Self studying anything seems to be rather hard, as there is not much help I can get when I get stuck. I have a specific question regarding metric spaces, and that is, how can we find the correct $\delta>0$ in terms of $\epsilon>0$, such that $B_{L_2}(x,\delta)\subset B_{L_\infty}(x,\epsilon)?$ I am able to do this for $B_{L_2}(x,\delta)\subset B_{L_1}(x,\epsilon)$, but the former case seems harder for me. Can anyone help me out please?
Notation clarifcation $($$B_d(a,r)$ is an open ball with center $a$, and $x$ such that $d(x,a)< r$. $L_2$ is the Euclidean metric and $L_\infty$ is the metric denoted by $d_{\infty}((x_1,....,x_n),(y_1,...,y_n))=\max\{|x_1-y_1|,....,|x_n-y_n|\}$, so $B_{L_\infty}(a,r) = \{x:d_{\infty}(x,a)<r\})$
We may let $\delta =\epsilon$ because $$y\in B_2(x,\epsilon)\implies\sum_{i=1}^n(x_i-y_i)^2<\epsilon^2\implies$$ $$\implies \forall i\in \{1,...,n\}\;( |x_i-y_i|<\epsilon)\implies$$ $$\implies y\in B_{\infty}(x,\epsilon).$$ This is the largest possible value for $\delta.$ Because if $\delta>\epsilon$ then the point $y=(y_1,...,y_n),$ where $y_1=x_1+\frac {\delta+\epsilon}{2}$ and $y_i=x_i$ for $i>2,$ is a member of $B_2(x,\delta)$ but not a member of $B_{\infty}(x,\epsilon).$
It may help to draw a picture of $B_2(x,r)$ and $B_{\infty}(x,r)$ with $n=2$ or $n=3.$
BTW. The reason for the notation $B_{\infty}(x,r)$ is that $d_p(x,y)=[\;\sum_i|x_i-y_i|^p\;]^{1/p}$ is a metric for $1\leq p\in \Bbb R,$ and $\lim_{p\to \infty}d_p(x,y)= \max_i|x_i-y_i|.$