The forum discussion centers on a problem in Real Analysis regarding the properties of open sets in ℝ. The participants prove that if A is a non-empty open subset of ℝ and both A and its complement are open, then A is unbounded above. They utilize proof by contradiction, showing that if a=sup(A) were in A, it would contradict the openness of A. Furthermore, they demonstrate that if B is defined as the set of elements in A less than or equal to a point in its complement, then B is non-empty and has a lower bound.
PREREQUISITESStudents and educators in mathematics, particularly those focused on real analysis and topology, as well as anyone seeking to deepen their understanding of the properties of open and closed sets in ℝ.
vela said:Why are you saying ##m## is inf(B)? The problem statement only says to show that ##m## bounds B from below and is greater than or equal to ##x##. It's not necessarily the greatest lower bound.