0
$\begingroup$

Let $S\subseteq R$ be nonempty and bounded above. Let $s$=sup $S$ and define

$$T=(3x\in R : x \in S)$$

Prove that sup $T$ exists and sup $T$=$3s$.

What i tried.

Clearly $T$ is non empty as it contains the element $3x$. $T$ is bounded above by $3S$ (explanation needed). And thus by the axiom of completeness sup $T$ exists.

Next to prove that sup $T$ =$3s$, we have that since the set $S$ is 3 times that of the set $T$ and that since $s$ lies in $S$ and is the lowest upper bound of the set $S$, then $3s$ also lies in $T$ and thus sup $T$=$3s$.

While i know intuitively how to solve the problem and at least have a rough idea of how to solve the problem, my explanation isn't clear enough. Could someone explain how do i solve the problem that is written more clearly. Thanks

$\endgroup$
  • $\begingroup$ And which element is $x$, exactly? You should say something like "$T$ is non-empty because $S$ is non-empty" $\endgroup$ – Omnomnomnom Feb 13 '17 at 17:38
1
$\begingroup$

First, $T$ is nonempty because there is an element $x \in S$.

Consider an elemental approach. If $a \in S$, then we know $a \le s$, so $3a \le 3s$. Therefore $3s$ is an upper bound for $T$.

Now, we show that there is no upper bound less than $3s$. If there is $t \in T$ such that $t < 3s$, then we know that $t/3 \in S$ and $t/3 < s$. Because $t/3$ is less than the least upper bound, there exists a number $y$ such that $y \in S$ and $t/3 < y \le s$. Then $t \in T < 3y \in T$, so $t$ is not an upper bound for $T$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.