Find the connected components of $ \ [0,1] \ $ with respect to the lower-limit topology $ \ \large T_{[,)} \ $ ,
where $ T_{[,)} \ $ contains the basic open sets of the form $ \ [a,b) \ $.
Also find continuous functions $ \ f: ([0,1] , T_{usual}) \to ([0,1], T_{lower-limit}) \ $
Answer:
At first to find the connected components of $ \ [0,1] \ $ in lower limit topology $ \ \large T_{[,) } \ $
Let $ A \subset [0,1] \ $ be non-empty and connected.
We will show that $ A \ $ is a singleton set.
For,
let $ \ a \in A \ $ , then
$ A=(([0,a) \cap A) \cup ([a,1] \cap A) \ $ , (disjoint union of open sets)
But as $ A \ $ is connected , at most one of the sets can be non-empty.
Now, since $ a \in ([a,1] \cap A ) \neq \phi \ $ , it follows that
$ [0,a) \cap A=\phi \ $
Hence if $ c \in A , \ then \ \ c \notin [0,a) \ $
Therefore, $ \ c \geq a \ $
Since $ a, c \ $ are arbitrarily chosen , we can have $ c \leq a \ $
Thus, $ a,c \in A \Rightarrow a=c \ $
This show that $ A \ $ is a singleton set.
Thus the connected components of $ [0,1] \ $ are singleton set in lower-limit topology.
Am I right ?
Hellp me out
Also what about the continuous functions ?
First, I think that is easier to show that if $A\subseteq [0,1]$ and $|A|\geq 2$ then $A$ is disconnected.
Second, if we want to find the continuous functions $f\colon ([0,1], \tau)\to([0,1],\mathcal{L}_S)$ (where $\tau$ denotes the usual topology and $\mathcal{L}_S$ denotes the lower limit topology) we need to use the previous result, i.e., the singletons are the only connect sets in $\mathcal{L}_S$. We know that a continuous image of a connected set is also connected. Then, if $f$ is continuous, then, $f[[0,1]]\subseteq [0,1]_{\mathcal{L}_S}$ is connected. Thus, $f[[0,1]]$ is a singleton. Take a fixed point $a\in [0,1]$. By the previous argument, $f[[0,1]]=\{a\}$, i.e., the function is constant.