# Homework Help: Sup(A) less than or equal to sup(B)

1. Sep 20, 2010

### The Captain

1. The problem statement, all variables and given/known data
Given A and B are sets of numbers, $$A \neq \left\{ \right\}$$, $$B$$ is bounded above, and $$A \subseteq B$$.

Explain why sup(A) and sup(B) exist and why $$sup(A) \leq sup(B)$$.

2. Relevant equations
$$\exists r \in \mathbb R \: : \: r \geq a \: \forall a \in A$$
$$\exists r \in \mathbb R \: : \: A \subset \left[ - \infty , r \right]$$

3. The attempt at a solution
If $$k \in B: k \geq s, \: \forall s \in S$$. Then $$k = sup(B)$$.

Last edited: Sep 21, 2010
2. Sep 20, 2010

### slider142

What is your attempted proof? Ie., where are you having trouble?

3. Sep 20, 2010

### The Captain

Proving $$sup(a) \leq sup(B)$$

4. Sep 20, 2010

### The Captain

Didn't think that it would have been that easy.

5. Sep 20, 2010

### Char. Limit

Not too much of a problem. Just don't do it again and you'll be fine.

6. Sep 20, 2010

### The Captain

So what you're saying is that I'm assuming there is an element in B, such that it is less than or equal to the sup(B), and then taking another element from A and because A is a subset of B, that the element in A is less than or equal to the sup(B)?

7. Sep 20, 2010

### JeSuisConf

Not quite (careful with how you phrase things). The supremum is the least upper bound on the set. But if any element of A is in B, and z=sup(B) is an upper bound on every element of B, then what can you say about the relationship between A and z?

8. Sep 20, 2010

### The Captain

Without proving it, just explaining it:

If some element of A is in B, and the sup(B) is the least upper bound of B, then sup(A) is less than or equal to the sup(B).

9. Sep 20, 2010

### JeSuisConf

Right, so you explained it correctly but that's not a proof because you just restated what you want to prove... instead you want to show the reasoning step by step, where each step follows logically from the other. You'll think the proof is simple once you get it yourself.

10. Sep 20, 2010

### The Captain

Suppose $$b=sup(B), \forall x \in B : x \leq b$$ and $$a=sup(A), \forall y \in A : y \leq a$$.

$$\exists m \in \mathbb R : m \geq b \ \forall b \in B, B \subset \left( - \infty , m \right]$$

Since $$A \subseteq B, \ \forall x \in A$$, then
$$y \leq a \leq x \leq b$$, and since $$a=sup(A)$$ and $$b=sup(B)$$, then $$y, x \in A$$ and $$y \leq x$$, then $$a \leq b$$.

Therefore: $$sup(A) \leq sup(B)$$

11. Sep 20, 2010

### snipez90

Hmmm, I think you can cut down on the number of things to define. Right now you have $sup(A) \leq x$ for all x in A, so that doesn't really make much sense.

Keep the b = sup(B) so for all x in B, $x\leq b$ part. Now take an arbitrary p in A. What does the hypothesis about set containment tell us about p?

12. Sep 20, 2010

### The Captain

Could I word it this way?

Suppose $$\exists b \in B : \ b=sup(B), \ \forall x \in B : x \leq b$$.
Then $$\exists a \in A: \ a=sup(A), \forall y \in A: \ y \leq a$$.

$$\exists m \in \mathbb R : m \geq b \ \forall b \in B, B \subset \left( - \infty , m \right]$$

Since $$A \subseteq B, \ \forall x \in A$$ and $$\forall x \in B$$ , then
since $$x \leq b$$ and $$x \geq a$$, therefore $$a \leq b$$.

$$sup(A) \leq sup(B)$$

13. Sep 20, 2010

### vela

Staff Emeritus
I just wanted to bring up a technical point. It won't affect your proof much. Your definition of sup seems a bit strange. Are you sure that sup(A) has to be an element of A? With your definition, if A=(0,1), sup(A) does not exist because there's no element in (0,1) that's greater than or equal to every element in (0,1).

14. Sep 21, 2010

### The Captain

Assume $$A = \left\{ 1- \frac{1}{n} : \ n \in \mathbb Z^{+} \right\}$$, prove sup(A) = 1.

If 1 is the least upperbound such that $$\forall \epsilon > 0, 1 - \epsilon$$ is not an upperbound of A, then

$$\exist a \in A: \ a \in \left[ 1 - \epsilon , 1 \right)$$, then
$$a = 1 - \frac{1}{n_{0}}$$ for some $$n_{0} \in \mathbb Z^{+}$$

$$1 - \frac{1}{n_{0}} \geq 1 - \epsilon$$
$$- \frac{1}{n_{0}} \geq - \epsilon$$
$$\frac{1}{n_0{0}} \leq \epsilon$$
$$n_{0} \geq \frac{1}{ \epsilon }$$

Therefore, $$sup(A) = 1$$

15. Sep 21, 2010

### The Captain

What if I rewrote the last part as:

Since $$A \subseteq B$$, $$\forall x \in B$$ and $$\forall y \in A$$ then
$$y \leq x$$ so $$a \leq x \leq b$$,
then $$a \leq b$$.

Therefore $$sup(A) \leq sup(B)$$.

16. Sep 21, 2010

### vela

Staff Emeritus
I might have misread your intent. I assumed the above is your definition for sup(B). What is your definition of the supremum? The definition is the key to this problem.
Note that 1 isn't an element of A. The supremum of a subset isn't necessarily in the subset. In many of your attempts at the proof so far, you seem to assume sup(A) is in A and sup(B) is in B.

Last edited: Sep 21, 2010