(Theorem 3.2.13) A set \(O\) is open if and only if \(O^c\) is closed. Likewise, a set \(F\) is closed if and only if \(F^c\) is open.


Proof

Let \(O \in \mathbf{R}\) be an open set. We need to prove that \(O^c\) is a closed set. To do so, we need to show that \(O^c\) contains all of its limit points. Let \(x\) be an arbitrary limit point of \(O^c\). Now suppose for the sake of contradiction that \(x \notin O^c\) so \(x \in O\). By the definition of an open set, this means that there exists some \(\delta\)-neighborhood, \(V_{\delta}(x)\), of \(x\) such that \(V_{\delta}(x) \subseteq O\). By the definition of a limit point, there exists a sequence \((a_n)\) such that \((a_n) \rightarrow x\) and \(a_n \in O^c\) and \(x\) is not a term of the sequence. This implies that for any \(\epsilon > 0\), we must have an \(N \in \mathbf{N}\) such that when \(n > N\), \(|a_n - x| \leq \epsilon\). But if we set \(\epsilon = \delta\), then \(a_n \in (x - \delta, x + \delta)\). This means that \((a_n)\) could not possibly converge since the terms in that interval are in \(O\) and not \(O^c\) because we earlier said that \(V_{\delta}(x) \subseteq O\). This is a contradiction and so \(x \in O\).

For the other direction. Suppose that \(O^c\) is closed. We need to prove that \(O\) is open. Then by the definition of an open set, given an element \(x \in O\), we need to produce an \(\epsilon\)-neighborhood of \(x\) such that \(V_{\epsilon}(x) \subseteq O\). Since we assumed that \(O^c\) is closed, then \(O^c\) must contains all of its limits points and so \(x\) is not a limit point of \(O^c\). The definition of a limit point states that if \(x\) was a limit point, then every \(\epsilon\)-neighborhood of \(x\) intersects \(O^c\) at some point other than \(x\) but \(x\) is not a limit point of \(O^c\) and so there exists some \(V_{\epsilon}(x)\) that doesn’t intersect \(O^c\) and therefore \(V_{\epsilon}(x) \in O\).

For the second statement, it follows from the fact that \((O^c)^c = O\). \(\blacksquare\)

Other Definitions and Properties


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