| dbp:proof
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	- By contradiction, suppose  is not connected. So it can be written as the union of two disjoint open sets, e.g. . Because  is connected, it must be entirely contained in one of these components, say , and thus  is contained in . Now we know that:
The two sets in the last union are disjoint and open in , so there is a separation of , contradicting the fact that  is connected. (en)
 
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