So I have been tasked with this current problem for an introductory Real analysis class. I believe that I have the logical framework down, I am just struggling with writing the proof, as my professor critiqued the proof as not using enough definitions. I am hoping someone can give me suggestions, or at least point me in the right direction as to what I can be doing to improve my proof writing skills.
Here is the question:
Let $S$ be a nonempty bounded subset of $\mathbb{R}$ and let $b < 0$. Show that $b\sup(S) = \inf(bS)$.
Here is my proof:
By definition of bounded, $S$ has an upper and lower bound, both of which are real numbers.
Let $\alpha$ denote the least upper bound of $S$, such that $\alpha \geq s, \forall s \in S$.
By definition, $\alpha$ is the supremum of $S$, so $\alpha = \sup(S)$.
Since $b < 0$, then $b\alpha \leq bs, \forall s \in S$.
That is to say, $b\alpha$ is the greatest lower bound of the set $bS=\{bs : s\in S\}$, and by definition, $b\alpha = \inf(bS)$.
Thus, $b\sup(S) = \inf(bS)$.
Thanks for the help!
Edit on 9/19/2019
I have revised my proof, and I hope that this is better.
Here goes:
Because S is a bounded nonempty subset of R, it is bounded above and below, and has both an infimum and supremum.
Let 𝛼 = sup(𝑆).
Then, by definition of supremum, 𝛼 ≥𝑠,∀𝑠 ∈ 𝑆.
Since 𝑏 < 0, then 𝑏𝛼 ≤ 𝑏𝑠,∀𝑠 ∈ 𝑆. That is to say 𝑏𝛼 is a lower bound of set 𝑏𝑆={𝑏𝑠 : 𝑠 ∈ 𝑆}.
Now let γ be any lower bound of bS. If x ∈ S, then γ ≤ bx.
Then, γ/b ≥ x, and thus, γ/b is an upper bound of s, so we have γ/b ≥ sup(S).
Therefore, γ ≤ b sup(S) and by definition of infimum, we can conclude that b sup(S) = inf(bS).
Is this looking any better? Thanks again for all of the help.
Best Answer
It seems that you are confusing the notions of upper bound and least upper bound, and similarly for lower bound and greatest lower bound.
An upper bound of a set $A$ is a number $U$ such that $U\geq a$ for all $a\in A$.
A least upper bound $L$ is an upper bound with an extra property: If $U$ is any upper bound of $A$, then $L\leq A$.
Similarly for lower bound and greatest lower bound.
So what you have done is proven that, for $\alpha=\sup(S)$, we have $b\alpha\leq bs$ for all $s\in S$. In other words, $b\alpha$ is a lower bound of $bS$.
But why is it the greatest lower bound? Take any lower bound $c$ of $bS$. Can you prove that $b\alpha\geq c$? (Hint: first prove that $c/b$ is an upper bound for $S$, and then use the fact that $\alpha$ is the least upper bound, so $\alpha\leq c/b$.)