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Because at each step the open middle third of each remaining intervals is removed, which always leaves the end points of the intervals at the earlier stage in the set, and as end points of the new intervals (one of the end points the other for each interval is created at each step).Explain why all the end points of the closed intervals that comprise $C_n$ are in the Cantor set $C_n$.
I understand why this is true but I don't know how to explain it.