# Proving that $\mu$ is $\sup S$

Suppose a set $S$ of real numbers is bounded and let $\mu$ be an upper bound for $S$. Show that $\mu$ is the least upper bound of $S$ $\Longleftrightarrow$ for every $\epsilon > 0$ there is an element of $S$ in the interval $[\mu - \epsilon, \mu]$.

My Work

($\Rightarrow$)

If there is no element of $S$ in the interval $[\mu - \epsilon, \mu]$, then $\mu - \epsilon$ could also be an upper bound for $S$, but since $\mu = \sup S$ and $\mu - \epsilon < \mu$ there is a contradiction.

($\Leftarrow$)

Considering the least upper bound of $S$, I need to show that it cannot be smaller than $\mu$. But I am not sure how to do this using the condition I am given.

Edit

By the definition of a supremum, if $\lambda$ is another upper bound of $S$, then if $\mu = \sup S \Rightarrow \mu \le \lambda$. So proof by contradiction, assuming that $\mu \ne \sup S$, does that mean that there is an element $\lambda \in [\mu - \epsilon, \mu], \lambda \notin S, \lambda < \mu$? How does this prove that there is no element of $S$ in $[\mu - \epsilon, \mu]$ for every $\epsilon > 0$?

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Assume that for some $\epsilon >0$, $\not{\exists} x \in [\mu -\epsilon , \mu]$, such that $x\in S$. This implies that $\exists \lambda < \mu$ where $\lambda \not \in S \implies \mu \ne \sup S$. So if $\mu = \sup S$ then has to be an element of S in the interval $[μ−ϵ,μ]$ – funktor Aug 14 '12 at 15:19
@BiditAcharya why does $\lambda \notin S \Longrightarrow \mu \ne \sup S$ – Moderat Aug 14 '12 at 15:23
$\lambda \not \in S$ means that $\lambda$ is a upper bound for $S$ which is less than $\mu$. If you remember, one of the properties of the least upper bound is that if there exists a quantity that is less than the least upper bound (say $l$), that quantity has to be in the set under consideration. Or else, $l$ can't be the least upper bound – funktor Aug 14 '12 at 15:28
I could elaborate more if you'd like to – funktor Aug 14 '12 at 15:37
Ah! So Since we assumed that there is no $x \in [\mu -\epsilon, \mu]$ and arrived at $\mu \ne \sup S$ that means that $\mu$ is the supremum? – Moderat Aug 14 '12 at 15:40

Recall the definition of supremum

An element $\mu$ is a supremum of a set $S$ if

$(1)$ It is an upper bound. That is, for $x$ in $S$, we have $x \leq \mu$.

$(2)$ If $\eta$ is any other upper bound, $\mu \leq \eta$.

Now the theorem you might want is

THEROEM $\mu$ is supremum of $S$ if and only if for each $\epsilon >0$, there is some $x \in S$ for which $$\mu-\epsilon \color{red}{<} x \leq \mu$$

Note the $<$ and not the $\leq$. You will see why this is so.

$(\Rightarrow)$ Suppose $\mu$ is a supremum. Then clearly for any $x\in S$, $x\leq \mu$. Now argue by contradiction. Suppose there were some $\epsilon \;>0$ such that $$\mu-\epsilon \geq x$$ for each element of $A$. This means that $x \leq \mu-\epsilon$ for each $x$. But that would mean $\mu-\epsilon$ is upper bound with $\mu-\epsilon \leq \mu$ which is impossible, since $\mu$ is the supremum.

I'm not sure about the $(\Leftarrow)$. I've always seen this used in one direction. The theorem is sometimes called the approximation theorem, and it says that given a bounded nonempty subset $S$ of the reals, there is a sequence of elements of $S$, call it $s_n$ such that $s_n \to \sup S$,and another sequence, call it $u_n$, such that $u_n \to \inf S$

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Thank you. This is clear to me now @PeterTamaroff – Moderat Aug 14 '12 at 17:11
(+1) Neat answer! – funktor Aug 14 '12 at 17:12
@JayeshBadwaik then why the $\Longleftrightarrow$? – Moderat Aug 14 '12 at 17:12
@jmi4 sorry, my bad, I deleted the comment. – Jayesh Badwaik Aug 14 '12 at 17:13

If $\mu$ is an upper bound and there is no element of $S$ in $(\mu-\varepsilon,\mu]$ then $\mu-\varepsilon$ is also an upper bound, so $\mu$ is not the least one.

