Prove that $\mathbb{R}$ is a complete metric space

Question

Prove that $(\mathbb{R},d)$ ; $d=|x-y|$ is a complete metric space.

The definition of complete metric we use is "$X$ is a complete metric space if every Cauchy sequence in $X$ is convergent in $X$".

Answer 1

I'll say enough to get you started.

Let $\{x_{n}\}$ be a Cauchy sequence in $\mathbb{R}.$ Then $\{x_{n}\}$ is bounded. To see this, we first choose $N$ so large that $n \geq N \Rightarrow |x_{n}-x_{N}|<1,$ which is possible by the Cauchy property. You should be able to show that $\{|x_{n}|\}$ is thus bounded by $$ M = \max\{|x_{1}|,\ldots,|x_{N-1}|,|x_{N}|+1\}$$ by using (a version of) the triangle inequality.

Now, the Bolzano-Weierstrass Theorem states that any bounded sequence has a convergent subsequence $\{x_{n_{k}}\}$ for $k = 1,2,\ldots.$ Let the limit of this subsequence be $\ell.$

Can you now use the Cauchy property to show that the limit of the sequence $\{x_{n}\}$ is also $\ell$?