Nonempty subset of reals which is bounded

[STEMucator]

Homework Statement

The problem : http://gyazo.com/aa487398b3658600b98deabca8086334

Homework Equations

The Attempt at a Solution

Let A be a nonempty subset of reals which is bounded above.

$("\Rightarrow")$ Assume $sup(A)$ exists, call it s.

Since s exists, we know $a ≤ s, \space \forall a \in A$ so that s is an upper bound for a.

So, $\forall ε > 0$ we know we can find $a \in A$ such that $a > s - ε$ because $s - ε$ is not an upper bound for A for any $ε$. This tells us there is some $a \in A$ which is an upper bound for A so that $a ≥ s$.

Therefore if sup(A) exists, it must satisfy (i) and (ii).

$("\Leftarrow")$ Assume (i) and (ii) both hold and recall A is a nonempty subset of reals which is bounded above.

So there is some number s which is an upper bound for A ( so s ≥ a ) and $\forall ε > 0$ we know we can find $a \in A$ such that $a > s - ε$ which tells us that $a + ε > s$ so that s is bounded above by $a + ε$

Hence (i) and (ii) together both imply that sup(A) = s.

As for the corresponding result for the infimum :

$inf(A) = s$
$⇔$
(i). s is a lower bound for A.
(ii). $\forall ε > 0, \exists a \in A \space | \space a < s - ε$

 

[christoff]

$\Rightarrow$: I don't quite follow you. s is definitely an upper bound; that's OK. However, you need to actually show how to find a, or at least prove it exists. s-ε is not an upper bound, I agree with you there. However, why does this mean

there is some a∈A which is an upper bound for A so that a≥s.

And how does this imply the existence of said epsilon?

As for $\Leftarrow$, I don't think you're quite there yet. What you have written is that given any $a\in A,$ there exists $\epsilon>0$ such that $s<a+\epsilon$. This doesn't mean that s (what should be the sup) is a least upper bound; you need to show that given any other upper bound for $A$, say $s'$, you have $s\leq s'$. I don't see how the previous statement directly implies this.

If you have another upper bound, you could go by contradiction, assuming that $s'<s$ (ie, s is not the sup). Then, consider the difference $s-s'>0$. This number could be very large, or it could be very small. How can you relate this quantity to the original problem? What do you know about $s$?

Hint below.

Spoiler

Hint: set $\epsilon=s-s'$

 

[STEMucator]

$\Rightarrow$: I don't quite follow you. s is definitely an upper bound; that's OK. However, you need to actually show how to find a, or at least prove it exists. s-ε is not an upper bound, I agree with you there. However, why does this mean And how does this imply the existence of said epsilon?

As for $\Leftarrow$, I don't think you're quite there yet. What you have written is that given any $a\in A,$ there exists $\epsilon>0$ such that $s<a+\epsilon$. This doesn't mean that s (what should be the sup) is a least upper bound; you need to show that given any other upper bound for $A$, say $s'$, you have $s\leq s'$. I don't see how the previous statement directly implies this.

If you have another upper bound, you could go by contradiction, assuming that $s'<s$ (ie, s is not the sup). Then, consider the difference $s-s'>0$. This number could be very large, or it could be very small. How can you relate this quantity to the original problem? What do you know about $s$?

Hint below.

Spoiler

Hint: set $\epsilon=s-s'$

When I write the arrow in such a way, I mean assume p and show it implies q. For example, assume sup(A) = s and show it implies (i) and (ii) hold.

Lets fix $\Rightarrow$ first I guess. I was given that : Let A be a nonempty subset of reals which is bounded above. That's why (i) was very quick and painless.

So the problem is with (ii)? I see where my wording went wrong in my first attempt, let me try again.

We must show $\forall ε > 0, \exists a \in A \space | \space a > s - ε$.

Since (i) was true, we know that $s - ε$ is not an upper bound for A. So we have that $s - ε ≤ a ≤ s$

So choosing $ε = s - a$ yields $a > s$. This show us that s is bounded above for some $a \in A$. Therefore (ii) holds.

 

[christoff]

Since (i) was true, we know that s−ε is not an upper bound for A. So we have that s−ε≤a≤s

In this sentence, what is $a$? Is it some element of $A$?

So choosing ε=s−a yields a>s.

