Developing Proof Writing Skills

Question

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 = sup().

Then, by definition of supremum, ≥,∀ ∈ .

Since < 0, then ≤ ,∀ ∈ . That is to say is a lower bound of set ={ : ∈ }.

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.

Answer 1

It'd be best to do this by strict definitions.

The definition of $m = \sup S$ is twofold:

i) $m$ is an upper bound of $S$. That is: for every $s\in S$, $s < m$.

ii) $m$ is the least upper bound. That is a) if $y < m$ then there $y$ is not an upper bound; which can be reworded as b) for all $y < m$ there is an $s \in S$ so that $y < s$.

Claim 1: $b\sup S$ is an lower bound of $bS$.

For every $k \in bS$ there is an $s \in S$ so that $k = bs$. And as $s \le \sup S$ and $b < 0$ then $k = bs \ge b\sup S$.

So $b\sup S$ is a lower bound of $bS$.

Claim 2: $b\sup S$ is the greatest lower bound.

For any $y > b\sup S$ then $\frac yb < \sup S$. So there is an $s\in S$ so that $\frac yb < s$. So $y > bs$. And $bs\in bS$. So $y$ is not a lower bound.

So $b\sup S = \inf bS$.

Answer 2

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 $\alpha\leq c/b$.)