Looks to me like you just have to write the definition of inf and sup.
a is not in E, but is greater than any other number less than any element of E.
Let a,b exist in the reals with a<b.
(i) for the open interval E=(a,b), prove that infE=a and supE=b.
(ii) for the closed interval F=[a,b], prove that the infF=a and infF=b.
2 answers
Can you say:
Since x in E then E is not empty and is a subset of the reals. Since there exist and x in E where a<x then a is a lower bound of E. then let lambda be a different lower bound in E then let a precede lambda. this can't occur because we stated that a is a lower bound of E thus lambda precedes a. and thus infE=a. prove supE=b similarly.
The for the closed set, since a and b are in E if there exist and x in E its less than or equal to b and greater than or equal to a. then assume like the one above that there is another element in E where its either preceded for is preceded by a or b?
Since x in E then E is not empty and is a subset of the reals. Since there exist and x in E where a<x then a is a lower bound of E. then let lambda be a different lower bound in E then let a precede lambda. this can't occur because we stated that a is a lower bound of E thus lambda precedes a. and thus infE=a. prove supE=b similarly.
The for the closed set, since a and b are in E if there exist and x in E its less than or equal to b and greater than or equal to a. then assume like the one above that there is another element in E where its either preceded for is preceded by a or b?
0 answers