MHB Proof of Supremum of $M$ Mapping into Itself

[ozkan12]

Let $f$ be a mapping of a metric space $M$ into itself. For $A\subset M$ let $\And(A)=sup\left\{d(a,b);a,b\in A\right\}$ and for each $x\in M$, let $O(x,n)=\left\{x,Tx,...{T}^{n}x\right\}$ $n=1,2,3...$

$O(x,\infty)=\left\{x,Tx,...\right\}$ Please prove that $\And[O(x,\infty)]=sup\left\{\And[O(x,n)]:n\in\Bbb{N}\right\}$...Thank you for your attention...Best wishes...

 

[Opalg]

Let $f$ be a mapping of a metric space $M$ into itself. For $A\subset M$ let $\And(A)=sup\left\{d(a,b);a,b\in A\right\}$ and for each $x\in M$, let $O(x,n)=\left\{x,Tx,...{T}^{n}x\right\}$ $n=1,2,3...$

$O(x,\infty)=\left\{x,Tx,...\right\}$ Please prove that $\And[O(x,\infty)]=sup\left\{\And[O(x,n)]:n\in\Bbb{N}\right\}$.

First comment: I assume that $f$ is the same as $T$ here?

Second comment: Informally, you should think of this function $\&(A)$ as being the diameter of the set $A$. It is the sup of the distance between any two points of $A$.

Now coming on to the sets $O(x,n)$ and $O(x,\infty)$, each set $O(x,n)$ is contained in $O(x,\infty)$, so the diameter of $O(x,n)$ is less than or equal to the diameter of $O(x,\infty)$. It follows that $\sup\left\{\And[O(x,n)]:n\in\Bbb{N}\right\} \leqslant \And[O(x,\infty)]$.

To prove the reverse inequality, notice that if you take any two points $T^px$ and $T^qx$ in $\And[O(x,\infty)]$, they must be contained in one of the sets $O(x,n)$ (namely, take $n$ to be whichever of $p$, $q$ is larger). So $d(T^px,T^qx) \leqslant \&[O(x,n)] \leqslant \sup\left\{\And[O(x,n)]:n\in\Bbb{N}\right\}$. Now take the sup over all $p$ and $q$ to conclude that $\And[O(x,\infty)] \leqslant \sup\left\{\And[O(x,n)]:n\in\Bbb{N}\right\}$.

 

[ozkan12]

Dear professor,

First of all, thank you so much...But I didnt understant second part...How we get conclude that $\And[O(x,\infty)]=sup\left\{\And[O(x,n)]:n\in\Bbb{N}\right\}$ by taking sup for all p and q... Thank you for your attention...Best wishes...:)

 

[Taryn1]

lol I read this thread title too fast and for a minute I thought it said 'proof of superman'!

 

[Evgeny.Makarov]

Suppose that $I$ is an index set and $f$ is a function that maps indices into numbers (or, more generally, to elements of an ordered set).

The definition of supremum has two clauses.

(1) $$f(i)\le \sup_{j\in I}f(j)$$ for all $i\in I$.

(2) If $f(i)\le s'$ for all $i\in I$, then $$\sup_{i\in I}f(i)\le s'$$.

Clause (1) bounds the supremum from below, and clause (2) bounds it from above.

Lemma 1. If $A\subseteq B\subseteq I$, then $\sup\limits_{i\in A}f(i)\le\sup\limits_{i\in B}f(i)$.

Proof. For all $i\in B$, including $i\in A$, we have $f(i)\le\sup\limits_{i\in B}f(i)$ by (1); therefore, $\sup\limits_{i\in A}f(i)\le\sup\limits_{i\in B}f(i)$ by (2).

Now I rewrite Opalg's proof in more detail.

Lemma 2. $\sup_{n\in\Bbb{N}}\And[O(x,n)] \le \And[O(x,\infty)]$.

Proof. $O(x,n)\subset O(x,\infty)$; therefore,
\[
\And[O(x,n)]=
\sup\limits_{a,b\in O(x,n)}d(a,b)\le
\sup\limits_{a,b\in O(x,\infty)}d(a,b)=
\And[O(x,\infty)]\qquad(3)
\]
by Lemma 1. In applying the lemma, we instantiate $I$ with $M^2$ (i.e., the set of all ordered pairs $\langle a,b\rangle$ where $a,b\in M$), $A$ with $O(x,n)^2$, $B$ with $O(x,\infty)^2$, and for $i=\langle a,b\rangle\in M$ we set $f(i)=d(a,b)$. Now by applying clause (2) of the definition above to (3), we get
\[
\sup\limits_{n\in\mathbb{N}}\And[O(x,n)]\le \And[O(x,\infty)],
\]
as required. In applying (2) we set $I=\mathbb{N}$ and $f(i)=\And[O(x,i)]$ for $i\in\mathbb{N}$.

Lemma 3. $\And[O(x,\infty)]\le\sup_{n\in\Bbb{N}}\And[O(x,n)]$.

Proof. As Opalg explains, for all $a,b\in O(x,\infty)$ there exists an $n\in\Bbb N$ such that $a,b\in O(x,n)$. Therefore, for all $a,b\in O(x,\infty)$ we have
\begin{align}
d(a,b)&\le\sup\limits_{a,b\in O(x,n)}d(a,b)&&\text{by (1)}\\
&=\And[O(x,n)]\\
&\le\sup_{n\in\Bbb N}\And[O(x,n)]&&\text{by (1)}
\end{align}
Now we apply (2) to $d(a,b)\le\sup\limits_{n\in\Bbb N}\And[O(x,n)]$ and get
\[
\sup_{a,b\in O(x,\infty)}\le\sup_{n\in\Bbb N}\And[O(x,n)].
\]

The required statement is the conjunction of Lemmas 2 and 3.

 

[ozkan12]

Dear Makarov,

You are very good, thank you for everything...This is very helpful for me...:) Thank you for your attention...Best wishes :)