$\sup(a + B) = a + \sup B$

Question

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.

Answer 1

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 $\sup(a +B) = a + \sup B$.

Answer 2

$$\sup\left\lbrace x + y\right\rbrace = \sup \left\lbrace x \right\rbrace + \sup \left\lbrace x \right\rbrace $$ $$\inf\left\lbrace x + y\right\rbrace = \inf \left\lbrace x \right\rbrace + \inf \left\lbrace x \right\rbrace $$ 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 > 0$ $ \space \exists (x_1 + y_1) \in \left\lbrace x + y\right\rbrace $ for which $$\inf \left\lbrace x \right\rbrace + \inf \left\lbrace y \right\rbrace \leqslant (x_1 + y_1) < \inf \left\lbrace x \right\rbrace + \inf \left\lbrace y \right\rbrace + \epsilon $$ This gives $$\inf\left\lbrace x + y\right\rbrace = \inf \left\lbrace x \right\rbrace + \inf \left\lbrace x \right\rbrace $$

Answer 3

Given $B$ is non-empty, $B$ is bounded above and $\sup B$ is the least upper bound of $B$ then

Claim 1: $a + B$ is non-empty.

Pf: This will be the layout of all the claims.

$B$ is non empty. So there exists a $b \in B$ so $a + b \in a + B$. So $a+B$ is not empty.

Claim 2: $a + B$ is bounded above.

Pf: $B$ is bounded above. so there exist $g$ so that $g \ge b$ for all $b \in B$.

Let $k = a + B$. Then $k = a + b$ for some $b \in B$. So $g \ge b$ so $a+q \ge a+b=k$. So $a+B$ is bounded above by $g$.

Claim 3: $a + \sup B$ is an upper bound for $a+B$.

Pf: Apply the argument of Claim to but apply $\sup B$ as the upper bound in use. If $k \in a+ B$ there is a $b$ sso taht $k =a+b$ and $\sup B \ge b$ so $a + \sup B \ge a + b = k$.

Claim 4: If $l < a + \sup B$ then $l$ is not an upper bound.

If $l < a + \sup B$ then $l - a < \sup B$ and so $l-a$ is not an upper bound of $B$. SO there exists a $b\in B$ so that $l-a < b$.

.... You can do this......