You cannot choose $\epsilon$; your epsilon must in some way allow you to choose $a$ such that $s-\epsilon<a$. If you mean "take $a=s-\epsilon$", then this is more correct, but in general it doesn't work because your $a$ might not actually be in your set $A$. Consider the following example: $A=[0,1],$ $\epsilon=2$. Then $s=sup(A)=1$, and your choice of a would give you $a=1-2=-1\notin A$.

This show us that s is bounded above for some a∈A. Therefore (ii) holds.

If s were bounded above by some element of A in full generality, then we would be in big trouble. Consider the open interval $A=(0,1)$. The supremum of this is $s=1$, but this is clearly not bounded by ANY element of A.

With these kind of existence results (and that's what it is, you need to show that something exists, and in this case, it is an element of your set $A$ that satisfies the given inequality), it is often much easier to go by contradiction. Let me get you started.

Assume that condition ii) fails. What does this mean? We have to negate the statement:
"$\forall \epsilon>0, \exists a\in A$ such that $s-\epsilon<a$."
The negation of this is:
"$\exists \hat\epsilon>0$ such that $\forall a\in A$, $s-\hat\epsilon\geq a$".
Here, I'm using $\hat\epsilon$ with the hat to emphasize that it's a "bad" epsilon. So, we have the inequality $s-\hat\epsilon\geq a$, where this inequality holds for ALL $a\in A$. This is a contradiction; can you see why?

Spoiler

Hint: remember that s is supposed to be the supremum, and that the "bad" epsilon is strictly positive; $\hat\epsilon\neq 0$

 

[STEMucator]

In this sentence, what is $a$? Is it some element of $A$?

You cannot choose $\epsilon$; your epsilon must in some way allow you to choose $a$ such that $s-\epsilon<a$. If you mean "take $a=s-\epsilon$", then this is more correct, but in general it doesn't work because your $a$ might not actually be in your set $A$. Consider the following example: $A=[0,1],$ $\epsilon=2$. Then $s=sup(A)=1$, and your choice of a would give you $a=1-2=-1\notin A$.
If s were bounded above by some element of A in full generality, then we would be in big trouble. Consider the open interval $A=(0,1)$. The supremum of this is $s=1$, but this is clearly not bounded by ANY element of A.

With these kind of existence results (and that's what it is, you need to show that something exists, and in this case, it is an element of your set $A$ that satisfies the given inequality), it is often much easier to go by contradiction. Let me get you started.

Assume that condition ii) fails. What does this mean? We have to negate the statement:
"$\forall \epsilon>0, \exists a\in A$ such that $s-\epsilon<a$."
The negation of this is:
"$\exists \hat\epsilon>0$ such that $\forall a\in A$, $s-\hat\epsilon\geq a$".
Here, I'm using $\hat\epsilon$ with the hat to emphasize that it's a "bad" epsilon. So, we have the inequality $s-\hat\epsilon\geq a$, where this inequality holds for ALL $a\in A$. This is a contradiction; can you see why?

Spoiler

Hint: remember that s is supposed to be the supremum, and that the "bad" epsilon is strictly positive; $\hat\epsilon\neq 0$

I see, so I can't choose $ε$ like I did in my previous post.

So negating the statement yields $\exists \hat\epsilon>0$ such that $\forall a\in A$, $s-\hat\epsilon\geq a$

Now, we know that s = sup(A) so that s is an upper bound for A, but $s-\hat\epsilon$ is not an upper bound for A so it cannot possibly bound every $a \in A$ which contradicts the assumption that $\exists \hat\epsilon>0$ such that $\forall a\in A$, $s-\hat\epsilon\geq a$.

Thus (ii) actually holds so $("\Rightarrow")$ holds.

Ill try fixing $("\Leftarrow")$ now.

Suppose (i) and (ii) both hold. We want to show that s = sup(A).

Suppose by contradiction that s ≠ sup(A) so that s is an upper bound for A. Suppose we have another upper bound for A, let's call it $s'$. We want $0 ≤ s - s'$ ( i.e s' is the supremum ).

Now take $ε = s - s'$ then (ii) yields $a > s'$, but this is a contradiction because s' is an upper bound for the elements of A.

So it must be the case that s = sup(A) holds so $("\Leftarrow")$ holds.

 

[christoff]

Now, we know that s = sup(A) so that s is an upper bound for A, but s−ϵ^ is not an upper bound for A so it cannot possibly bound every a∈A

Very close, but there's a bit missing. You say that $s-\hat\epsilon$ is not an upper bound for A. Here, you seem to be assuming what you want to prove. The reason I say this is that, in fact, by assuming that the statement was false (ie. by attempting to go by contradiction), we came up with another upper bound for $A$, namely $bound=s-\hat\epsilon\geq a$ for all $a\in A$.

