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 $a<b$. 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 $c<b$. 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 $c<b$, there exists $d \in (c,c+r) \cap [a,b]$ such that $d \in A_1$. Hence, $c<d$. 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.

  • 1
    $\begingroup$ "$A$ 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 $a,b\in I$. $\endgroup$ Nov 19, 2022 at 14:36
  • $\begingroup$ The second sentence of your attempt seems suspicious: Are you assuming, as already proved, that $\mathbb R$ itself is connected? $\endgroup$
    – Lee Mosher
    Nov 19, 2022 at 15:06
  • $\begingroup$ @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 $X$. $\endgroup$
    – user1089451
    Nov 19, 2022 at 15:08
  • 5
    $\begingroup$ @SamuelAdrianAntz Your argument is circular. Yes, path-connected spaces are connected, but the proof of this fact requires to know that intervals are connected. $\endgroup$
    – Paul Frost
    Nov 19, 2022 at 15:40
  • 1
    $\begingroup$ @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 $c$.) $\endgroup$ Nov 19, 2022 at 21:46

1 Answer 1


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.


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