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Let $M = w_0w_1... \in \{0,1\}^*$. For any computable function $f$ define $M_f = w_{f(0)}w_{f(1)}...$

Let for any computable requirement function $f$ there is continuous computable mapping between $M_f$ and $M$ (we can reestablish $M$ by its any computable subsequence)

Is it possible that $M$ is non-computable?

upd: I mean that $g$ is continuos if for any $x$ and $y$ that $x$ is begin of $y$ $g(x)$ is begin of $g(y)$

Let $M = w_0w_1... \in \{0,1\}^*$. For any computable function $f$ define $M_f = w_{f(0)}w_{f(1)}...$

Let for any computable strictly increasing function $f$ there is continuous computable mapping between $M_f$ and $M$ (we can reestablish $M$ by its any computable subsequence)

Is it possible that $M$ is non-computable?

upd: I mean that $g$ is continuos if for any $x$ and $y$ that $x$ is begin of $y$ $g(x)$ is begin of $g(y)$

Let $M = w_0w_1... \in \{0,1\}^*$. For any computable function $f$ define $M_f = w_{f(0)}w_{f(1)}...$

Let for any computable function $f$ with infinitely range of values there is continuous computable mapping between $M_f$ and $M$ (we can reestablish $M$ by its any computable subsequence)

Is it possible that $M$ is non-computable?

upd: I mean that $g$ is continuos if for any $x$ and $y$ that $x$ is begin of $y$ $g(x)$ is begin of $g(y)$

Let $M = w_0w_1... \in \{0,1\}^*$. For any computable function $f$ define $M_f = w_{f(0)}w_{f(1)}...$

Let for any computable requirement function $f$ there is continuous computable mapping between $M_f$ and $M$ (we can reestablish $M$ by its any computable subsequence)

Is it possible that $M$ is non-computable?

upd: I mean that $g$ is continuos if for any $x$ and $y$ that $x$ is begin of $y$ $g(x)$ is begin of $g(y)$