Can anyone check my proof involving least-upper-bounds?

[Eclair_de_XII]

Homework Statement

"If $x=sup(S)$, show that for each $\epsilon > 0$, there exists $a∈S$ such that $x-\epsilon < a ≤ x$"

Homework Equations

$x=sup(S)$ would denote the least upper bound for $S$

The Attempt at a Solution

"First, we consider the case where $x=sup(S)∈S$. Then $x=max(S)$, and so there exists an $a∈S$, namely $a=x=max(S)$, such that $x-\epsilon<a≤x$."

"Next we consider the case where $x=sup(S)∉S$. Then because $x$ is the least upper bound for $S$, the interval $(x-\epsilon,x)$ is guaranteed to be a non-empty subset of $S$. Therefore, there exists $a∈S$ such that $a∈(x-\epsilon,x)⊂(x-\epsilon,x]$."

 

[LCKurtz]

Homework Statement

"If $x=sup(S)$, show that for each $\epsilon > 0$, there exists $a∈S$ such that $x-\epsilon < a ≤ x$"

Homework Equations

The Attempt at a Solution

"First, we consider the case where $x=sup(S)∈S$. Then $x=max(S)$, and so there exists an $a∈S$, namely $a=x=max(S)$, such that $x-\epsilon<a≤x$."

"Next we consider the case where $x=sup(S)∉S$. Then because $x$ is the least upper bound for $S$, the interval $(x-\epsilon,x)$ is guaranteed to be a non-empty subset of $S$. Therefore, there exists $a∈S$ such that $a∈(x-\epsilon,x)⊂(x-\epsilon,x]$."

You are just saying the result is true because it is "guaranteed". What are you given for a definition of sup(S)? Whatever it is, you have to use it in your proof. You should give the definition of sup in your "relevant equations" section.

 

[Eclair_de_XII]

What are you given for a definition of sup(S)?

I am told that it is the least upper bound for S, which may or may not be a member of S itself.

I changed my wording so that it now says: "Since $x$ is a least upper bound for $S$, then because $x-\epsilon<x$, $x-\epsilon∈S$. Therefore, $(x-\epsilon,x)$ is a non-empty subset of $S$. Thus, we can conclude that there exists an $a∈S$ such that $a∈(x-\epsilon,x)⊂(x-\epsilon,x]$."

 

[PeroK]

I am told that it is the least upper bound for S, which may or may not be a member of S itself.

I changed my wording so that it now says: "Since $x$ is a least upper bound for $S$, then because $x-\epsilon<x$, $x-\epsilon∈S$. Therefore, $(x-\epsilon,x)$ is a non-empty subset of $S$. Thus, we can conclude that there exists an $a∈S$ such that $a∈(x-\epsilon,x)⊂(x-\epsilon,x]$."

You're making this all far too complicated. You can do both cases together. Hint: Is $x -\epsilon$ an upper bound for $S$?

 

[Eclair_de_XII]

Hint: Is $x -\epsilon$ an upper bound for SS?

I don't think so; I mean, there might still be elements of $S$ higher than $x-\epsilon$ but lower than $x$.

 

[PeroK]

I don't think so; I mean, there might still be elements of $S$ higher than $x-\epsilon$ but lower than $x$.

$x- \epsilon < x$ so it can't be an upper bound, as $x$ is the least upper bound.

 

[LCKurtz]

I am told that it is the least upper bound for S, which may or may not be a member of S itself.

I changed my wording so that it now says: "Since $x$ is a least upper bound for $S$, then because $x-\epsilon<x$, $x-\epsilon∈S$.

That does not follow. Take $\epsilon = \frac 1 2$. $4$ is the least upper bound of $S= \{1,2,3,4\}$ and $4-\epsilon = 3.5 < 4$ but $3.5$ is not in $S$.

 

[Eclair_de_XII]

That does not follow. Take $\epsilon = \frac {1}{ 2}$. $4$ is the least upper bound of $S= \{1,2,3,4\}$ and $4-\epsilon = 3.5 < 4$ but $3.5$ is not in $S$.

What about if I considered my first case? This one, I mean:

"First, we consider the case where $x=sup(S)∈S$. Then $x=max(S)$, and so there exists an $a∈S$, namely $a=x=max(S)$, such that $x−ϵ<a≤x$.

 

[PeroK]

What about if I considered my first case? This one, I mean:

That is correct, as far as it goes.

 

[Eclair_de_XII]

Well, thanks for the feedback, everyone. I'll let you know if my teacher will accept my proof or not, by the end of next Thursday.

 

[LCKurtz]

Why don't you show us your proof and get our comments first?

