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**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?