Having established separation criteria for connected spaces, the first major test case is the real line. The connected subsets of the real line are exactly intervals, and this theorem explains why gaps destroy connectedness. Intervals and connected subsets of the real line are central because many later results reduce a general connectedness question to a real-valued continuous function. Students should watch the role of the least upper bound property in the converse direction. The focus keyword intervals and connected subsets of the real line appears here as the bridge between metric topology and the intermediate value theorem. :::definition[Interval] Let $I\subseteq\mathbb{R}$. Then $I$ is called an $interval$ if whenever $x,y\in I$ and $x<z<y$, then $z\in I$. ::: :::theorem[Connected Subsets of the Real Line] Let $I$ be a subset of $\mathbb{R}$ with the usual metric. Then $I$ is connected if and only if $I$ is an interval. ::: :::proof Given that $I\subseteq\mathbb{R}$. To prove that $I$ is connected if and only if $I$ is an interval. First assume that $I$ is connected. To prove that $I$ is an interval, let $x,y\in I$ and $x<z<y$. If possible let $z\notin I$. Define $$\begin{equation*} A=I\cap(-\infty,z),\qquad B=I\cap(z,\infty). \end{equation*}$$ Then $x\in A$ and $y\in B$, so both sets are nonempty. They are disjoint, open in the subspace $I$, and $I=A\cup B$. Therefore $I$ is disconnected. A contradiction. Hence $z\in I$. Thus $I$ is an interval. Conversely, assume that $I$ is an interval. Suppose, if possible, that $I$ is disconnected. Then there exist nonempty separated subsets $A$ and $B$ such that $I=A\cup B$. Choose $a\in A$ and $b\in B$. Without loss of generality, assume $a<b$. Since $I$ is an interval, $[a,b]\subseteq I$. Let $$\begin{equation*} c=\sup(A\cap[a,b]). \end{equation*}$$ Then $c\in[a,b]\subseteq I$. By the definition of supremum, $c\in\overline{A}$. Since $A$ and $B$ are separated, $\overline{A}\cap B=\varnothing$. Therefore $c\notin B$. Since $I=A\cup B$, we get $c\in A$. Again, $A\cap\overline{B}=\varnothing$, so $c\notin\overline{B}$. Therefore there exists $\varepsilon>0$ such that $$\begin{equation*} (c-\varepsilon,c+\varepsilon)\cap B=\varnothing. \end{equation*}$$ Choose $t$ such that $c<t<b$ and $t\in(c-\varepsilon,c+\varepsilon)$. Then $t\notin B$. Also $t\notin A$, because $c$ is the supremum of $A\cap[a,b]$. Hence $t\notin I$. But $a<t<b$ and $[a,b]\subseteq I$, so $t\in I$. A contradiction. Hence $I$ is connected. $\square$ ::: :::corollary The real line $\mathbb{R}$ is connected. ::: :::proof Given that $\mathbb{R}=(-\infty,\infty)$ is an interval. To prove that $\mathbb{R}$ is connected. By the theorem on connected subsets of the real line, every interval in $\mathbb{R}$ is connected. Therefore $\mathbb{R}$ is connected. $\square$ ::: :::corollary The only subsets of $\mathbb{R}$ that are both open and closed in $\mathbb{R}$ are $\varnothing$ and $\mathbb{R}$. ::: :::proof Given that $\mathbb{R}$ is connected. To prove that the only clopen subsets of $\mathbb{R}$ are $\varnothing$ and $\mathbb{R}$. Suppose that $A\subseteq\mathbb{R}$ is both open and closed. If $A\neq\varnothing$ and $A\neq\mathbb{R}$, then $A$ is a nonempty proper clopen subset of $\mathbb{R}$. By the disconnectedness criterion, $\mathbb{R}$ would be disconnected. This contradicts the connectedness of $\mathbb{R}$. Therefore $A=\varnothing$ or $A=\mathbb{R}$. $\square$ ::: :::example[Open and Closed Intervals] The sets $(a,b)$, $[a,b]$, $(a,b]$, $[a,b)$, $(-\infty,b)$, $(-\infty,b]$, $(a,\infty)$, $[a,\infty)$, and $\mathbb{R}$ are intervals. Hence each of them is connected with the usual metric. $\square$ ::: :::example[A Set with a Gap] Let $S=[0,1]\cup[2,3]$. Since $0,2\in S$ but $3/2\notin S$, the set $S$ is not an interval. Therefore $S$ is disconnected as a subspace of $\mathbb{R}$. $\square$ ::: :::remark A subset of the real line is connected exactly when it has no gap between any two of its points. This is why intervals and connected subsets of the real line are the same objects in the usual metric. ::: :::solved-problem Show that $\mathbb{R}\setminus\{0\}$ is disconnected. ::: :::solution Given that $\mathbb{R}\setminus\{0\}=(-\infty,0)\cup(0,\infty)$. To prove that $\mathbb{R}\setminus\{0\}$ is disconnected. The set $\mathbb{R}\setminus\{0\}$ is not an interval, because $-1,1\in\mathbb{R}\setminus\{0\}$ but $0\notin\mathbb{R}\setminus\{0\}$. By the theorem on connected subsets of the real line, it is not connected. Therefore $\mathbb{R}\setminus\{0\}$ is disconnected. $\square$ ::: :::exercise [1] State the interval criterion for connected subsets of $\mathbb{R}$. [2] Decide whether $(0,1)\cup(1,2)$ is connected. [3] Explain why $\mathbb{R}$ has no nonempty proper clopen subset. [4] Show that $[0,\infty)$ is connected. ::: :::answer [1] A subset of $\mathbb{R}$ is connected if and only if it is an interval. [2] It is disconnected because it has the gap point $1$. [3] Such a subset would disconnect $\mathbb{R}$. [4] The set $[0,\infty)$ is an interval, and hence it is connected. ::: :::faq[Frequently Asked Questions] Q: Why is the supremum needed in the proof? A: It locates the boundary between two alleged separated pieces of an interval. Q: Are all connected subsets of $\mathbb{R}$ open intervals? A: No. Closed, half-open, unbounded, and singleton intervals are also connected. Q: Does this theorem hold in every metric space? A: No. It is a special characterization of subsets of the real line. ::: :::call-to-action[Continue to Two-Point Tests] subtitle: Use intervals and connected subsets of the real line to understand how continuous maps detect separation. button: Next Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/two-point-tests-for-connectedness button-ghost: Previous Lesson | /functional-analysis/metric-spaces/connectedness-and-path-structure-in-metric-spaces/separation-criteria-for-connected-spaces :::