Prove that T : K $\rightarrow$ $l_{\infty}$ is an isometry on K

Question

Prove that T : K $\rightarrow$ $\ell_\infty$ is an isometry on K, but T has no fixed points, where $K=\{x \in \ell_\infty | \ \lim x_n =1\}$ and T(x)=(0, $x_1 , x_2 ,..)$ for $x = (x_1 , x_2,...)$.

Is an exercise by N L Carothers book, real analysis. Chapter 8 exercise 70

I've already shown that K is closed subset of $\ \ell_\infty $ and that $\ T(K) \subset K$.

I have some ideas but am a bit confused with the symbols. Any help will be respectful.

Answer 1

It is obvious from the definition of norm that $T$ is an isometry. [ $\sup \{0,|x_1|,|x_2|,...\}=\sup \{|x_1|,|x_2|,...\}$].

If $Tx=x$ where $x=(x_1,x_2,...)$ we get $x_1=0, x_2=x_2,x_3=x_2,...$ which implies $x_n=0$ for all $n$. But then $\lim x_n \neq 1$ so there is no fixed point for $T$.