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

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In a complete metric space $(X,d)$, the iteration sequence is defined as ${x}_{n}=T{x}_{n-1}={T}^{n}{x}_{0}$, where ${x}_{0}$ is an arbitrary point in $X$. The discussion clarifies that for an arbitrary sequence $\{y_n\}$ in $X$, one cannot assert that ${y}_{n}=T{y}_{n-1}$ unless $\{y_n\}$ is specifically defined as an iteration sequence. Thus, the conclusion is that the arbitrary nature of $\{y_n\}$ prevents any definitive relationship with the operator $T$.

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ozkan12
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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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