Infimum and Supremum proof I tried to prove: $\inf(A-B) = \inf(A) - \sup(B)$. I was hoping someone could check my proof and perhaps give me some tips, I didn't quite get the ending right. Thanks in advance!
First, I show that $\inf(A-B) = \inf(A) + \inf(-B)$:
Suppose $x \in A$ and $y \in B$. Then $-y \in -B$. 
$x \geq \inf(A)$ and $-y \geq \inf(-B)$. This gives us $x-y \geq \inf(A) + \inf(-B)$. So $\inf(A) + \inf(-B)$ is a lower-bound and $\inf(A-B) \geq \inf(A) + \inf(-B)$
Now, suppose $\epsilon>0$. Then $\exists_{x\in A, -y\in -B}$ such that $x < \inf(A) - \frac{\epsilon}{2}$ and $-y < \inf(B) - \frac{\epsilon}{2}$
So $x-y < \inf(A) + \inf(-B) - \epsilon$. This indicates that $\inf(A) + \inf(-B) \geq \inf(A-B)$. 
So $\inf(A) + \inf(-B) = \inf(A-B)$. 
So now I have to prove that $\inf(-B) = -\sup(B)$. My attempt:
Suppose $y\in B$. Then $y \leq \sup(B)$ this gives $-y \geq -\sup(B)$. 
Also $-y \geq \inf(-B)$ by definition. So $\inf(-B) = -\sup(B)$?
 A: If I'm not mistaken, the first portion is correct. Concerning the inf(−B) = −sup(B) portion, consider -sup(B). If you can show that -sup(B) is a lower bound of -B, perhaps by first considering sup(B), the solution should follow by combining a similar result on the -inf(-B) being an upper bound of B, i.e., eventually showing -sup(B) $\leqslant$ inf(-B) and -sup(B) $\geqslant$ inf(-B). Note my work assumes B is a nonempty bounded subset of R.
A: Your proof of the first part looks correct to me, but the second part proof seems partial.
Just for comparison, here is essentially the same proof in a 'calculational' style, and avoiding $\epsilon$'s.$\newcommand{\calc}{\begin{align} \quad &}\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }\newcommand{\endcalc}{\end{align}}\newcommand{\ref}[1]{\text{(#1)}}\newcommand{\inf}[1]{\text{inf}(#1)}\newcommand{\sup}[1]{\text{sup}(#1)}\newcommand{\then}{\Rightarrow}\newcommand{\when}{\Leftarrow}\newcommand{\true}{\text{true}}\newcommand{\false}{\text{false}}$
(Notation remarks: I assume we are working in $\;\mathbb R\;$.  Just like in your question, I use 'lifted' notation, so that e.g., $\;A + B\;$ is the set of all $\;a+b\;$ where $\;a \in A\;$ and $\;b \in B\;$.)
Your approach of splitting the proof in two parts is a good one, since it separates two different concerns: we should prove that for any $\;A,B\;$ with a lower bound,
\begin{align}
\tag{1} & \inf{A + B} = \inf{A} + \inf{B} \\
\tag{2} & \inf{A} = {- \sup{-A}} \\
\end{align}
In this answer I will focus on $\ref{1}$.
But first, let's make clear what properties of $\;\inf{\cdots}\;$ we will use.  It seems simplest to just use the definition: given a set $\;A\;$ which has a lower bound, $\;\inf{A}\;$ is the unique real number for which
\begin{align}
\tag{3a} & \langle \forall x : x \in A : \inf{A} \leq x \rangle \\
\tag{3b} & \langle \forall z :: \langle \forall x : x \in A : z \leq x \rangle \;\then\; z \leq \inf{A} \rangle \\
\end{align}
(The first of these says that $\;\inf{A}\;$ is a lower bound of $\;A\;$, and the second is that it is the greatest lower bound.) And because of the uniqueness, to prove something of the form $\;\inf{A} = E\;$ it suffices to prove both $\;\langle \forall x : x \in A : E \leq x \rangle\;$ and $\;\langle \forall z :: \langle \forall x : x \in A : z \leq x \rangle \;\then\; z \leq E \rangle\;$.

