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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?

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    $\begingroup$ Your proof is fine. To show that $3a$ being a UB for $B$ implies that $a$ is a UB for $A$, suppose not. Then there's an element $x \in A$ that's greater than $a$. So $3x > 3a$. And since $x$ is in $A$, $3x$ is in $B$. So $3a$ is not a UB for $B$. This contradicts the supposition, so the supposition is false. $\endgroup$ – John Hughes Dec 15 '14 at 14:19
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    $\begingroup$ Related. $\endgroup$ – Git Gud Dec 15 '14 at 14:28
  • $\begingroup$ Thanks @GitGud, I've actually wanted to try and attempt the general case but it seemed too difficult for me $\endgroup$ – mr eyeglasses Dec 15 '14 at 14:38
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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 $3a\le 3\cdot \sup A=15$.

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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

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

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While $C=${$3$}, it defines $B=C\cdot A={\{x \cdot y \mid x \in C, y \in A\}}$

remembering the theorem: $$\sup(A \cdot B)= \sup(A) \cdot \sup(B)$$

so:$\hspace{2cm}$ $sup(B)=sup(C \cdot A)=sup(C) \cdot sup(A)=3 \cdot 5=15$

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