1. topology; connectedness

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

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

Let $f$ be continuous.
Then $f^{-1}(U), f^{-1}(V)$ are both open in $[0,1]$.
Also $S=U \cup V \Rightarrow f^{-1}(U) \cup f^{-1}(V)=[0,1]$.
Again $U \cap V = \Phi \Rightarrow f^{-1}(U) \cap f^{-1}(V)=\Phi$. Also $0 \in f^{-1}(U), 1 \in f^{-1}(V)$ hence these are non empty.
The above suggests that $[0,1]$ 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 $S$ be a disconnected space, say $S=U \cup V$, where $U \cap V = \Phi$ and $U, V$ are both open and non-empty. If $x \in U$ and $y \in V$, prove that there can be no continuous function $f:[0,1] \rightarrow S$ such that $f(0)=x$ and $f(1)=y$.
Let $f$ be continuous.
Then $f^{-1}(U), f^{-1}(V)$ are both open in $[0,1]$.
Also $S=U \cup V \Rightarrow f^{-1}(U) \cup f^{-1}(V)=[0,1]$.
Again $U \cap V = \Phi \Rightarrow f^{-1}(U) \cap f^{-1}(V)=\Phi$. Also $0 \in f^{-1}(U), 1 \in f^{-1}(V)$ hence these are non empty.
The above suggests that $[0,1]$ 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 $x$ and $y$ 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?