# 3 part sup/inf question

1. Apr 26, 2013

### Zondrina

1. The problem statement, all variables and given/known data

The question : http://gyazo.com/e2dfc19ce0bc2ff857019edfc3f42edb

2. Relevant equations

Supremum and infimum axioms.

3. The attempt at a solution

Suppose $A = \{p,q \space | \space p,q \in ℝ, \space p ≠ q\}$ is a bounded subset of $ℝ$.

(a) Since $A$ is bounded, we know it is bounded above by p, q or ∞ and we also know it is bounded below by p, q or -∞

W.L.O.G, lets assume that $p > q$. So $p$ is an upper bound for $A$, but $A$ is also bounded above by $∞$. Since $p < ∞$, we may write $sup(A) = p < ∞$.

Similarly, $q$ is a lower bound for $A$, but $A$ is also bounded below by $-∞$. Since $p > -∞$, we may write $inf(A) = q > -∞$.

Putting these together we observe $-∞ < inf(A) < sup(A) < ∞$.

(b) Suppose $B$ is a nonempty subset of $A$.

Since $inf(A)$ is a lower bound of $A$, we have $inf(A) ≤ a, \space \forall a \in A$.

In particular, $inf(A) ≤ b, \space \forall b \in B$ since $B \subseteq A$. Thus $inf(A)$ is a lower bound of $B$, so $inf(A) ≤ inf(B)$.

Picking any $b \in B$ shows that $inf(B) ≤ b ≤ sup(B)$, which gives the second inequality.

Finally, since $sup(A)$ is an upper bound of $A$, we know $\space a ≤ sup(A), \space \forall a \in A$.

Hence $b ≤ sup(A) \forall b \in B \Rightarrow sup(A)$ is an upper bound of $B$. Thus $sup(B) ≤ sup(A)$ since $sup(B)$ is the least upper bound of $B$.

Therefore putting these all together, we have our desired inequality.

(c) Suppose $B = \{ b \in ℝ \space | \space b ≥ a, \space \forall a \in A \}$

Since $A ≠ ∅$ and is bounded, we know $sup(A)$ exists so $B ≠ ∅$. ( Don't think this is right ).

Suppose $a \in A$. Since $a ≤ b, \space \forall b \in B$, we know $b$ is a lower bound of $B$, so $B$ is bounded below.

I'm having a bit of trouble proving $inf(B) = sup(A)$ at this point.

2. Apr 27, 2013

### Mandelbroth

I thought this looked interesting like an interesting question.
Let's start with this: why do you doubt that $B\neq \varnothing$?

Here's my hint for your trouble spot:
To show that $\inf(B) = \sup(A)$, I'd show that $A\cap B \neq \varnothing$. How do you think you might go about doing that?

3. Apr 27, 2013

### micromass

Staff Emeritus
I'm having serious difficulties grasping your notation? Why would $A$ have that form you write?? All you know is that $A$ is a very arbitrary set that contains at least two elements. You seem to think that $A$ is a set with exactly two elements.

4. Apr 27, 2013

### micromass

Staff Emeritus
Well, since it isn't true, I think he would have quite some difficulties proving it.

5. Apr 27, 2013

### kramer733

This is from NL Carother's Real analysis. Question 2 of chapter 1

6. Apr 28, 2013

### Mandelbroth

If one element of A and one element of B are the same (namely, the infimum of B and the supremum of A) then sets A and B share at least one element. Thus, the intersection of A and B is nonempty. What am I missing that would make that incorrect?

7. Apr 28, 2013

### micromass

Staff Emeritus
Supremum of $A$ is not necessarily in $A$.
Take $A=(0,1)$. Then $A\cap B = \emptyset$.

8. Apr 28, 2013

### Zondrina

Now that I feel slightly more comfortable with the material, I'll attempt this again.

Let A be a bounded subset of the reals containing at least two points.

(a) Prove -∞ < inf(A) < sup(A) < ∞.

Since A is a bounded subset of reals, we know $m ≤ a ≤ M, \space \forall a \in A$

Since A is nonempty ( since it contains at least 2 points ) and it is bounded above and below, we know sup(A) and inf(A) both exist. So we get $m ≤ inf(A) ≤ a ≤ sup(A) ≤ M$.

I'm not sure how ±∞ comes into play here, not sure if I'm allowed to be assuming things.

9. Apr 28, 2013

### LCKurtz

More precisely: Since A is a bounded subset of reals, we know there exist real numbers m and M such that $m ≤ a ≤ M, \space \forall a \in A$

Once you have the existence of m and M, that last inequality shows the sup and inf aren't infinite doesn't it?

10. Apr 28, 2013

### Zondrina

Yes this is what I was thinking. Since m ≤ inf(A) ≤ a and a ≤ sup(A) ≤ M, we know inf(A) and sup(A) must be finite and therefore bounded below and above by -∞ and +∞ respectively.

11. Apr 28, 2013

### LCKurtz

My point was that you need to include the red statement in your argument. You have to say where the m and M come from. And that last sentence I struck out is pointless. You aren't trying to show that "therefore it is bounded by $\pm\infty$".

12. Apr 28, 2013

### Zondrina

Ohh I see now. I just really need to confirm something now then as it's been confusing me for a while now.

When we say a set of real numbers is bounded above, does it mean a < M or a ≤ M? If it's a < M then we get m < inf(A) ≤ a ≤ sup(A) < M do we not ( a > m or a ≥ m )?

13. Apr 28, 2013

### micromass

Staff Emeritus
Those two are equivalent. Try to prove it.

14. Apr 28, 2013

### LCKurtz

The statement that the set A of real numbers is bounded above means that there exists a real number M such that, for all $a\in A$, $a\le M$. It doesn't matter whether you say $a\le M$ or $a<M$ in that definition although $\le$ is usually used. They describe the same sets.

The M and m given by the definition of boundedness have nothing to do with the sup and inf. They are just bounds for the set. They might be extra big or extra small. They might accidentally be the sup or inf. They are just a couple of bounds.

15. Apr 28, 2013

### Zondrina

The endpoints don't matter since we can get arbitrarily close to them. It's just I have a hard time digesting it no matter how true I know it is.

I see why adding that phrase in red solidifies the argument though LC, since it proves that sup(A) and inf(A) can't possibly be infinite as they are bounded.

With that in mind I think my part (b) looks okay, so I'll see if I can get part (c) done.

(c) Suppose $B = \{b \in ℝ \space | \space b ≥ a, \space \forall a \in A \}$

Using part (a), we know m ≤ inf(A) ≤ a ≤ sup(A) ≤ M and we know that $M \in B$ so that $B ≠ ∅$. Since sup(A) exists, we know it must be a lower bound for the set of all upper bounds, hence sup(A) is a lower bound for B. We know any $a \in A$ is a lower bound for for B, but a ≤ sup(A), so by the inf axiom using these facts we can write inf(B) = sup(A) as desired.

16. Apr 28, 2013

### LCKurtz

I agree your part b looks OK. Probably your best argument I have seen so far.

Agreed.

That last line isn't convincing. "using these facts" isn't a clear argument. Since sup(A) is an upper bound for A, it is an element of B. And you have shown above that sup(A) is a lower bound for B. Use those to show inf(B) = sup(A).