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Around the 4 minute mark the lecturer makes this statement, but I am not convinced this is true. I accept that

(1) if a set is closed, its complement is open.

but consider the converse.

Consider an open ball [itex]S[/itex] of some arbitrary radius centered at the origin (in whatever dimension [itex]d[/itex] you want). [itex]S[/itex] is trivially subset of itself, and its complement (in [itex]S[/itex] that is) is [itex]S^c = S \setminus S[/itex] is the empty set [itex]ø[/itex]. But the empty set is open, implying [itex]S[/itex] is closed by (1), and this is a contradiction since we started with choosing [itex]S[/itex] as open.

What am I missing here? Does it have to do with choosing the complement in [itex]S[/itex] rather than the complement in [itex]\mathbb{R}^d[/itex]?

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# The a set is open iff its complement is closed?

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