----- > [!proposition] Proposition. ([[linear continuums in the order topology and the intervals they contain are connected]]) > If $L$ is a [[linear continuum]] in the [[order topology]], then $L$ is [[connected]], and so are [[interval|intervals]] and [[ray|rays]] in $L$. > [!proposition] Corollary. > $\mathbb{R}$ in the [[standard topology on the real line|standard topology]] is [[connected]], as are the [[interval|intervals]] and [[ray|rays]] in $\mathbb{R}.$ > [!proof]- Proof. ([[linear continuums in the order topology and the intervals they contain are connected]]) > It suffices to show that any [[order-convex set|order-convex subspace]] $Y$ of $L$ is [[connected]], for all these items, including $L$ are convex subspaces of $L$. Suppose there exists a [[separation of a topological space|separation]] $A \sqcup B$ of $Y$. Choose $a \in A$ and $b \in B$; WLOG $a<b$. The [[interval]] $[a,b]$ of points in $L$ is contained in $Y$ by convexity; hence, $[a,b]$ is the union of disjoint sets $A_{0}=A \cap [a,b], \ \ B_{0}=B \cap[a,b].$ Let $c=\sup A_{0}$. We claim $c \notin A_{0}$ and $c \notin B_{0}$; proceed by way of contradiction. > **Case 1.** Suppose $c \in B_{0}$. Then either $c=b$ (can't happen because then $B_{0}=\emptyset$) or we have $a<c<b$. 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 $B_{0}$, contradicting the LUB property of $c$. > **Case 2.** Suppose $c \in A_{0}$. Since $A_{0}$ is open in $[a,b]$, there is some interval of the form $[c, d)$ in $A_{0}$, contradicting that $c$ is an upper bound of the set. ----- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ```