0

Can someone let me know whether the following theorem is written correctly? I based it off Apostal.

Let $S$ and $T$ be nonempty subsets of $\Bbb{R}$ such that $s\geq t$ for all $s$ in $S$ and $t$ in $T$. Then if $S$ has an infimum, then $T$ has an infimum and $\operatorname{inf}(S)\geq \operatorname{inf}(T)$.

1 Answers1

1

It's false. Consider $S=[0,1]$ and $T=(-\infty,-20]$. Then, clearly holds that $s\geq t$ for all $s$ in $S$ and $t\in T$. Moreover, $\inf(S)=0$ but $T$ doesn't have an infimum.

Carlos Jiménez
  • 3,895
  • 1
    I see! So did I write the comparison property for infimum wrong or does such a property exist for infimum? I just replaced some terms from the comparison property for supremum... –  Jul 16 '18 at 04:11
  • 1
    I think that the correct form for your 'theorem' is «Let $S$ and $T$ be nonempty subsets of $\mathbb{R}$ such that $s\geq t$ for all $s$ in $S$ and $t$ in $T$. Then if $T$ has an infimum then $S$ has an infimum and $\inf(S)\geq\inf(T)$». Note that the infimum of $T$ is a lower bound for $S$ by hypothesis. Then, there exists $\inf(S)$ and by definition of infimum then we have that $\inf(S)\geq\inf(T)$ – Carlos Jiménez Jul 16 '18 at 04:19