Equivalent metrics in $\ell ^\infty$ space

Question

$$\ell^{\infty}=\{z=(z_n):{\sup}_{n}|z_n|<\infty\} \ and\ x,y\in \ell^{\infty}$$ $$d_{\infty}(x,y)=sup_{n}|x_n-y_n| $$ and $$d(x,y)=\sum_{n=1}^{\infty}{\frac{1}{2^n}.{\frac{|x_n-y_n|}{1+|x_n-y_n|}}}$$ show that these metrics is not equivalent metrics. How can I do this without using topology

Answer 1

$(x^{(k)}=(1,1,...1,0,0,...): \forall n\geq k; x_n^k=0)\subset \ell ^\infty$ $$\lim_{k \to \infty}{x^k}=(1,1,.....)\in \ell ^ \infty$$ $$x^k \nrightarrow^{d_\infty} (1,1,1,....)$$ We must show $d_\infty(x^k,(1,1,...))\geq \epsilon_0 \ \exists \epsilon_0 >0$ $$Let's\ see \ this.$$ $Take \ \epsilon_0=\frac{1}{2} \ \exists N \in \Bbb{N} \forall k>N ; d_{\infty}(x^k,(1,1,...))=sup_{n}{|x_n-1|}=1\geq\epsilon_0=\frac{1}{2}$ $$x^k \rightarrow^{d} (1,1,1,....)$$ $Given \ \epsilon>0 \ take \ \ N>log_{(\frac{1}{2})}^{2.\epsilon} \ and \ \forall \ k\geq N $ $$d(x^k,(1,1,....))=\sum_{n=1}^{\infty}{\frac{1}{2^n}}{\frac{|x_{n}^{k}-1|}{1+|x_{n}^{k}-1|}}=\sum_{n=1}^{k}{\frac{1}{2^n}}{\frac{|x_{n}^{k}-1|}{1+|x_{n}^{k}-1|}}+\sum_{n=k+1}^{\infty}{\frac{1}{2^n}}{\frac{|x_{n}^{k}-1|}{1+|x_{n}^{k}-1|}}=\sum_{n=k+1}^{\infty}{\frac{1}{2^n}}{\frac{1}{2}}=\sum_{n=1}^{\infty}{\frac{1}{2^{n+1}}}-\sum_{n=1}^{k+1}{\frac{1}{2^{n+1}}}<\frac{1}{2}.\frac{1}{1-\frac{1}{2}}-\sum_{n=1}^{k+1}{\frac{1}{2^{n+1}}}<1-\sum_{n=1}^{k+1}{\frac{1}{2^{n+1}}}=1-{\frac{1}{2}}.\frac{1-(\frac{1}{2})^{k+1}}{1-(\frac{1}{2})}=(\frac{1}{2})^{k+1}<(2^{-1-N})<\epsilon$$ Because $$\sum_{n=1}^{k}{\frac{1}{2^n}}{\frac{|(1=x_{n}^{k})-1|}{1+|x_{n}^{k}-1|}}=\sum_{n=1}^{k}{\frac{1}{2^n}}{\frac{|1-1|}{1+|1-1|}}=\sum^{n=1}_{k}{\frac{1}{2^k}}.0=0$$ and this means $x^{k} \rightarrow ^ {d} (1,1,...)$ then $d_{\infty}\neq d$