MHB Can We Say "${y}_{n}=T{y}_{n-1}" in a Complete Metric Space?

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In a complete metric space (X, d), an iteration sequence is defined as xn = T(xn-1) = T^n(x0), where x0 is any point in X. The discussion raises the question of whether an arbitrary sequence yn can be expressed as yn = T(yn-1). It is clarified that if yn is arbitrary and T is predefined, one cannot conclude that yn = T(yn-1). If such a relationship held true, yn would qualify as an iteration sequence, contradicting its arbitrary nature. Therefore, the assertion that yn = T(yn-1) does not hold.
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Let $\left(X,d\right)$ be a complete metric space, and in this space we define iteration sequence by ${x}_{n}=T{x}_{n-1}={T}^{n}{x}_{0}$ ${x}_{0}$ is arbitrary point in $X$...Also, let $\left\{{y}_{n}\right\}$ be a arbitrary sequence in $X$ but $\left\{{y}_{n}\right\}$ is not iteration sequence...İn this case, Can we say that ${y}_{n}=T{y}_{n-1}$ ?

Thank you for your attention...:)
 
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If $\{y_n\}$ is an arbitrary sequence, and $T$ is already given, there's definitely no way you could conclude that $y_n=Ty_{n-1}$. If $y_n=Ty_{n-1}$ were true (which it isn't), then $\{y_n\}$ would be an iteration sequence.
 

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