The set G = {[Math] (x, y) \in R^{2} : (x, y) \neq (1, 0)[/Math]} is an open set in [Math]R^{2}[/Math]
Proof 1: An open subset, by definition, of R is a subset E of R such that for every x in E there exists [Math]\epsilon > 0[/Math] such that [Math]B_{\epsilon}(x)[/Math] is contained in E. We will prove this definition with regards to the set G.
Let J = [Math]R^{2} / G[/Math] which implies j is a single point (1, 0) in [Math]R^{2}[/Math] and thus [Math]J \in R^{2}[/Math]. This means that G is everything in [Math]R^{2}[/Math], but the single point (1, 0).
Let a metric space be defined by [Math](R^{2}, d)[/Math] with [Math]z \in G[/Math] where [Math]r \in R^{1}[/Math] and [Math]0 < r < d(j, z)[/Math]. We can define the open ball [Math]B_{r}(z)[/Math] to be the set {[Math]a \in R^{2} | d(a, z) < r[/Math]}. Since [Math]d(j, z) > 0[/Math] then let d = [Math]d(j, z)[/Math]. We can conclude that [Math]d(a, z) < d(j, z) = d[/Math] thus all the points of [Math]B_{r}(z)[/Math] are contained in G. We have shown and proved that G is an open set.
Proof 1: An open subset, by definition, of R is a subset E of R such that for every x in E there exists [Math]\epsilon > 0[/Math] such that [Math]B_{\epsilon}(x)[/Math] is contained in E. We will prove this definition with regards to the set G.
Let J = [Math]R^{2} / G[/Math] which implies j is a single point (1, 0) in [Math]R^{2}[/Math] and thus [Math]J \in R^{2}[/Math]. This means that G is everything in [Math]R^{2}[/Math], but the single point (1, 0).
Let a metric space be defined by [Math](R^{2}, d)[/Math] with [Math]z \in G[/Math] where [Math]r \in R^{1}[/Math] and [Math]0 < r < d(j, z)[/Math]. We can define the open ball [Math]B_{r}(z)[/Math] to be the set {[Math]a \in R^{2} | d(a, z) < r[/Math]}. Since [Math]d(j, z) > 0[/Math] then let d = [Math]d(j, z)[/Math]. We can conclude that [Math]d(a, z) < d(j, z) = d[/Math] thus all the points of [Math]B_{r}(z)[/Math] are contained in G. We have shown and proved that G is an open set.
