Let $X= \left \{(x_i)_{i \geq 1} : x_i \in \{0,1 \}\ \text {for all}\ i \geq 1 \right \}$ with the metric $d \left ( (x_i),(y_i) \right ) = \sum\limits_{i \geq 1} \frac {|x_i-y_i|} {2^i}.$ Let $f: X \longrightarrow [0,1]$ be the function defined by $f(x_i)_{i \geq 1} = \sum\limits_{i \geq 1 } \frac {x_i}{2^i}.$ Then which one of the following statement/s is/are correct?
$(1)$ $f$ is onto.
$(2)$ $f$ is open.
Any help will be highly appreciated. Thank you very much.
The issue of whether $f$ is onto does not depend on the metric assigned.
Clearly the restriction of $f$ to the range $[0,1)$ is onto because and real number in $[0,1)$ can be represented as a (possibly unending) sequence of bits, and that is a member of $X$. So the onto issue comes down to whether there is any $x \in X$ such that $f(X) = 1$. And indeed the member $(x_i)_{i \geq 1 1} : \forall i x_i = 1$ has the property that $f(x) = \sum_{i \geq 1} 2^{-i} = 1$. Therefore, $f$ is onto.
The issue of whether $f$ is open becomes the question of whether the image of every open set under the peculiar metric $d$ is also an is an open set under the familiar metric defined by $f$. The latter open sets are the ordinary familiar open sets within $(0,1)$.