# Suprenum Question

1. Nov 30, 2005

### Bob19

Hello

I have two non-empty sets A and B wich is bounded above by R.

Then I'm tasked with proving that

$$sup(A \cup B) = max(sup A, sup B)$$

which supposedly means that $$sup(A \cup B)$$ is the largest of the two numbers sup A and sup B.

Can this then be written as $$sup(A) < sup(A \cup B)$$ and $$sup(B) < sup(A \cup B)$$ ???

Can this then be proven by showing that $$sup(A) < sup(A \cup B)$$ is true?

Or am I totally on the wrong path here??

/Bob

Last edited: Nov 30, 2005
2. Nov 30, 2005

### matt grime

correct your tex and just verify the definitions of sup.

3. Nov 30, 2005

### Bob19

My definition of Supremum is a follows:
every non-empty, bounded above subset of R has a smallest upper bound.

Then $$sup(A \cup B)$$ has a larger smallest upper bound than sup(A) and sup(B) according to the definition of Supremum ???
Does this prove the given argument in my first post?
/Bob

p.s. If my idear is true, can this then be proven by taking a number z, which I then prove $$z \in sup(A \cup B)$$ but $$z \notin sup(A)$$ and $$z \notin sup(B)$$ ???

/Bob

Last edited: Nov 30, 2005
4. Nov 30, 2005

### matt grime

why are you treating sup as a set (and taking elements in it?). Sup is not a set, it is an element of R.

Sup of a set is the least upper bound (when it exists)

obivously the least upper bound of AuB is the max of the least upper bounds, but you need to verify it, ie show it is an upper bound, and show it is the least upper bound. The first is easy, the second slightly harder.

5. Nov 30, 2005

### Bob19

Okay those two aspects then prove the argument that

sup(AuB) = max(sup A, sup B) ????

/Bob

6. Nov 30, 2005

### matt grime

As I explained to someone else earlier tonight, I can easily see the answer becuase of experience, YOU need to demonstrate that you understand the answer by not having to let me fill in any blanks. If you don't see that an argument proves something then YOU need to do some work to rectify that, not me.