$\sup(a + B) = a + \sup B$ I believe my proof of this simple fact is fine, but after a few false starts, I was hoping that someone could look this over. In particular, I am interested in whether there is an alternate proof.


  For a real number $a$ and non-empty subset of reals $B$, define $a + B = \{a + b : b \in B\}$. Show that if $B$ is bounded above, then $\sup(a + B) = a + \sup B$.


My attempt: 


  Fix $a \in \mathbb{R}$, take $B \subset \mathbb{R}$ to be nonempty and bounded above, and define
  $$a + B = \{a + b : b \in B\}.$$
  Since $B$ is nonempty and bounded above, the least-upper-bound axiom guarantees the existence of $\sup B$. For any $b \in B$, we have
  $$b \leq \sup B,$$
  which implies
  $$a + b \leq a + \sup B.$$
  As this is true for any $b \in B$, it follows that $a + \sup B$ is an upper bound of $a + B$, and hence $\sup(a + B)$ exists, by the completeness axiom, since $B \neq \emptyset$ implies immediately that $a + B \neq \emptyset$. I claim that $a + \sup B$ is in fact the least upper bound of $a + B$. As we have already shown it to be an upper bound, it suffices to demonstrate that $a + \sup B$ is the least of the upper bounds. Let $\gamma$ be an upper bound of $a + B$. Hence, for any $b \in B$,
  $$a + b \leq \gamma,$$
  which implies that
  $$b \leq \gamma - a.$$
  As this holds for all $b \in B$, $\gamma - a$ is an upper bound of $B$. Hence, by the definition of supremum, 
  $$\gamma - a \geq \sup B,$$
  which implies that
  $$\gamma \geq a + \sup B,$$ 
  as desired. 


I tried to write the proof initially be showing that $\sup(a + B) \leq a + \sup B$ and $\sup(a + B) \geq a + \sup B$, but didn't have any luck. If there is a trick to it, I would be interested in hearing it.
What you've done looks correct to me, but I think we can rework it more concisely using exactly the strategy that you suggest at the end of the question. Note that both suprema exist since the sets are non-empty.

First direction: Let $\lambda \in a + B$. Then $\lambda = a + b$ for some $b \in B$. Since a supremum is an upper bound, $b \leq \sup B$, so $\lambda \leq a + \sup B$. Since $\lambda \in a + B$ was arbitrary, $a + \sup B$ is an upper bound for $a + B$, hence $\sup(a + B) \leq a + \sup B$.

At this point it might be worth pausing to try the other direction yourself - the idea is similar, so it would be a good test of understanding.

Other direction: Let $b \in B$. Then $a + b \in a + B$, and since a supremum is an upper bound, $a + b \leq \sup(a + B)$. Rearranging, $b \leq \sup(a + B) - a$, so $\sup(a + B) - a$ is an upper bound on $B$, and hence $\sup B \leq \sup(a + B) - a$, and it follows that $\sup(a + B) \geq a + \sup B$.

Conclusion: It follows immediately that $\sup(a +B) = a + \sup B$.
$$\sup\left\lbrace x + y\right\rbrace = \sup \left\lbrace x \right\rbrace + \sup \left\lbrace x \right\rbrace $$
$$\inf\left\lbrace x + y\right\rbrace = \inf \left\lbrace x \right\rbrace + \inf \left\lbrace x \right\rbrace $$
For non negative numbers same properties we have for multiplication.

Proof for second:

$ \exists \inf \left\lbrace x \right\rbrace $ and $\exists \inf \left\lbrace y \right\rbrace \Rightarrow  \exists \inf\left\lbrace x + y\right\rbrace $ and $\inf \left\lbrace x \right\rbrace + \inf \left\lbrace x \right\rbrace \leqslant x + y$.

For $ \forall \epsilon &gt; 0$ $ \space \exists (x_1 + y_1) \in \left\lbrace x + y\right\rbrace  $ for which $$\inf \left\lbrace x \right\rbrace + \inf \left\lbrace y \right\rbrace \leqslant (x_1 + y_1) &lt; \inf \left\lbrace x \right\rbrace + \inf \left\lbrace y \right\rbrace + \epsilon $$
This gives 
$$\inf\left\lbrace x + y\right\rbrace = \inf \left\lbrace x \right\rbrace + \inf \left\lbrace x \right\rbrace $$