2

This is what I have done, but I am not sure if it is correct.

$s = \sup(cA) \Rightarrow s \ge -ca, \, \forall a \in A$ and $\nexists t \in \mathbb{R} : t \ge -ca, t < s, \forall a \in A$

Restated with the sign applied to the inequality:

$\dfrac{s}{c}\le a, \forall a \in A$ and $\nexists t \in \mathbb{R} : \dfrac{t}{c}\le a, \forall a \in A, t < s \Rightarrow \dfrac{s}{c} = \inf(A)$

$\dfrac{s}{c} = \inf(A) \Rightarrow s = \sup(cA) = c \cdot \inf(A)$

Is this correct?

Martin Sleziak
  • 56,060
tmaric
  • 345
  • 1
    Note that if $s = sup(cA)$ then $s \ge ca$ for all $a \in A$ – Zestylemonzi Sep 29 '15 at 09:03
  • 1
    Hint: $sup(A) = - inf(-A)$, where $-A = {-a : a \in A }$ –  Sep 29 '15 at 09:24
  • @Vader: thanks, but is the proof above valid? – tmaric Sep 29 '15 at 09:26
  • 1
    I don't think so. If $s>ca,\ \forall a \in A$ then it is not true that $s>-ca,\ \forall a \in A$ –  Sep 29 '15 at 09:29
  • 1
    These post might be useful in relation to your problem: http://math.stackexchange.com/questions/488022/how-to-prove-infs-sup-s and http://math.stackexchange.com/questions/947035/how-to-show-sup-a-infa (If you know this, you can reduce the problem to the case $c>0$.) – Martin Sleziak Sep 29 '15 at 11:30

1 Answers1

2

Your attempt at a proof is, at the least, insufficient. I suggest you to use less symbols and to spell out what you need to prove.

Let $r=\inf(A)$; you want to prove

  1. $cr\ge x$, for all $x\in cA$
  2. for all $\varepsilon>0$, there exists $y\in cA$ such that $y>cr-\varepsilon$.

First condition. Since $r=\inf(A)$, we know that $r\le a$, for all $a\in A$; if $x\in cA$, then $x=ca$, for some $a\in A$; from $r\le a$ and $c<0$, it follows $cr\ge ca=x$.

Second condition. There exists $b\in A$ such that $r-\dfrac{\varepsilon}{c}>b$, because $-\varepsilon/c>0$. Then $$ c\left(r-\frac{\varepsilon}{c}\right)<cb $$ which is the same as $$ cr-\varepsilon < cb $$ and we can take $y=cb\in cA$.

egreg
  • 244,946