Can someone point out the holes in my proof using well ordering, archimedean and completeness to prove for a given set E, inf(E) = -sup(E)

Let, $$E\subset \mathbb R$$ be a non empty bounded above set. Define $$-E = \{ -x : x \in E \}.$$ Then, $\operatorname{inf} (-E) = -\operatorname{sup}(E)$.

Proof - It follows from the completeness property that $$\exists x_o < x , \forall x_o \in E$$

such that, $$\operatorname{sup}(E) = x \tag {1} \label{1}$$

Since, it is already given that the set is bounded, it must have a lower bound. $$\exists x_o < x, \forall x_o \in E,$$ we have shown already.

Now, $$\forall x_o>x ,\exists x_o \notin E$$

such that, $$\operatorname{inf}(E) = -x \tag {2} \label{2}$$

From (1) and (2), we get -

$$\operatorname{inf}(E) = -x = -\operatorname{sup}(E)$$

Hence, proved. (black dot to the far left)

I know there might be many logic blunders. I hope to solidify my logic and understanding through the feedback! :D

• 1) ∃xo<x∀xo∈E this notation is wrong. You mean $\exists x; \forall x_o\in E x_o < x$. But this a only notation. – fleablood Dec 5 '16 at 0:50
• 2)"Since, it is already given that the set is bounded, it must have a lower bound. " It was not given that the set was bounded. It was only given that the set was bounded above. – fleablood Dec 5 '16 at 0:51
• 3) "Since, it is already given that the set is bounded, it must have a lower bound. ∃xo<x,∀xo∈E," That's an upper bound not a lower bound. – fleablood Dec 5 '16 at 0:53
• 4)"Now, ∀xo>x∃xo∉E such that, inf(E)=−x" in english. "Anything larger than the least upper bound is not in E, therefore the greatest lower bound is the negative of the least upper bound".... ???HUH???? I can't even begin to understand why you would conclude that. if E = [1,2] then if x > 2, x isnt in [1,2]. How could this possibly imply $\inf [1,2] = -2$? – fleablood Dec 5 '16 at 1:03
• Did you mean -E in some of the places you wrote E. I can't really follow your logical notation. It seems like you might be trying to say "if x0 > x then x0 not in E and therfore if x1 < -x then x1 not in -E and so -x is a lower bound. But you never specifically stated the implication so it is hard to tell if that was what you meant to say. – fleablood Dec 5 '16 at 1:09

If you have that $\exists x_{0}$ such that $\forall x\in E$, $x\le x_{0}=sup(E)$ if you multiply by $-1$ the inequality you have that $y=-x\ge -x_{0}=-sup(E)$, $\forall x\in E$, $\forall y\in -E$, you have that $inf(-E)\ge -sup(E)$. So you have to prove that they are equals. (Hint: prove the other inequality).

• I think you're misinterpreting what the OP means - when they write "$\exists x_0<x$," I believe they mean "$\exists x>x_0$." – Noah Schweber Dec 5 '16 at 0:46
• Yes, I think I understand that, but I change the notation for a better understanding. – Skullgreymon Dec 5 '16 at 0:50

You seem to be using $x$ as both the supremum and infimum of $E$. You can't do this, and indeed you can't even assume $E$ is bounded below; all you know is that $E$ is bounded above! Also, your conclusion doesn't involve $-E$ at all, whereas you're supposed to be proving a statement about $-E$ and $E$.

EDIT: In particular, your conclusion that $-x=\inf(E)$ is completely unjustified, and indeed could be false: suppose $E=(1, 2)$. Then $x=2$. Is $\inf(E)=-2$?

Here's a hint: suppose $x=\sup E$. Is $-x$ a lower bound for $-E$? Is it the greatest lower bound?

Also, your quantification is mostly backwards: to write "There is an element bigger than everything in $E$," you write $$\exists x\forall x_0\in E(x>x_0).$$

You need to prove: $\inf (-E) = - \sup E$.

For instance if $E = (-\infty,-5]$ then $-E = [5, \infty)$. $\sup E = -5$. As it so happens, $\inf -E = 5 = -\sup E$. (What are the odds?)

========

Outline of what you want to do:

Since $E$ is bounded above we know $\sup E$ exists.

The definition of $\sup$ is that:

1) $\sup E \ge x$ for all $x \in E$.

2) If $y < \sup E$, $y$ is not an upper bound. i.e. if $y < \sup E$ than there exists an $x \in E$ so that $y < x$.

Now $a < b \iff -a > -b$ and $x \in E \iff -x \in -E$.

So we can restate statement 1) and 2) using $-x,-y, -\sup E,$ and $-E$ rather instead of $x,y \sup E$ and $E$.

The restatement of 1) and 2) will satisfy the definition of $\inf -E$.