# Thread: topology; connectedness

1. ## topology; connectedness

Let $\displaystyle S$ be a disconnected space, say $\displaystyle S=U \cup V$, where $\displaystyle U \cap V = \Phi$ and $\displaystyle U, V$ are both open and non-empty. If$\displaystyle x \in U$ and $\displaystyle y \in V$, prove that there can be no continuous function $\displaystyle f:[0,1] \rightarrow S$ such that $\displaystyle f(0)=x$ and $\displaystyle f(1)=y$.

I want to know HOW important is the "such that" thing.
Here is my attempt:

Let $\displaystyle f$ be continuous.
Then $\displaystyle f^{-1}(U), f^{-1}(V)$ are both open in $\displaystyle [0,1]$.
Also $\displaystyle S=U \cup V \Rightarrow f^{-1}(U) \cup f^{-1}(V)=[0,1]$.
Again $\displaystyle U \cap V = \Phi \Rightarrow f^{-1}(U) \cap f^{-1}(V)=\Phi$. Also $\displaystyle 0 \in f^{-1}(U), 1 \in f^{-1}(V)$ hence these are non empty.
The above suggests that $\displaystyle [0,1]$ is disconnected which is wrong hence we have a contradiction.
Have i missed something or is the proof correct?

2. ## Re: topology; connectedness

Originally Posted by abhishekkgp
Let $\displaystyle S$ be a disconnected space, say $\displaystyle S=U \cup V$, where $\displaystyle U \cap V = \Phi$ and $\displaystyle U, V$ are both open and non-empty. If$\displaystyle x \in U$ and $\displaystyle y \in V$, prove that there can be no continuous function $\displaystyle f:[0,1] \rightarrow S$ such that $\displaystyle f(0)=x$ and $\displaystyle f(1)=y$.
Let $\displaystyle f$ be continuous.
Then $\displaystyle f^{-1}(U), f^{-1}(V)$ are both open in $\displaystyle [0,1]$.
Also $\displaystyle S=U \cup V \Rightarrow f^{-1}(U) \cup f^{-1}(V)=[0,1]$.
Again $\displaystyle U \cap V = \Phi \Rightarrow f^{-1}(U) \cap f^{-1}(V)=\Phi$. Also $\displaystyle 0 \in f^{-1}(U), 1 \in f^{-1}(V)$ hence these are non empty.
The above suggests that $\displaystyle [0,1]$ is disconnected which is wrong hence we have a contradiction.
Have i missed something or is the proof correct?
Well that is close to a complete proof.
We know that the continuous image of a connected set is connected.

3. ## Re: topology; connectedness

Originally Posted by Plato
Well that is close to a complete proof.
We know that the continuous image of a connected set is connected.
I was worried about my proof because $\displaystyle x$ and $\displaystyle y$ are images of the END POINTS of [0,1] while my proof would work even if x and y were images of ANY two distinct points in [0,1].
My proof's right, right?