Now, let's prove $\ref{1}$.  As I said, because of uniqueness it suffices to prove both $\ref{1a}$ and $\ref{1b}$:
\begin{align}
\tag{1a} & \langle \forall x : x \in A + B : \inf{A} + \inf{B} \leq x \rangle \\
\tag{1b} & \langle \forall z :: \langle \forall x : x \in A + B : z \leq x \rangle \;\then\; z \leq \inf{A} + \inf{B} \rangle \\
\end{align}
Expanding the definition of $\;A + B\;$, we can rewrite these as
\begin{align}
\tag{4a} & \langle \forall a,b : a \in A \land b \in B : \inf{A} + \inf{B} \leq a+b \rangle \\
\tag{4b} & \langle \forall z :: \langle \forall a,b : a \in A \land b \in B : z \leq a+b \rangle \;\then\; z \leq \inf{A} + \inf{B} \rangle \\
\end{align}
Now $\ref{4a}$ directly follows from two applications of $\ref{3a}$: for any $\;a,b\;$,
$$\calc
    \inf{A} + \inf{B} \leq a+b
\op\when\hint{arithmetic}
    \inf{A} \leq a \;\land\; \inf{B} \leq b
\op\when\hint{using $\ref{3a}$ twice, with $\;x := a\;$ and $\;x := b\;$}
    a \in A \;\land\; b \in B
\endcalc$$
Proving $\ref{4b}$ takes slightly longer, but is still pretty direct from two applications of $\ref{3b}$: for any $\;z\;$,
$$\calc
    z \leq \inf{A}+\inf{B}
\op=\hint{arithmetic -- rewrite to prepare for $\ref{3b}$}
    z-\inf{B} \leq \inf{A}
\op\when\hint{by $\ref{0b}$}
    \langle \forall a : a \in A : z-\inf{B} \leq a \rangle
\op=\hint{arithmetic -- rewrite to prepare for $\ref{3b}$}
    \langle \forall a : a \in A : z-a \leq \inf{B} \rangle
\op\when\hint{by $\ref{0b}$}
    \langle \forall a : a \in A : \langle \forall b : b \in B : z-a \leq b \rangle \rangle
\op=\hint{arithmetic; logic: merge quantifications}
    \langle \forall a,b : a \in A \land b \in B : z \leq a+b \rangle
\endcalc$$
This completes the proof of $\ref{1}$.

Now before proving $\ref 2$, let's again make explicit the basic properties that we will use, which are very similar to $\ref{3a}$ and $\ref{3b}$: for any $\;C\;$ with an upper bound, $\;\sup{C}\;$ is uniquely determined by
\begin{align}
\tag{5a} & \langle \forall x : x \in C : x \leq \sup{C} \rangle \\
\tag{5b} & \langle \forall z :: \langle \forall x : x \in C : x \leq z \rangle \;\then\; \sup{C} \leq z \rangle \\
\end{align}

Again because of uniqueness, for $\ref 2$ it suffices to prove both $\ref{2a}$ and $\ref{2b}$:
\begin{align}
\tag{2a} & \langle \forall x : x \in A : {- \sup{-A}} \leq x \rangle \\
\tag{2b} & \langle \forall z :: \langle \forall x : x \in A : z \leq x \rangle \;\then\; z \leq {- \sup{-A}} \rangle \\
\end{align}
Let's see how we can use the properties of $\;\sup{\cdots}\;$ to prove these: for any $\;x\;$,
$$\calc
    {- \sup{-A}} \leq x
\op=\hint{arithmetic -- to prepare for $\ref{5a}$}
    {-x} \leq \sup{-A}
\op\when\hint{using $\ref{5a}$}
    {-x} \in {-A}
\op=\hint{definition of $\;{-A}\;$}
    x \in A
\endcalc$$
and also for any $\;z\;$,
$$\calc
    z \leq {- \sup{-A}}
\op=\hint{arithmetic -- to prepare for $\ref{5b}$}
    \sup{-A} \leq {-z}
\op\when\hint{using $\ref{5b}$}
    \langle \forall x : x \in {-A} : x \leq {-z} \rangle
\op=\hint{substitute $\;x:={-x}\;$, using the definition of $\;{-A}\;$; simplify using arithmetic}
    \langle \forall x : x \in A : z \leq x \rangle
\endcalc$$
These prove $\ref{2a}$ and $\ref{2b}$, respectively, completing the proof of $\ref{2}$.