However, $s$ is the supremum, so it is a least upper bound; what this means is the following: If $b$ is an upper bound for $A$, then $sup(A)\leq b$.

So since $bound=s-\hat\epsilon$ is an upper bound, the above tells us that we should have $$sup(A)=s\leq s-\hat\epsilon.$$ But this is a contradiction, because $\hat\epsilon$ is positive (subtract $s$ from both sides and you will obtain $0\leq -\hat\epsilon$, which implies $\hat\epsilon<0$, a contradiction).

Suppose by contradiction that s ≠ sup(A) so that s is an upper bound for A. Suppose we have another upper bound for A, let's call it s′. We want 0≤s−s′ ( i.e s' is the supremum ).

So you are saying that $s'\neq s$ is the "actual" supremum? OK.

We want 0≤s−s′

Actually, we are given that inequality because we are assuming that s' is the supremum (so in particular, this isn't what we "want", it's what we already have have). Also, we need 0<s-s', since we need that epsilon to be strictly positive. To be clear, we always have this particular inequality because we are assuming that s≠s' (and this plus s' being a supremum gives us 0<s-s'). It isn't a big problem, but I thought I would just mention it.

The rest is ok. Good work :)

 

[STEMucator]

Very close, but there's a bit missing. You say that $s-\hat\epsilon$ is not an upper bound for A. Here, you seem to be assuming what you want to prove. The reason I say this is that, in fact, by assuming that the statement was false (ie. by attempting to go by contradiction), we came up with another upper bound for $A$, namely $bound=s-\hat\epsilon\geq a$ for all $a\in A$.

However, $s$ is the supremum, so it is a least upper bound; what this means is the following: If $b$ is an upper bound for $A$, then $sup(A)\leq b$.

So since $bound=s-\hat\epsilon$ is an upper bound, the above tells us that we should have $$sup(A)=s\leq s-\hat\epsilon.$$ But this is a contradiction, because $\hat\epsilon$ is positive (subtract $s$ from both sides and you will obtain $0\leq -\hat\epsilon$, which implies $\hat\epsilon<0$, a contradiction).So you are saying that $s'\neq s$ is the "actual" supremum? OK.Actually, we are given that inequality because we are assuming that s' is the supremum (so in particular, this isn't what we "want", it's what we already have have). Also, we need 0<s-s', since we need that epsilon to be strictly positive. To be clear, we always have this particular inequality because we are assuming that s≠s' (and this plus s' being a supremum gives us 0<s-s'). It isn't a big problem, but I thought I would just mention it.

The rest is ok. Good work :)

Yes this makes much more sense. I was actually lingering on the notion that 0 ≤ s-s' because it included the possibility of $ε = 0$. I'll clean up everything into one post for the sake of completeness.

$\Rightarrow$ Assume sup(A) = s, we must show that (i) and (ii) hold.

Since sup(A) = s, we know that $a ≤ s, \space \forall a \in A$ so that s is an upper bound for A. Hence (i) is satisfied.

For (ii), let's assume the contrary that $\exists ε > 0 \space | \space \forall a \in A, \space a ≤ s - ε$. This means that $s - ε$ is an upper bound $\forall a \in A$. Hence we have the inequality $s ≤ s - ε$, but this is a contradiction because we know that $ε > 0$. Therefore it must be the case that (ii) holds so that $\Rightarrow$ holds.

$\Leftarrow$ Suppose (i) and (ii) both hold, we must show sup(A) = s.

Suppose by way of contradiction that $sup(A) ≠ s$ so that s is just an upper bound for A. Since A is a nonempty subset of the reals which is bounded above, it has a supremum, let's call it $s'$ so that $a ≤ s', \space \forall a \in A$. Hence we have the inequality $s' < s$ or equivalently $0 < s - s'$ because s is an upper bound for A.

Since (ii) is assumed to hold, we know for $ε = s - s'$ we have $a > s'$ which is a contradiction because we just stated that s' = sup(A). Therefore it must be the case that sup(A) = s so that $\Leftarrow$ holds.

$∴ sup(A) = s ⇔ (i) and (ii) hold$.

The corresponding result for the infimum :

$inf(A) = s$
$⇔$
(i). s is a lower bound for A.
(ii). $\forall ε > 0, \exists a \in A \space | \space a < s - ε$