 

[Eclair_de_XII]

It looks something like:

"First, we consider the case where $x=sup(S)∈S$. Then $x=max(S)$, and so there exists an $a∈S$--namely $a=x=max(S)$--such that $a∈(x-\epsilon,x]⊂S$.

Next, we consider the case where $x=sup(S)∉S$. Then, because $x$ is the least upper bound for $S$, and because $x-\epsilon<x=sup(S)$, it is implied that $x-\epsilon∈S$. Thus, $(x-\epsilon,x)$ is a non-empty subset of $S$. And so, it follows that there exists an $a∈S$ such that $a∈(x-\epsilon,x)⊂(x-\epsilon,x]⊂S$."

 

[LCKurtz]

It looks something like:

"First, we consider the case where $x=sup(S)∈S$. Then $x=max(S)$, and so there exists an $a∈S$--namely $a=x=max(S)$--such that $a∈(x-\epsilon,x]⊂S$.

Next, we consider the case where $x=sup(S)∉S$. Then, because $x$ is the least upper bound for $S$, and because $x-\epsilon<x=sup(S)$, it is implied that $x-\epsilon∈S$. Thus, $(x-\epsilon,x)$ is a non-empty subset of $S$. And so, it follows that there exists an $a∈S$ such that $a∈(x-\epsilon,x)⊂(x-\epsilon,x]⊂S$."

Didn't you understand the counterexample I gave to that implication in post #7?

 

[Eclair_de_XII]

That counterexample doesn't apply just to the first paragraph, then, does it?

 

[LCKurtz]

It never did apply to the first paragraph. I showed you that you can't claim $x - \epsilon \in S$ in the second paragraph. You have to find another argument that there is $a\in S$ in $(x-\epsilon, x)$, which is all you need to prove. My example in post #7 also shows you can't say $(x-\epsilon,x]\subset S$ at the end. Luckily you just need to show the $a$ thing.

 

[Eclair_de_XII]

So $(x-\epsilon,x)$ needn't necessarily be a subset of $S$, then?

 

[PeroK]

So $(x-\epsilon,x)$ needn't necessarily be a subset of $S$, then?

There are sets that are not intervals and do not contain any intervals. A subset of the rationals would do.

Also, epsilon is any positive number, not just a "small" one. So, $x- \epsilon$ might be less than any member of $S$. E.g. $S = (0,1)$, $x =1$ and $\epsilon =10$.

 

[Eclair_de_XII]

Okay, so here's my new argument: If $x=sup(S)$, then it follows that there exists an element $a∈S$ such that $a≤x$. I mean, since $x$ is the least upper bound for some non-empty set S, then there has to be some element $a∈S$ less than $x$. But how can I guarantee that there will exist an $a>x-\epsilon$ for each $\epsilon>0$?

 

[PeroK]

Okay, so here's my new argument: If $x=sup(S)$, then it follows that there exists an element $a∈S$ such that $a≤x$. I mean, since $x$ is the least upper bound for some non-empty set S, then there has to be some element $a∈S$ less than $x$. But how can I guarantee that there will exist an $a>x-\epsilon$ for each $\epsilon>0$?

Is $x-\epsilon$ an upper bound for $S$?

If not, what does that mean?

 

[Eclair_de_XII]

Well... since $x-\epsilon$ is not an upper bound, then $x-\epsilon$ must lie somewhere within $S$, or it must be some lower bound of $S$... $x-\epsilon$ is not necessarily an element of $S$, as has been stated before. But I'm thinking that the fact that $x-\epsilon$ is either a lower bound or lies somewhere in the set $S$ (but is not necessarily an element of $S$) means that the interval $(x-\epsilon,x]$ is non-empty and must contain some element $a∈S$.

 

[LCKurtz]

Okay, so here's my new argument: If $x=sup(S)$, then it follows that there exists an element $a∈S$ such that $a≤x$. I mean, since $x$ is the least upper bound for some non-empty set S, then there has to be some element $a∈S$ less than $x$.

That is false. $S$ might have just one member.

 

[LCKurtz]

Well... since $x-\epsilon$ is not an upper bound, then $x-\epsilon$ must lie somewhere within $S$, or it must be some lower bound of $S$... $x-\epsilon$ is not necessarily an element of $S$, as has been stated before. But I'm thinking that the fact that $x-\epsilon$ is either a lower bound or lies somewhere in the set $S$ (but is not necessarily an element of $S$) means that the interval $(x-\epsilon,x]$ is non-empty and must contain some element $a∈S$.

How can x lie "somewhere within $S$" but not an element of $S$? x is either an element of $S$ or it isn't. And there is nothing given about lower bounds in this problem.
You have never actually stated what the definition of "least upper bound" is. You need to know exactly what that means and use it to get the result you are trying to prove.

