[Munkres Topology] 24. Connected Subspaces of the Real Line

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Definition. A simply ordered set $L$ having more than one element is called a linear continuum if the following hold:

(1) $L$ has the least upper bound property.

(2) If $x<y$, there exists $z$ such that $x<z<y$.

Theorem 24.1. If $L$ is a linear continuum in the order topology, then $L$ is connected, and so are intervals and rays in $L$.

Proof. Recall that a subspace $Y$ of $L$ is said to be convex if for every pair of points $a,b$ of $Y$ with $a<b$, the entire interval $[a,b]$ of points of $L$ lies in $Y$. We prove that if $Y$ is a convex subspace of $L$, then $Y$ is connected.

[Convex$\rightarrow$connected] Suppose that $Y$ is the union of the disjoint nonempty sets $A$ and $B$, each of which is open in $Y$. Choose $a\in A$ and b\in B$; suppose for convenience that$a<b$. The interval$[a,b]$of points of$L$is contained in$Y$. Hence$[a,b]$is the union of the disjoint sets$$A_0=A \cap [a,b]\quad \mbox{and}\quad B_0=B\cap [a,b],$$each of which is open in$[a,b]$in the subspace topology, which is the same as the order topology. The sets$A_0$and$B_0$are nonempty bacause$a\in A_0$and$b\in B_0$. Thus,$A_0$and$B_0$constitute a separation of$[a,b]$.

Let $c=\sup A_0$. We show that $c$ belongs neither to $A_0$ nor to $B_0$. which contradicts the fact that $[a,b]$ is the union of $A_0$ and $B_0$.

Case 1. Suppose that $c\in B_0$. Then $c \ne a$, so either $c=b$ or $a<c<b$. In either case , it follows from the fact that $B_0$ is open in $[a,b]$ that there is some interval of the form $(d,c]$ contained in $B_0$. If $c=b$, we have a contradiction at once, for $d$ is a smaller upper bound on $A_0$ than $c$(since $(d,c] \notin A_0$). If $c<b$, we note that $(c,b]$ does not interrsect $A_0$ (because $c$ is an upper bound on $A_0$). Then

$$(d,b]=(a,c]\cup (c,b]$$ does not intersect $A_0$. Again, $d$ is a smaller upper bound on $A_0$ than $c$, contrary to construction. See Figure 24.1.

Case 2. Suppose that $c\in A_0$. Then $c\ne b$, so either $c=a$ or $a<c<b$. Because $A_0$ is open in $[a,b]$, there must be some interval of the form $[c,e)$ contained in $A_0$. See Figure 24.2. Because of order property (2) of the linear continuum $L$, we can choose a point $z$ of $L$ such that $c<z<e$. Then $z\in A_0$, contrary to the fact that $c$ is an upper bound for $A_0$.

Corollary 24.2. The real line $\mathbb{R}$ is connected and so are intervals and rays in $\mathbb{R}$.

Theorem 24.3 (Intermediate value theorem). Let $f:X\rightarrow Y$ be a continuous map, where $X$ is a connected space and $Y$ is an ordered set in the order topology. If $a$ and $b$ are two points of $X$ and if $r$ is a point of $Y$ lying between $f(a)$ and $f(b)$, then there exists a point $c$ of $X$ such that $f(c)=r$.

Proof. Assume the hypotheses of the theorem. The sets

$$A=f(X)\cap (-\infty,r)\quad \mbox{and}\quad B=f(X)\cap (r,+\infty)$$ are disjoint, and they are nonempty because one contains $f(a)$ and the other contains $f(b)$. Each is open in $f(X)$, being the intersection of an open ray in $Y$ with $f(X)$. In there we no point $c$ of $X$ such that $f(c)=r$, then $f(X)$ would be the union of the sets $A$ and $B$. Then $A$ and $B$ would constitute a separation of $f(X)$, contradicting the fact that the image of a connected space under a continuous map is connected.

Definition. Given points $x$ and $y$ of the space $X$, a path in $X$ from $x$ to $y$ is a continuous map $f:[a,b]\rightarrow X$ of some closed interval in the real line into $X$, such that $f(a)=x$ and $f(b)=y$. A space $X$ is said to be path connected if every pair of points of $X$ can be jointed by a path in $X$.

It is easy to see that a path-connected space $X$ is connected. Suppose $X=A\cup B$ is a separation of $X$. Let $f:[a,b]\rightarrow X$ be any path in $X$. Being the continuous image of a connected set, the set $f([a,b])$ is connected, so that it lies entirely in either to the assumption that $X$ is path connected.