The discussion focuses on proving that an open interval I, represented as (a, b), is an open set in the context of real analysis. The proof utilizes the concept of neighborhoods, specifically defining r as the minimum of b-x and x-a for any x within the interval. This approach confirms that (x-r, x+r) is a subset of (a, b), establishing x as an interior point of the interval. Additionally, the relationship between open and closed sets is acknowledged, emphasizing that if one set is open, its complement is closed.
PREREQUISITESMathematics students, particularly those studying real analysis and topology, as well as educators seeking to deepen their understanding of set properties and proofs.
gauss^2 said:You have really only changed the problem from showing that one set is open to showing that two are closed; you'll need to use neighborhoods either way. Much simpler to just take [itex]x \in (a,b)[/itex] (assuming [itex]a < b[/itex]), take [itex]r[/itex] to be the minimum of [itex]b-x[/itex] and [itex]x-a[/itex] so that [itex]x - r >= x - (x-a) = a[/itex] and [itex]x + r <= x + (b-x) = b[/itex]. Then [itex](x-r, x+r) \subset (a,b)[/itex] holds so that [itex]x[/itex] is an interior point of [itex](a,b)[/itex].