Answer to Question #217035 in Differential Geometry | Topology for Prathibha Rose

Question

Prove that a subspace of a real line with the usual topology is connected if and only if it is an interval

Answer 1

• First suppose that $X\subseteq \mathbb R$ is not an interval, let us prove that it is not connected. By definition, an intervals are the convex subsets of $\mathbb{R}$, i.e. $(\forall x,y \in I)(\forall z)\;(x\leq z\leq y)\rightarrow (z\in I)$. Therefore, if $X$ is not an interval, we should have $\exists x,y \in X\; \exists z\notin X \; x\leq z \leq y$. Let us then consider two open sets $U:= (-\infty; z)$ and $V:= (z;+\infty)$. First of all, $X\subseteq U\cup V=\mathbb R\setminus \{z\}$. Secondly, $x\in U\cap X \text{ and } y\in V\cap X$, so $U\cap X, V\cap X$ are not empty. Finally, $U\cap V =\emptyset$. Therefore, $X$ is not connected.
• Now suppose that $I$ is an interval and we shall prove that it is connected. There are several ways to do so, I will mention some of them. First of all, we can use an alternative characterization of connectedness : $X$ is connected if for every continuous function $f:X\to \{0;1\}$, $f$ is constant. Suppose that $f:I\to \{0; 1 \}$ is continuous, then if $f\neq const$, we can use an intermediate value theorem to conclude that $f$ takes values between and 1, which is impossible, as we have specified the range of $f$ to be $\{ 0; 1\}$. Therefore, $I$ is connected. Another proof could be arguing that $I$ is path connected (for any $x,y$ there is a path $\gamma (t)=(1-t)x+ty$ between them), but the implication "path connected implies connected" is using the fact that $[0;1]$ is connected, so this is partially a circular argument. Finally, we could prove directly that $[0;1]$ (or any other interval) is connected, using the intrinsic properties of $\mathbb R$, but this proof would just repeat the proof of the intermediate value theorem.

From these two points we conclude that the connected subsets of the real line are exactly the intervals.