Question

TOPOLOGY

Find the connected components of   $\left [ 0,1 \right ]$ with the topology   $T_{\left [ ,\right )}$ and describe every continuous   $f:\left ( \left [ 0,1 \right ],T_{usual} \right )\rightarrow \left ( \left [ 0,1 \right ] ,T_{[,)})$ .

Explain it.

Answer 1

$\large \\ \textbf{Connected \ components \ of \ [0,1] \ in \ lower - limit \ topology:} \\ Let \ \ A \subseteq [0,1] \ be \ connected \ components \ of \ [0,1] \ in \ T_{[,)} \ topology . \\ If \ A \ be \ empty , then \ A \ is \ trivially \ connected \ components \ of \ [0,1] . \\ Let \ A \ be \ non-empty , \ i.e., \ A \neq \phi . \\ Let \ a \in A , then \ A \ is \ the \ disjoint \ union \ of \ two \ relatively \ open \ sets \\ in \ A \ given \ by \ [0,a) \cap A \ and \ [a,1] \cap A , \\ i.e., \ A=\left ( [0,a ) \cap A \right ) \cup \left ( [a,1] \cap A \right ) , \\ Since \ A \ is \ connected \ and \ non-empty , \ one \ of \ the \ open \ sets \ must \\ be \ empty. \\ Now \ as \ \ a \in [a,1] \cap A \neq \phi , \ it \ follows \ that \ [0,a) \cap A =\phi. \\ Thus \ if \ \ r \in A , \ then \ r \notin [0,a) . \\ This \ implies \ that \ {\color{Blue} r \geq a}$

$\large \\ But \ since \ a \ and \ r \ are \ chosen \ arbitrarily , their \ roles \ can \ be \ interchanged \\ so \ that \ we \ have \ \ {\color{Blue} a \geq r} \\ Therefore, \\ a , r \in A \ \Rightarrow a=r \\ In \ other \ words , \\ A \ is \ a \ singleton \ set. \\ \\ Hence \ the \ only \ connected \ components \ of \ [0,1] \ in \ lower-limit \\ topology \ are \ {\color{Blue} empty \ set \ and \ singleton \ sets. }$

$\large \\ \textbf{Next}, \\ \\ There \ only \ continuous \ functions \\ f : \left ( [0,1] , T_{usual} \right ) \to \left ( [0,1], T_{[,)} \right ) \ are \ the \ {\color{Blue} constant \ functions.}$