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

• The Captain
In summary: That's not necessarily true.Also, the definition of the supremum is given in the first post. It's not helpful to just say "sup(A)=1" and not explain why it is. You have to explain why 1 is the least upper bound.
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 $$k = sup(B)$$.

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

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

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

JeSuisConf said:
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.

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)?

The Captain said:
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?

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).

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.

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

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?

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 $$a \leq b$$.

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

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).

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, $$sup(A) = 1$$

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

The Captain said:
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.
The Captain said:
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.

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## What does "Sup(A) less than or equal to sup(B)" mean?

This statement means that the supremum (or least upper bound) of set A is less than or equal to the supremum of set B. In other words, the highest possible value in set A is either equal to or lower than the highest possible value in set B.

## How is "Sup(A) less than or equal to sup(B)" different from "Sup(A) less than sup(B)"?

The first statement, "Sup(A) less than or equal to sup(B)", allows for the supremum of set A to be equal to the supremum of set B. The second statement, "Sup(A) less than sup(B)", only allows for the supremum of set A to be strictly less than the supremum of set B.

## What does it mean for two sets to have equal suprema?

If two sets have equal suprema, it means that the highest possible value in both sets is the same. In other words, both sets have the same upper bound. This does not necessarily mean that the two sets are identical, as they could have different elements and still have equal suprema.

## Can the supremum of a set be negative?

Yes, the supremum of a set can be negative. The supremum of a set is simply the highest possible value in that set, regardless of whether it is positive or negative. For example, the set {-3, -2, -1} has a supremum of -1.

## What is the importance of knowing "Sup(A) less than or equal to sup(B)"?

Knowing that the supremum of set A is less than or equal to the supremum of set B can be useful in understanding the relationship between two sets. It can also help in proving the existence of certain values within a set, such as the least upper bound or the greatest lower bound. Additionally, this knowledge can be applied in various mathematical and scientific contexts, such as in the study of limits and inequalities.

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