Problem:
This exercise is for those with some knowledge of topology. It will be used in Theorem 6.7.7. Suppose $ M $ is connected. Show that a subset $ Z \subseteq M $ which is both open and closed must be either $ M $ or $ \emptyset $.
Solution:
Suppose the contrary that $ \emptyset \neq A \subsetneq M $ is both open and closed, now we can write $ A \cup (M - A) = M $ where $ A \neq \emptyset $ is both open and closed. By definition, $ M - A \neq \emptyset $ is also open because $ A $ is closed.
But $ M $ is connected, a connected set cannot be written as a disjoint union of two non-empty open sets, so we have reached a contradiction that proved the required proposition.
This exercise is for those with some knowledge of topology. It will be used in Theorem 6.7.7. Suppose $ M $ is connected. Show that a subset $ Z \subseteq M $ which is both open and closed must be either $ M $ or $ \emptyset $.
Solution:
Suppose the contrary that $ \emptyset \neq A \subsetneq M $ is both open and closed, now we can write $ A \cup (M - A) = M $ where $ A \neq \emptyset $ is both open and closed. By definition, $ M - A \neq \emptyset $ is also open because $ A $ is closed.
But $ M $ is connected, a connected set cannot be written as a disjoint union of two non-empty open sets, so we have reached a contradiction that proved the required proposition.
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