Another Su/pInf problem

[daon]

I think I got most of it, but am stuck at a particular point.

"If A $\displaystyle \subseteq$ B then SupA $\displaystyle \le$ SupB and InfB $\displaystyle \le$InfA."

I broke it down as follows:


If SupA $\displaystyle \in$ A then SupA $\displaystyle \in$ B since for all A $\displaystyle \in$ A, a $\displaystyle \in$ B. By the definition of SupB, SupA $\displaystyle \le$ SupB.

If SupA $\displaystyle \notin$ A then:
. . .If SupA $\displaystyle \in$ B then
. . .. . .by defn of SupB, SupA $\displaystyle \le$ SupB
. . .Else we have SupA $\displaystyle \notin$ B, so
. . .. . .I want to say SupA=SupB, but I can't think of a justification other than my intuition



Similarly, for Inf:


If InfA $\displaystyle \in$ A then InfA $\displaystyle \in$ B since for all A $\displaystyle \in$ A, a $\displaystyle \in$ B. By the definition of InfB, InfB $\displaystyle \le$ InfA.

If InfA $\displaystyle \notin$ A then:
. . .If InfA $\displaystyle \in$ B then
. . .. . .by defn of InfB, InfB $\displaystyle \le$ InfA
. . .Else we have InfA $\displaystyle \notin$ B, so
. . .. . .Like above I want to say InfA=InfB, but I can't think of a valid justification


Thank you,
Daon

 

[JakeD]

I think I got most of it, but am stuck at a particular point.

"If A $\displaystyle \subseteq$ B then SupA $\displaystyle \le$ SupB and InfB $\displaystyle \le$InfA."

I broke it down as follows:


If SupA $\displaystyle \in$ A then SupA $\displaystyle \in$ B since for all A $\displaystyle \in$ A, a $\displaystyle \in$ B. By the definition of SupB, SupA $\displaystyle \le$ SupB.

If SupA $\displaystyle \notin$ A then:
. . .If SupA $\displaystyle \in$ B then
. . .. . .by defn of SupB, SupA $\displaystyle \le$ SupB
. . .Else we have SupA $\displaystyle \notin$ B, so
. . .. . .I want to say SupA=SupB, but I can't think of a justification other than my intuition

Let A = (0,1), B = A U {2}. Sup A = 1 $\displaystyle \notin$ A and $\displaystyle \notin$ B but SupA = 1 $\displaystyle \not =$ Sup B = 2.

My proof: When A $\displaystyle \subset$ B, any upper bound b for B is an upper bound for A because a $\displaystyle \in$ A implies a $\displaystyle \in$ B so a $\displaystyle \le$ b. So b = SupB is an upper bound for A. Since SupA is the least upper bound for A, SupA $\displaystyle \le$ SupB.



Similarly, for Inf:


If InfA $\displaystyle \in$ A then InfA $\displaystyle \in$ B since for all A $\displaystyle \in$ A, a $\displaystyle \in$ B. By the definition of InfB, InfB $\displaystyle \le$ InfA.

If InfA $\displaystyle \notin$ A then:
. . .If InfA $\displaystyle \in$ B then
. . .. . .by defn of InfB, InfB $\displaystyle \le$ InfA
. . .Else we have InfA $\displaystyle \notin$ B, so
. . .. . .Like above I want to say InfA=InfB, but I can't think of a valid justification


Thank you,
Daon

 

[daon]

It seems so obvious now! Thank you for clearing that up for me.