 

[Eclair_de_XII]

So according to Wikipedia:

"A real number $x$ is the least upper bound for a non-empty set of real numbers $S$ if $x≥s$ for all $s∈S$, and $x≤y$ for every upper bound $y$ of $S$."

I'm sorry, but I'm clueless as to how to use the fact that $x$ is an upper bound that is equal to or less than every upper bound $y$ of $S$ in my proof. Oh, and there's also the least-upper-bound property, which states that: "Any non-empty set of real numbers $S$ that is bounded from above must have a least upper bound $x∈ℝ$."

 

[PeroK]

So according to Wikipedia:

"A real number $x$ is the least upper bound for a non-empty set of real numbers $S$ if $x≥s$ for all $s∈S$, and $x≤y$ for every upper bound $y$ of $S$."

I'm sorry, but I'm clueless as to how to use the fact that $x$ is an upper bound that is equal to or less than every upper bound $y$ of S in my proof. Oh, and there's also the least-upper bound property, which states that: "Any non-empty set of real numbers $S$ that is bounded from above must have a least upper bound $x∈ℝ$."

I'll be honest, it's difficult to help because you don't seem to be able to grasp the material. This sort of rigorous maths is difficult for most people - to begin with at least.

In any case you need to do one more thing:

Give your definition of an upper bound. Don't look anything up. Try to write your own definition for $y$ is an upper bound for the set $S$

After you've done this, look up the definition.

Does the definition agree with yours? If not, why not? Do you understand what an upper bound is?

 

[Eclair_de_XII]

Try to write your own definition for $y$ is an upper bound for the set $S$

I'll try...

Upper bound: "A real number $y$ is said to be an upper bound for a set $S$ if $y≥s$ for all $s∈S$."

 

[PeroK]

I'll try...

Upper bound: "A real number $y$ is said to be an upper bound for a set $S$ if $y≥s$ for all $s∈S$."

Okay, good. Now, what about defining what it means if $y$ is not an upper bound for $S$.

 

[Eclair_de_XII]

Let's see... "A real number $y$ is not an upper bound for $S$ iff there exists an $s∈S$ such that $s>y$."

I think that $x=sup(S)$ is like the maximum value for a set $S$, $max(S)$, except we include the possibility that $x∉S$.

 

[PeroK]

Let's see... "A real number $y$ is not an upper bound for $S$ iff there exists an $s∈S$ such that $s>y$."

Okay, good. Now, we need to look at the least upper bound, also known as $Sup(S)$. You looked up the definition, which was:

"A real number $x$ is the least upper bound for a non-empty set of real numbers $S$ if $x≥s$ for all $s∈S$, and $x≤y$ for every upper bound $y$ of $S$."

That defintion is sometimes useful, but there is another equivalent definition, which is:

"A real number $x$ is the least upper bound for a non-empty set of real numbers $S$ if $x≥s$ for all $s∈S$, and if $y < x$ then $y$ is not an upper bound of $S$."

This second definition is, in fact, more useful for your current proof.

Now, we go back to the question I asked twice before. If $x = Sup(S)$, then is $x - \epsilon$ an upper bound for $S$?

 

[Eclair_de_XII]

Well, since $x-\epsilon<x$, then by the second definition, it is not an upper bound of $S$.

 

[PeroK]

Well, since $x-\epsilon<x$, then by the second definition, it is not an upper bound of $S$.

Good. Now, go back to your post:

"A real number y is not an upper bound for S iff there exists an s∈S such that s>y."

And put the two together, with $y$ replaced by $x- \epsilon$

 

[Eclair_de_XII]

So, $x-\epsilon$ is not an upper bound for $S$ because there exists an $a∈S$ such that $a>x-\epsilon$.

...Oh, so because $x-\epsilon$ is not an upper bound for $S$, then it is implied that there must exist an element $a∈S$ greater than $x-\epsilon$?

 

[PeroK]

So, $x-\epsilon$ is not an upper bound for $S$ because there exists an $a∈S$ such that $a>x-\epsilon$.

...Oh, so because $x-\epsilon$ is not an upper bound for $S$, then it is implied that there must exist an element $a∈S$ greater than $x-\epsilon$?

Yes!

 

[Eclair_de_XII]

Thanks for bearing with me and helping me finish my homework. I'm kind of apprehensive about how I'll handle the rest of the class, to be honest...

 

[LCKurtz]

So, $x-\epsilon$ is not an upper bound for $S$ because there exists an $a∈S$ such that $a>x-\epsilon$.

...Oh, so because $x-\epsilon$ is not an upper bound for $S$, then it is implied that there must exist an element $a∈S$ greater than $x-\epsilon$?

I like the fact that you italicized the word "must" above. That's the first inkling you have given that you actually get it. Good for you. That's what this forum is here for.