Completeness of R2 with Taxicab Norm

[gtfitzpatrick]

Homework Statement

Given R is complete, prove that R² is complete with the taxicab norm

The Attempt at a Solution

you know that ,x_k $$\rightarrow$$ x , y_k $$\rightarrow$$ y

Then, given $$\epsilon$$, choose N_x and N_y so that $$\left|x_n - x_m\left|$$ and $$\left|y_n - y_m\left|$$ are less than $$\epsilon/2$$ respectively, whenever m,n $$\geq$$ N = $$\left|N_x\left|+\left|N_y\left|$$.

Then d(($$\ x_n,y_n$$),($$\ x_m,y_m$$)) = $$\sqrt{(x_n - x_m)^2 + (y_n - y_m)^2}$$ $$\leq$$ $$\sqrt{(\epsilon^2 /4) + (\epsilon^2 /4)}$$ = $$\epsilon/2$$ < $$\epsilon$$

i've modified an answer from another question here, i think this work but I am not sure...

 

[Tinyboss]

Let $$x_n,y_n$$ be Cauchy sequences in R, then you know they have limits, x,y, elements of R.

Given epsilon>0, choose Nx such that $$|x_n-x|<\varepsilon/2$$ for all n>Nx, and choose Ny the same way. Then let N=max(Nx,Ny).

Then for n>N, you have:
$$d((x_n,y_n),(x,y))=|x_n-x|+|y_n-y|<\varepsilon.$$
(Remember it specified the taxicab norm, not Euclidean!)

Now you know that $$(x_n,y_n)$$ converges to (x,y), and that (x,y) is actually in R2. So R2 is complete.