Proof of Supremum of $M$ Mapping into Itself

In summary, we proved that $\And[O(x,\infty)]=sup\left\{\And[O(x,n)]:n\in\Bbb{N}\right\}$ by showing that the supremum of the diameters of the sets $O(x,n)$ is equal to the diameter of the set $O(x,\infty)$, and vice versa. This was done using the two clauses of the definition of supremum, and by showing that each set $O(x,n)$ is contained in $O(x,\infty)$. This completes the proof.
  • #1
ozkan12
149
0
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...
 
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  • #2
ozkan12 said:
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 $\And[O(x,\infty)] \leqslant \sup\left\{\And[O(x,n)]:n\in\Bbb{N}\right\}$.
 
  • #3
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...:)
 
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  • #4
lol I read this thread title too fast and for a minute I thought it said 'proof of superman'!
 
  • #5
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) \(\displaystyle f(i)\le \sup_{j\in I}f(j)\) for all $i\in I$.

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

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.
 
  • #6
Dear Makarov,

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

FAQ: Proof of Supremum of $M$ Mapping into Itself

What is "Proof of Supremum of M Mapping into Itself"?

"Proof of Supremum of M Mapping into Itself" is a mathematical concept that involves proving that the supremum (or least upper bound) of a set M is also a member of that set, meaning that the set is closed under the supremum operation. This concept is often used in analysis and calculus.

Why is it important to prove the supremum of a set mapping into itself?

Proving the supremum of a set mapping into itself is important because it allows for the use of certain mathematical tools and techniques that require the set to be closed under the supremum operation. It also helps to ensure the accuracy and validity of mathematical proofs and calculations.

What are some common techniques used in proving the supremum of a set mapping into itself?

There are several common techniques used in proving the supremum of a set mapping into itself, including the use of the least upper bound property, the definition of supremum, and the use of mathematical induction. Other techniques may also be used depending on the specific set and problem being analyzed.

Can the supremum of a set mapping into itself be proven for all types of sets?

Yes, the supremum of a set mapping into itself can be proven for all types of sets, including finite and infinite sets, as long as the set has a well-defined supremum and is closed under the supremum operation.

What are some real-world applications of proving the supremum of a set mapping into itself?

The concept of proving the supremum of a set mapping into itself has various applications in fields such as economics, physics, and engineering. For example, it can be used to analyze the stability of financial systems, determine optimal solutions in game theory, and find maximum or minimum values in optimization problems.

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