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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 ?

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    $\begingroup$ The continuous image of a connected space … $\endgroup$ – Daniel Fischer Jan 22 '18 at 18:52
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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.

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