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We begin by recalling the formal definition of the least upper bound of an ordered set.

Defintion: For a set $A$ (ordered), $\alpha= \sup A$ iff

• $\alpha$ is an upper bound to $A$
• If $\exists \beta \lt \alpha$ then, $\beta$ is not an upper bound of $A$

Assume that for some $ϵ>0$, $\nexists x∈[μ−ϵ,μ]$, such that $x∈S$. Since $S \subset \mathbb R$ there can't exist an interval $[\varphi_1,\varphi_2]=\emptyset$, this implies that $∃λ<μ$ where $λ \notin S \implies \mu \ne \sup S$. Now if $λ∉S$, this means that $λ$ is, but another upper bound for $S$ which is less than $μ$. Note the second property of $\sup A$ in the definition. We claim that $\mu= \sup A$ the fact that $\lambda \lt \mu$ suggest other wise. It would thus be erroneous to claim that $\mu$ is the least upper bound for $S$ when we just showed that there is another upper bound that is even less than $\mu$.

So if $μ= \sup S$ then has to be an element of $S$ in the interval $[μ−ϵ,μ]$

Now, the question says that $μ$ is the supremum iff there is an element of $S$ in the interval $[μ−ϵ,μ]$. But what we just did was started out by assuming that there is no element of $S$ in the interval $[μ−ϵ,μ]$ and proved that if this is the case, then $μ≠\sup S$. So, if $μ=\sup S$, there has to be an element of $S$ in the interval $[μ−ϵ,μ]$.

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Your definition seems to be somewhat strange. The set $\{1,2\}$ has the supremum $2$, but $1.5<2$ doesn't imply it is in the set. – Pedro Tamaroff Aug 14 '12 at 16:30
I am sorry, I don't follow – funktor Aug 14 '12 at 16:36
but $1.5$ is not an upper bound – funktor Aug 14 '12 at 16:37
You said "If there exists $\beta < \alpha$", then $\beta \in A$. That is not the defintion of the supremum of a set. What you want is "If $\beta$ is another upper bound, then $\alpha \leq \beta$. You seem to be using the strict order $<$ when supremums are defined with the weak order $\leq$. – Pedro Tamaroff Aug 14 '12 at 16:39
My bad! I wrote the definition wrong. I did not realize. – funktor Aug 14 '12 at 16:41

The sequence isn't really needed. Note that $\mu$ is an upper bound, so we need to show that nothing smaller than $\mu$ is an upper bound. That is, for all $z < \mu$ show that there is an element $x$ of $S$ such that $z < x \leq \mu$. The condition you are given should help here.

(Also, in the forward direction, how do you know that $\mu - \frac{\epsilon}{2}$ belongs to $S$?)

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Ok I found my mistake there thank you @ArthurFischer – Moderat Aug 14 '12 at 14:58

In $\Rightarrow$, if you're taking the closed interval, I don't see why you use $\mu-\epsilon/2$ instead of just $\mu-\epsilon$. Either way, it's not entirely correct: it could be that neither of these points is a member of $S$ (if, for example, $S=(\mu-\epsilon/2,\mu)$, or $S=\{\mu-2^{-n}\vert n\in \bf N \}$ and $\epsilon$ is a power of $2$). You need to choose the points a little more carefully.

For $\Leftarrow$, it would be best to take the least upper bound of $S$, and show that it can't be smaller than $\mu$.

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($\Rightarrow$) Suppose there exists an $\epsilon > 0$ so that for all elements $s \in S$, $\mu - \epsilon > s$. Then, we have contradicted the assumption that $\mu = \sup(S)$.

($\Leftarrow$) Suppose $\mu \neq \sup(A)$. Then, let $\gamma$ be an upper bound smaller than $\mu$. Let $0 < \epsilon < \mu- \gamma$. We see that no element of $S$ satisfies $\mu - \epsilon < s$, a contradiction.

Or you could prove it directly by noticing if $\gamma < \mu$, then $\epsilon = \mu - \gamma$ implies there exists $s \in S$ so that $\gamma = \mu - \epsilon < s$ so that $\gamma$ is not an upper bound.

We conclude the theorem.

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