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.