Prove that $\sup{\alpha A} = \alpha \sup{A}$ 
Prove that if $\alpha > 0$, then $\sup{\alpha A} = \alpha \sup{A}$ and $\inf{\alpha A} =\alpha \inf{A}$.

I think this can easily be proven using induction using the fact that $\sup(A+B) = \sup{A}+\sup{B}$ and $\inf(A+B) = \inf{A}+\inf{B}$. So we prove it for all $\alpha \neq 0$. Is this the way to approach solving this?
 A: Induction has nothing to do with this.
Hint: the map $x\mapsto \alpha x$ is increasing if $\alpha>0$.
Further hints as spoilers.
Spoiler 1

 Let $s=\sup A$; then, for each $x\in A$, $x\le s$; therefore, for each $x\in A$, $\alpha x\le\alpha s$. Hence $\alpha s$ is an upper bound for $\alpha A$

Spoiler 2

 Suppose $t$ is an upper bound for $\alpha A$. Then $\alpha^{-1}t$ is an upper bound for $A$ (why?). Therefore $\alpha^{-1}t\ge s$ and so $t\ge \alpha s$.

A: First note that $\sup A = -\infty$ if and only if $A=\emptyset$. Hence, this case is trivial as $\alpha \emptyset = \emptyset$.
Let us assume that $A$ is bounded, i.e. $\sup A < \infty$.
Let $a\in A$, then holds
$$ \alpha a \leq \alpha \sup A. $$
Hence, $\sup \alpha A \leq \alpha \sup A.$ 
Let now $\epsilon >0$, then there exists $a_{\epsilon}\in A$ such that
$$ \sup A - \epsilon \leq a_{\epsilon}.$$
Therefore,
$$ \alpha \sup A - \alpha \epsilon \leq \alpha a_{\epsilon} \leq \sup \alpha A.$$
Taking the limit for $\epsilon \rightarrow 0$ gives us
$$ \alpha \sup A \leq \sup \alpha A. $$
Putting the two inequalities together yields
$$ \alpha \sup = \sup \alpha A. $$
In the case where $\sup A = \infty$, there exists a sequence $(x_n)_{n\in \mathbb{N}}\subseteq A$ such that $x_n \rightarrow \infty$. However, then $(\alpha x_n)_{n\in \mathbb{N}} \subseteq \alpha A$ and $\alpha x_n \rightarrow \infty$, thus $\sup \alpha A= \infty$ and this proves the claim.
A: Let's assume $$\alpha \sup A \gt \sup \alpha A$$ 
By definition, there is $a \in A$, such that $$\alpha a \gt \sup \alpha A$$ 
this can not be true because we know that $$\alpha a \le \sup \alpha A$$
I think you can use similar approach for other proofs.
