Sup(A) less than or equal to sup(B)

[The Captain]

Homework Statement

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

Homework Equations

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

The Attempt at a Solution

If $$k \in B: k \geq s, \: \forall s \in S$$. Then $$k = sup(B)$$.

 

[slider142]

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

 

[The Captain]

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

 

[The Captain]

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

 

[Char. Limit]

Ah ok, I will not give complete answers like that in this forum. I didn't realize.

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

 

[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)?

 

[JeSuisConf]

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)?

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?

 

[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).

 

[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.

 

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

 

[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?

 

[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 $$a \leq b$$.

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

 

[vela]

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).

 

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

 

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

 

[vela]

If $$k \in B: k \geq s, \: \forall s \in S$$. Then $$k = sup(B)$$.

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.

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

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.