Prove $(\Bbb R^{n},d_{p})$ forms a complete metric space

So I know that in order for a metric space to be complete I must show that every Cauchy sequence in $\Bbb R^{n}$ converges on $\Bbb R^{n}$


So here's my attempt and I'll show where I get a bitty iffy: Firstly we define $$d_p( \vec{a} , \vec{b} )= (\sum \limits_{i=1}^{n} \lvert a_i - b_i\rvert^p)^{\frac{1}{p}} $$ For all $p\geq1$

Let $(\vec{x_n})$ be any sequences of vectors satisfying $$\lim_{l,m\rightarrow\infty}d_p(\vec{x_l} , \vec{x_m})=0 $$

(i.e. it is cauchy). Now the aim is to show that there is a vector $\vec{x} \in \Bbb R^{n} $ such that

$$\lim_{n\rightarrow\infty}d_p(\vec{x_n} , \vec{x})=0 $$

Now I claimed that, for each $i= 1, ..., n $ the components sequence $(x_{n_i})$ of our Cauchy sequence $(\vec{x_n})$ satisfies

$$\lim_{l,m\rightarrow\infty}|x_{l_i}-x_{m_i}|=0 $$ For each $i= 1, ..., n $ we have $$|x_{l_i}- x_{m_i}|^p \leq |x_{l_1}- x_{m_1}|^p+ |x_{l_2}- x_{m_2}|^p+ ...+|x_{l_n}- x_{m_n}|^p $$ $$ |x_{l_i}, x_{m_i}|\leq\left(\sum \limits_{i=1}^{n} |x_{l_i}- x_{m_i}|^p\right)^{\frac{1}{p}} = d_p(\vec{x_l} , \vec{x_m}) $$

Now consider the comparison $$ 0\leq |x_{l_i}- x_{m_i}|\leq d_p(\vec{x_l} , \vec{x_m}) $$ And taking the limits of everything we get $$ 0\leq \lim_{l,m\rightarrow\infty}|x_{l_i}- x_{m_i}|\leq \lim_{l,m\rightarrow\infty}d_p(\vec{x_l} , \vec{x_m}) = 0 $$ Thus by the Squeeze Theorem we know $ \lim_{l,m\rightarrow\infty}|x_{l_i}- x_{m_i}| = 0$

Now here's where I'm not entirely sure what to do however I thought that if I could just say that since I know $(\Bbb R,d)$ is complete $$\lim_{n\rightarrow\infty}|(x_{n_i} - x_i)|=0 $$ And we can rewrite $$\lim_{n\rightarrow\infty}d_p(\vec{x_n} - \vec{x})= \lim_{n\rightarrow\infty} \left(\sum \limits_{i=1}^{n} |x_{n_i}- x_{i}|^p\right)^{\frac{1}{p}} $$ $$ = \left(\sum \limits_{i=1}^{n} \lim_{n\rightarrow\infty} |x_{n_i}- x_{i}|^p\right)^{\frac{1}{p}}= 0$$ Thus I believe we can now say $(\Bbb R^{n},d_{p})$ is complete.

  • Sorry if this was a bit hard to follow but I would appreciate if anyone could tell me if I have proved it well enough if not anything I could add?


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