Let $\mathbb{X}$ be a topological space. Show that $\mathbb{X}$ is connected if and only if every continuous function $f:\mathbb{X} \rightarrow \{0, 1\}$ is constant.
Solution
$( \Rightarrow)$ Suppose that $\mathbb{X}$ is connected and let $f$ be a function. If $f$ is not constant , then the two sets $f^{-1}\left ( \left \{ 0 \right \} \right ), \; f^{-1}\left ( \left \{ 1 \right \} \right )$ are non empty and divide $\mathbb{X}$ which is a contradiction since $\mathbb{X}$ is connected.
$(\Leftarrow)$ Suppose that $\mathbb{X} =A \cup B$ where $A, B$ are two non empty open subsets of $\mathbb{X}$. Consider the function $f:\mathbb{X} \rightarrow \{0, 1\}$ which takes the value $0$ on $A$ and $1$ on $B$. We can immediately see that this function is continuous, which is a contradiction from the hypothesis. Therefore there does not exist a partition and $\mathbb{X}$ is connected.
The exercise is actually equivelant to the exercise Algebra and Topology.
Solution
$( \Rightarrow)$ Suppose that $\mathbb{X}$ is connected and let $f$ be a function. If $f$ is not constant , then the two sets $f^{-1}\left ( \left \{ 0 \right \} \right ), \; f^{-1}\left ( \left \{ 1 \right \} \right )$ are non empty and divide $\mathbb{X}$ which is a contradiction since $\mathbb{X}$ is connected.
$(\Leftarrow)$ Suppose that $\mathbb{X} =A \cup B$ where $A, B$ are two non empty open subsets of $\mathbb{X}$. Consider the function $f:\mathbb{X} \rightarrow \{0, 1\}$ which takes the value $0$ on $A$ and $1$ on $B$. We can immediately see that this function is continuous, which is a contradiction from the hypothesis. Therefore there does not exist a partition and $\mathbb{X}$ is connected.
The exercise is actually equivelant to the exercise Algebra and Topology.
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