discontinuity of characteristic function
Proof. Let be a discontinuity point of . Then any neighborhood (http://planetmath.org/Neighborhood) of contains the points and such that and . Thus and , whence is a boundary point of .
If, on the contrary, is a boundary point of and an arbitrary neighborhood of , it follows that contains both points belonging to and points not belonging to . So we have in the points and such that and . This means that cannot be continuous at the point (N.B. that one does not need to know the value ).
|Title||discontinuity of characteristic function|
|Date of creation||2015-02-03 21:23:33|
|Last modified on||2015-02-03 21:23:33|
|Last modified by||pahio (2872)|