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For $A\subset \mathbb{R}$ and $\lambda \in \mathbb{R}$ let's define: $$ \lambda A = \{\lambda a: a\in A\} $$ I have to prove that for $\lambda<0$ and bounded $A$ we have $\inf(\lambda A)=\lambda \sup(A)$ and I'm not sure if my proof is correct.

First, let's notice that $\forall a\in A$ we have: $$ a\leq \sup(A) \iff \lambda a \geq \lambda \sup(A) \hspace{0.3cm} \text{(because $\lambda$ is negative)}. $$ If the above holds $\forall \lambda a \in \lambda A$ then: $$ \inf(\lambda A) \geq \lambda \sup(A). $$

Now for the other inequality, $\forall \lambda a \in \lambda A$ we have: $$ \lambda a \geq \inf(\lambda A), $$ so: $$ a\leq \frac{1}{\lambda}\inf(\lambda A) $$ and because above holds $\forall a\in A$, we can write: $$ \sup(A)\leq \frac{1}{\lambda} \inf(\lambda A) $$ which is same as: $$ \lambda \sup(A) \geq \inf(\lambda A) $$ Thus we showed that: $$ \lambda \sup(A) \leq \inf(\lambda A) \leq \lambda \sup(A) $$ so $$ \lambda \sup(A) = \inf(\lambda A). $$

Can I prove it this way or is there anything missing? Thanks in advance for any help!

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    $\begingroup$ I see no mistake... $\endgroup$ – user302982 Feb 9 '16 at 15:27
  • $\begingroup$ I am using here equivalence relation, namely: $\forall \lambda a\in \lambda A \iff \forall a \in A$ and it follows from definition of the set $\lambda A$, right? $\endgroup$ – Mat Dyl Feb 9 '16 at 15:48
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First, here is my direct answer to your question: this proof is structurally fine, but its wording can be improved on two points. First, $\;\forall \lambda a \in \lambda A\;$ is formally incorrect and also confusing: you just mean $\;\forall a \in A\;$. Second, "If the above holds $\;\forall \ldots\;$" should be "Since the latter inequality holds $\;\forall\ldots\;$".$ \newcommand{\calc}{\begin{align} \quad &} \newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}} \newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} } \newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & } \newcommand{\endcalc}{\end{align}} \newcommand{\ref}[1]{\text{(#1)}}\newcommand{\inf}[1]{\text{inf}(#1)} \newcommand{\sup}[1]{\text{sup}(#1)}\newcommand{\then}{\Rightarrow} \newcommand{\when}{\Leftarrow}\newcommand{\true}{\text{true}} \newcommand{\false}{\text{false}} $

The step that you are trying to express by $\;\forall \lambda a \in \lambda A\;$ is, I think, to use the definition of $\;\lambda A \;$ to get from $$ \langle \forall a : a \in A : \lambda a \ge \ldots \rangle $$ to the equivalent $$ \langle \forall b : b \in \lambda A : b \ge \ldots \rangle $$ (I'm using slightly different logic notations from here on, which I find easier to work with. Hopefully those are not too confusing...)

Below, I'm referring to this step simply as "expand definition of $\;\lambda A\;$". But there is actually a bit of logic complexity hidden in that step, which I can perhaps most easily illustrate by going from the latter to the former: $$\calc \langle \forall b : b \in \lambda A : b \ge \ldots \rangle \op=\hint{definition of $\;\lambda A\;$} \langle \forall b : \langle \exists a : a \in A : b= \lambda a \rangle : b \ge \ldots \rangle \op{\tag{*} =}\hint{logic} \langle \forall b :: \langle \forall a: a \in A \;\land\; b = \lambda a : b \ge \ldots \rangle\rangle \op=\hint{logic: exchange quantifiers} \langle \forall a: a \in A : \langle \forall b : b = \lambda a : b \ge \ldots \rangle \rangle \op=\hint{logic: one-point rule} \langle \forall a: a \in A : \lambda a \ge \ldots \rangle \endcalc$$ Here $\ref{*}$ uses the fact that $\;\langle \exists a :: P(a) \rangle \then Q\;$ is equivalent to $\;\langle \forall a :: P(a) \then Q \rangle\;$ if $\;Q\;$ does not contain a free $\;a\;$.


Now as an illustration of how helpful it is to have simple definitions, and to make them explicit, here is an alternative proof.

The definitions which you (implicitly) used for $\;\sup{A}\;$ and $\;\inf{B}\;$ are that \begin{align} & \langle \forall a : a \in A : a \le \sup{A} \rangle \\ & \langle \forall a : a \in A : a \le z \rangle \;\then\; \sup{A} \le z \\ \end{align} and \begin{align} & \langle \forall b : b \in B : b \ge \inf{B} \rangle \\ & \langle \forall b : b \in B : b \ge z \rangle \;\then\; \inf{B} \ge z \\ \end{align} for any $\;z\;$, and for any set $\;A\;$ which has a lower bound and $\;B\;$ with an upper bound.

Instead, I propose to use the following equivalent but simpler definitions: for all $\;z\;$, $$ \tag{1} \sup{A} \leq z \;\equiv\; \langle \forall a : a \in A : a \leq z \rangle $$ and $$ \tag{2} z \leq \inf{B} \;\equiv\; \langle \forall b : b \in B : z \leq b \rangle $$

And as a proof principle, the following is very useful when dealing with upper and lower bounds: $$ \tag{3} x = y \;\equiv\; \langle \forall z :: z \le x \;\equiv\; z \le y \rangle $$ which says that two numbers are equal iff they have the same lower bounds.


Now, to prove $$ \tag{4} \inf{\lambda A} \;=\; \lambda \sup{A} $$ for $\;\lambda < 0\;$ and a lower-bounded $\;A\;$, we calculate as follows for any $\;z\;$: $$\calc z \le \inf{\lambda A} \op=\hint{'expand' $\;\inf{\cdots}\;$ using definition $\ref{2}$} \langle \forall b : b \in \lambda A : z \leq b \rangle \op=\hint{expand definition of $\;\lambda A\;$} \langle \forall a : a \in A : z \leq \lambda a \rangle \op= \hints{arithmetic: divide by negative $\;\lambda\;$} \hints{-- to prepare for introducing $\;\sup{\cdots}\;$ through $\ref{1}$,} \hint{since we are working towards the RHS of $\ref{4}$} \langle \forall a : a \in A : a \leq z / \lambda \rangle \op=\hint{introduce $\;\sup{\cdots}\;$ by definition $\ref{1}$} \sup{A} \leq z / \lambda \op=\hint{arithmetic: multiply by negative $\;\lambda\;$} z \leq \lambda \sup{A} \endcalc$$ By principle $\ref{3}$, this proves $\ref{4}$.

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