# Find the connected components of $\ [0,1] \$ with respect to the lower-limit topology $\ \large T_{[,)} \$

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}) \$

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 ?

• The continuous image of a connected space … – Daniel Fischer Jan 22 '18 at 18:52

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.