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

  • $\begingroup$ 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

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 $T$.


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