Let $A$ be non-empty and bounded above. and let $λ ∈ \mathbb{R}$. We define $λA = \{y ∈ \mathbb{R} : ∃x ∈A , y = λx\}$. Do we have $\sup(λA) = λ \sup(A)$?

If we define $\sup A = c$, then $\sup(\lambda A) = y = \lambda c$.

$\lambda c = \lambda \sup A = \lambda \sup A$.

Is this an adequate proof, have I made any mistakes?

  • 1
    $\begingroup$ Related. $\endgroup$ – Git Gud Mar 25 '15 at 11:18

Your statement is wrong if $\lambda \lt 0$.

Even for $\lambda \ge 0$, to make it a proof it would be helpful if you used the definition of $\sup$.


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