If $\sup A = 5$ and $B = \left\{ 3a \mid a \in A \right\}$ then $\sup B = 15$

Question

Prove that if $A \subset \mathbb{R}$, $\sup A = 5$, and $B = \left\{ 3a \mid a \in A \right\}$, then $\sup = 15$.

I tried to do contradiction by assuming the hypothesis and that there is a number $< 15$ that is the supremum of $B$. Then $3a < 15$ is the $\sup B$, and then we divide both sides by $3$ to get $a < 5$. To use contradiction for this particular case, I would have to show that if $3a$ is an upper bound for $B$, then that implies that $a$ is an upper bound for $A$ (so that I can say that $a < 5$ is an upper bound for $A$, which is a contradiction to the hypothesis that $5$ is the least upper bound for $A$). But I don't know how to show that $3a$ being an upper bound for $B$ implies that $a$ is an upper bound for $A$. Or am I going about this proof incorrectly?

Answer 1

Since $\sup A=5$ then $a_n\to 5$ for some $a_n\in A$. But since $3a_n\in B$ and $3a_n\to 3\cdot 5=15$, therefore $\sup B \ge 15$. For $\sup B \le 15$ notice that any element of $B$ has the form $3a,a\in A$ and $3a\le 3\cdot \sup A=15$.

Answer 2

We have

$$\forall a \in A,\; a\le \sup A\implies \forall a \in A,\; 3a\le 3\sup A $$

and $$\forall \epsilon>0,\; \exists a\in A \;|\; \sup A-\frac\epsilon 3< a\implies 3\sup A-\epsilon<3a$$

so we have

$$3\sup A=\sup\{3a\;|\; a\in A\}$$