How to prove that something is the supremum of a set?

  • Thread starter Thread starter AdrianZ
  • Start date Start date
  • Tags Tags
    Set Supremum
Click For Summary
To prove that 2 + 2√2 is the supremum of the set A = {2x + √2y : 0 < x < 1, -1 < y < 2}, one must show that it is an upper bound and the least upper bound. The initial inequality -√2 < 2x + √2y < 2 + 2√2 confirms that 2 + 2√2 is an upper bound. Since A is bounded, the supremum exists, and the set of upper bounds U = {a in R : a ≥ 2 + 2√2} indicates that 2 + 2√2 is the least upper bound. However, the discussion reflects uncertainty about formalizing the proof, despite understanding the concept of supremum.
AdrianZ
Messages
318
Reaction score
0

Homework Statement


well, the problem asks me to find the supremum(lub) of the set A={2x+sqrt(2)y : 0<x<1 , -1<y<2}. It's easy to show that for any x and y given in the defined domain, we have: -sqrt(2) < 2x+sqrt(2)y< 1+2sqrt(2). well, from this inequality, It's clearly seen that 1+2sqrt(2) is an upper bound for the set. but the question is how can I show that this is the supremum of the set or in other words how can I show that this is the least element of the set of the upper bounds of the set A? I know what a supremum means and stuff like that but I don't know what I should precisely do when a problem asks me to find the supremum of the set.

The Attempt at a Solution



-sqrt(2) < a < 1 +2sqrt(2). for any a in A.
 
Physics news on Phys.org
Did you mistype this? It is easy to see that 1+ 2\sqrt{2} is NOT an upper bound for the set \{2x+ \sqrt{2}y | 0&lt; x&lt; 1, -1&lt; y&lt; 1\} because that set contains 1.99+ 1.99\sqrt{2} which is larger that 1+ 2\sqrt{2}.
I assume you meant 2+ 2\sqrt{2}. (Or else the set is x+ y\sqrt{2}, not 2x+ y\sqrt{2}.)
 
Yea, thanks for noting that. I mistakenly typed 1+sqrt(2)2 instead of 2+2sqrt(2).
Now How can I show that's the least upper bound of the set?
 
You might note that the function has no critical points and thus is increasing or decreasing everywhere, then note that it is increasing, and then note that the upper endpoint is f(1,2) \scriptsize{\text{ which is }} \normalsize 2+2\sqrt2

but this is just a thought, I don't have experience proving something is a supremum of a set. I do know that a function is a set, and this is a function, so it seems reasonable.

Oh, extra thought:
Since the set contains a largest value, the largest value is the supremum. I would include that.
 
Last edited:
ArcanaNoir said:
Since the set contains a largest value, the largest value is the supremum. I would include that.
The sup isn't in A the way A is defined.
 
ArcanaNoir said:
but this is just a thought, I don't have experience proving something is a supremum of a set. I do know that a function is a set, and this is a function, so it seems reasonable.

Yea, that's my problem as well. I know what the supremum of A is. that part is obvious. but I don't know how to write it in a formal way.

vela said:
The sup isn't in A the way A is defined.
True.

well, this is what I've thought of. we know that because A is a bounded subset of R the supremum exists. since for any a in A we have: -sqrt(2)<2x+sqrt(2)y<2+2sqrt(2) then if a>=2+2sqrt(2) or a<=-sqrt(2) then a is not a member of A. therefore the set U={a in R: a>=2+2sqrt(2)} is the set of the upper bounds. by definition of U, It's clear that 2+sqrt(2) is the minimum of the set and therefore is the least upper bound or the supremum that we want.

but I'm still not so satisfied with all these explanations!
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 2 ·
Replies
2
Views
813
  • · Replies 8 ·
Replies
8
Views
8K
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K