# Is the metric space of polynomial of degree $\leq1$ complete with the following metric?

I must determine if the following metric space is complete

$$\mathbb R_1[x]\times \mathbb R_1[x]\rightarrow \mathbb R: d(p(x),q(x))=max\{|p(0)-q(0)|,|p(1)-q(1)|\}$$ where
$$\mathbb R_1[x]=\{p(x)=(ax+b)| a,b \in \mathbb R \}$$ is the subset of polynomials of degree $$\leq 1$$ Is $$(\mathbb R_1[x],d)$$complete?

my try:

Let $$\{p(x)\}_n$$ be a Cauchy sequence

by definition

$$\forall \epsilon, \exists n_0, \forall n,m \geq n_0$$ $$d(p_n(x),p_m(x))< \epsilon$$

$$d(p_n(x),p_m(x))=max\{|p_n(0)-p_m(0)|,|p_n(1)-q_m(1)|\}$$ since the max should be greater than any of the arguments: $$|p_n(0)-p_m(0)|\leq max\{|p_n(0)-p_m(0)|,|p_n(1)-q_m(1)|\}<\epsilon$$ and $$|p_n(1)-p_m(q)|\leq max\{|p_n(0)-p_m(0)|,|p_n(1)-q_m(1)|\}<\epsilon$$ so(I am not sure) this implies that $$p_n(1)=a_n$$ and $$p_n(0)=b_n$$ are Cauchy sequences

Does this make sense? At this part I am stuck, how does this help me(if it is correct) to conclude if the space is complete or not? I should show somehow the Cauchy sequence is convergent

• thanks, i just fixed that typo – J.C.VegaO Feb 7 at 10:26
• Your ideas are correct, though: $p_n(0)$ and $p_n(1)$ are Cauchy sequences, so they have limits $p(0)$ and $p(1)$ in $\mathbb{R}$. Hint: Does this give you a candidate for a limit of the $p_n$? – WoolierThanThou Feb 7 at 10:32
• If the coefficients converge, so does to polynomial to the one with the limit coefficients – J.C.VegaO Feb 7 at 10:50

Consider the map$$\begin{array}{ccc}(\mathbb R_1[x],d)&\longrightarrow&\mathbb R^2\\p(x)&\mapsto&\bigl(p(0),p(1)\bigr).\end{array}$$It is a bijection and it becomes an isometry if we consider on $$\mathbb R^2$$ the metric $$d_1$$ defined by$$d_\infty\bigl((x_1,y_1),(x_2,y_2)\bigr)=\max\bigl\{\lvert x_1-x_2\rvert,\lvert y_1-y_2\rvert\bigr\}.$$But $$(\mathbb R^2,d_\infty)$$ is complete (in fact, $$\mathbb R^n$$ endowed with a metric induced from any norm is complete) and therefore $$(\mathbb R_1[x],d)$$ is complete too.