclude that any interval $$C \subseteq \Bbb R$$ is connected. Q. E. D.

Does this approach correct? Thanks in advanced.

• "$A$ and $B$ are open in $C$" suddenly becomes "$A$ and $B$ are closed in $C$" later. By the way, a more simple proof would be to use, that path-connected spaced are connected as it's easier to show, that any interval is path-connected. Just take $\gamma\colon[0,1]\rightarrow I,t\mapsto a+(b-a)t$ for points $a,b\in I$. Nov 19, 2022 at 14:36
• The second sentence of your attempt seems suspicious: Are you assuming, as already proved, that $\mathbb R$ itself is connected? Nov 19, 2022 at 15:06
• @LeeMosher Yes, I already proved that the only clopen set in $\Bbb R$ are $\emptyset$ and $\Bbb R$ itself and use the theorem which said $X$ is connected iff the only clopen set in $X$ are $\emptyset$ and $X$.
– user1089451
Nov 19, 2022 at 15:08
• @SamuelAdrianAntz Your argument is circular. Yes, path-connected spaces are connected, but the proof of this fact requires to know that intervals are connected. Nov 19, 2022 at 15:40
• @PitikKepit Since $c=\sup A_1$ is the supremum, there has to be a sequence in $A_1$ converging to it, which directly implies $c\in\operatorname{Cl}(A_1)$. (Note, that if $c$ is an isolated point of $A_1$, it's trivial, that $c$ is in $\operatorname{Cl}(A_1)$ and the sequence is just constant $c$.) Nov 19, 2022 at 21:46

Your proof is correct. But there is no need to consider $$C = \mathbb R$$ separately, your proof also works in this case.
In a comment you say you already proved that the only clopen set in $$\mathbb R$$ are $$∅$$ and $$\mathbb R$$ itself which means that $$\mathbb R$$ is connected. If you know this, you also know that all open intervals $$C$$ are connected because they are homeomorphic to $$\mathbb R$$. But then also each $$C'$$ with $$C \subset C' \subset \overline C$$ is connected, and this gives you all intervals.