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Proof 2: We know a set X is defined to be closed if and only if its complement is open. Let \(\displaystyle J\) = \(\displaystyle R^{2} / G\) which implies \(\displaystyle J\) is a single point (1, 0) in \(\displaystyle R^{2}\). We can prove \(\displaystyle G\) is open if we prove J is closed.

We can use the definition \(\displaystyle \bar A = A \cup A'\) where \(\displaystyle \bar A\) is the closure of \(\displaystyle A\) and \(\displaystyle A'\) is the set of all limit points of \(\displaystyle A\). This means that the closure of a set \(\displaystyle A\) is the union of \(\displaystyle A\) with its boundary points. The closure of a set \(\displaystyle A\) is, also, the smallest closed set containing \(\displaystyle A\).

This implies that \(\displaystyle \bar A\) is closed and we can show J is closed by proving that it is equal to its closure. Let us set \(\displaystyle \bar J = J \cup J'\). By definition of a limit point, let \(\displaystyle b \in S\). A point \(\displaystyle x \in S\), \(\displaystyle x \neq b\) is a limit point of \(\displaystyle b\) if and only if every open neighborhood of \(\displaystyle x\) contains \(\displaystyle b\). Except \(\displaystyle J\) is a singular point and has no other elements \(\displaystyle x\). This implies that \(\displaystyle J\) has no limit points and \(\displaystyle J' = \{\}\), the null set.

This means that \(\displaystyle \bar J = J \cup J' = J \cup \{\} = J\). We now know that J is equal to its closure and thus J is closed. In turn, this means that G is open.