Thread: Question on Bounded Set

1. Question on Bounded Set

This is a rather easy problem but I'm not very sure if my proof is correct.

Please check for any lapses in logic.

Thank you in advance.

Show that for any subset A of R, sup A is unique.

pf) Suppose two sup A exist. b and b1

Def) b is a element in Ua.
b<= a for all element a that are in set Ua

b1 is a element in Ua
b1<=a for all element a that are in Ua

i) b<b1<a
Only b can be sup A. Contradiction

ii) b1<b<a
only b1 can be sup A. Contradiction.

Therefore b=b1.

2. Re: Question on Bounded Set

Originally Posted by rokman54
Show that for any subset A of R, sup A is unique.

pf) Suppose two sup A exist. b and b1
Stylistically, it is better to call them b1 and b2, or b and b'.

Originally Posted by rokman54
Def) b is a element in Ua.
b<= a for all element a that are in set Ua
You have not defined Ua. I assume it is the set of all upper bounds of A. Then why is the subscript lowercase a?

Originally Posted by rokman54
b1 is a element in Ua
b1<=a for all element a that are in Ua

i) b<b1<a
I assume you are considering two cases: b < b1 and b1 < b (b = b1 is impossible by assumption). Then you should say so. The problem is that the a you are using here is undefined: I have no idea what it is.

Originally Posted by rokman54
Only b can be sup A. Contradiction

ii) b1<b<a
only b1 can be sup A. Contradiction.
Same remark about a.

3. Re: Question on Bounded Set

Thanks for your response.

pf) Suppose two sup A exist. b1 and b2

Def) b1 is a element in UA.
b1<= a for all element a that are in set UA (UA is the set of all upper bounds for A)

b1 is a element in UA
b1<=a for all element a that are in UA

b2 is a element in UA
b2 <=a for all element a that are in UA

i) b1<b2<a
Only b1 can be sup A. Contradiction

ii) b2<b1<a
only b2 can be sup A. Contradiction.

Therefore b1=b2.

4. Re: Question on Bounded Set

Originally Posted by rokman54
b1 is a element in UA
b1<=a for all element a that are in UA

b2 is a element in UA
b2 <=a for all element a that are in UA

i) b1<b2<a
Only b1 can be sup A. Contradiction

ii) b2<b1<a
only b2 can be sup A. Contradiction.

Therefore b1=b2.
Again, writing just i) and ii) does not say what the relationship between these two subproofs are. Maybe they represent two steps that are supposed to be done consecutively. In this proof, this is not the case: i) and ii) are two alternatives. You should say so for the sake of the reader.

And you still use a in "b2<b1<a" without defining or introducing it. Imagine I say, "Every person in this room is a winner of a math Olympiad. Look at this person". You would naturally ask, "Which person?" This is because the scope of the first word "person" is restricted to the first sentence. In more detail, the first sentence says, "For every person p in this room, p is a math Olympiad winner". The phrase "For every p" introduces p; it makes p range over all people in the room. The second reference to p is legitimate; it occurs in the scope of "For every p". However, this scope ends with the end of the sentence. If I refer to p in the following sentence, it would be an error because p is no longer defined.

Similarly, in this problem, when you say "b1<=a for all element a that are in UA", the scope of a is this sentence only. After the sentence ends, references to a are not legitimate. Therefore, saying b1<b2<a is an error because it is not clear which a you have in mind. If you want to say something about all a ∈ UA, you need either repeat the phrase "For all a ∈ UA" or say, "Consider an arbitrary a ∈ UA. Then ...".

This was a long technical remark. More to the point, since the problem is easy, it would be better to provide more explanation to why b1 < b2 implies that b2 is not a sup A.

Usually the fact that b1 = b2 is proved directly (i.e., not by contradiction). Since b2 ∈ UA and b1 <= a for all a ∈ UA, we have b1 <= b2. Similarly, b2 <= b1, and since <= is antisymmetric, b1 = b2.

5. Re: Question on Bounded Set

Thank you for your help.