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- Thread starter ianchenmu
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- Jan 29, 2012

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$\Rightarrow)$ Continuous images of connected sets are connected. If $f:E\to \{0,1\}$ is continuous, $f(E)\subset \{0,1\}$ is connected so, $f(E)=\{0\}$ or $f(E)=\{1\}$. This means that $f$ is constant.Let $E∈\mathbb{R}^{n}$ be a non-empty subset. Prove that $E$ is connected if and only if any continuous function $f : E→${$0$, $1$} is constant.

$\Leftarrow)$ If $E$ is not connected, there exist non empty open subsets $U,V$ of $E$ (open relative to $E$) such that $U\cap V=\emptyset$ and $U\cup V=E$. Clearly the function $f:E\to \{0,1\}$ defined by $f(U)=\{1\}$ and $f(V)=\{0\}$ is continuous and no constant (contradiction).