ordered topology Let $X$ be a linearly ordered topological space with order relation $\leq$. Let $R$ be a equivalence relation on X and suppose that for each x in X, the equivalence class $[x]$ is closed in the ordered topology and convex (i.e. $xRy$ and $x\leq z\leq y$ implies that $xRz$). Let $Y$ be the quotient space 


*

*Prove that there is a natural ordering of $Y$ such that the quotient topology Y     is same as the order topology on $Y$.

*Consider the cantor set C and the relation $xRy$ if the interval $(x,y)$ in the     canter set is empty
i. Prove that in this case the resulting order space $C/R$ is dense in itself, order complete separable with a minimum and a maximum
ii.    Conclude that the Cantor set can be mapped onto the interval by a continuous at most two to one map.
 A: EXTENDED HINTS:
For the first problem:


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*Suppose that $u,v\in X$, $u\le v$, and $u\not Rv$. Show that if $wRu$ and $xRv$, then $w<x$. 

*Let $q:X\to Y$ be the quotient map. Suppose that $u,v\in X$, $q(u)\ne q(v)$, and $u\le v$. Use (1) to show that if $q(w)=q(u)$ and $q(x)=q(v)$, then $w<x$. 

*For $y,z\in Y$ define $y\preceq z$ if and only if there are $u,v\in X$ such that $u\le v$, $q(u)=y$, and $q(v)=z$; use (2) to show that $\preceq$ is well-defined.

*Finally, show that $\preceq$ is a linear order on $Y$; this should be very easy.
For the second problem, let $q:C\to C/R$ be the quotient map.
(i) This is completely straightforward. Some things to prove: 


*

*If $D$ is a countable dense subset of $C$, then $q[D]$ is a countable dense subset of $C/R$, so $C/R$ is separable.  

*$q(0)$ and $q(1)$ are the endpoints of $C/R$ under the ordering defined in the first problem.  

*Suppose that $x,y\in C$ with $x<y$; then either $(x,y)=\varnothing$ and $q(x)=q(y)$, or there is some $z\in(x,y)$ such that $q(x)<q(z)<q(y)$. Use this to see that $C/R$ is densely ordered.  

*For completeness: if $A$ is a non-empty subset of $C/R$, and $a=\sup_C q^{-1}[A]$, then $q(a)=\sum_{C/R}A$.


(ii) This is immediate from (i): $q$ is continuous and at most two-to-one, and the properties of $C/R$ mentioned in (i) characterize $[0,1]$ up to homeomorphism.
