Understanding a step in a solution of a problem

Question

The problem and its solution are given below:

But I do not understand why $\{z_{n}\} \subset K$ could anyone explain this for me please?

Answer 1

$f$ is linear so $f(x+y) = f(x) + f(y)$ and $f(x + ay) = f(x) + af(y)$.

And so

$f(z_n)= f(x_1 - \color{blue}{f(x_1)}\frac {x_n}{\color{blue}{f(x_n)}})$and as $f$ is linear, if we treat $\color{blue}{f(x_1)}$ and $\color{blue}{f(x_n)}$ constants we get

$f(z_n)= f(x_1 - \color{blue}{f(x_1)}\frac {x_n}{\color{blue}{f(x_n)}})=$

$f(x_1) - f(\color{blue}{ \frac{f(x_1)}{f(x_n)}}x_n)=$

$f(x_1) - \color{blue}{ \frac{f(x_1)}{f(x_n)}}f(x_n)=$

$f(x_1) - \color{blue}{f(x_1)}\frac {f(x_n)}{\color{blue}{f(x_n)}} = f(x_1)-f(x_1) = 0$

$z_n \in f^{-1}(0) = K$.

Answer 2

For each $n\geq 1$, we have that $$\begin{align} f(z_n) &= f\Big( x_1 - f(x_1)\frac{x_n}{f(x_n)} \Big) \\ &= f(x_1) - f(x_1)\frac{f(x_n)}{f(x_n)} = 0 \end{align}$$ so, for each $n\geq 1$, $f(z_n) = 0$, that is, $z_n \in f^{-1}(\{0\})$.