I have the following metric space:
$(M,\left\Vert \cdot\right\Vert_2 ), M=\{(x,\frac{1}{x})\mid x\in(0,\infty)\}$
I need to show that it is complete, therefore showing that each cauchy sequence converges is sufficient.
So let $(x_{n},y_{n})$ Be a cauchy sequence.
$\forall \varepsilon >0 $ there exists $N\in\mathbb{N}$ such that for all $m,n>N$ we have $d((x_{n},y_{n}),(x_{m},y_{m}))<\varepsilon.$
Denote $(x_{n},y_{n})=(x_{1},\frac{1}{x_{1}})$ and $(x_{m},y_{m})=(x_{2},\frac{1}{x_{2}})$
now $d((x_{n},y_{n}),(x_{m},y_{m}))<\varepsilon\iff(\sum\left|x_{1}-x_{2}\right|^{2}+\left|y_{1}-y_{2}\right|^{2})^{1/2}<\varepsilon$
How should I proceed, how can I formally show that it converges?
Hope this helps you.
The metric space is a subspace from $\mathbb{R}^{2}$ which is complete with $\parallel\cdot\parallel_{2}$, so it is sufficient to prove thar $M$ is closed.
In order to prove it, say that $\lbrace (x_{n}, \frac{1}{x_{n}})\rbrace_{n=1}^{\infty}$ is a sequence in $M$ such that it converges to an arbitrary point $(x, y)$. To be $M$ closed means that that limit lies into $M$ for that arbitrary sequence. It's a matter for seeing that $x>0$ since $\lbrace x_{n}\rbrace_{n=1}^{\infty}$ is a positive sequence converging to $x$.
On the other hand, the sequence $\lbrace \frac{1}{x_{n}}\rbrace_{n=1}^{\infty}$ converges to $\frac{1}{x}$ as a result for $\lbrace x_{n}\rbrace_{n=1}^{\infty}$ converging to $x$. By limit's uniqueness $y=\frac{1}{x}$. So indeed the arbitrary sequence in $M$ converges in $M$. Hence $M$ is closed.
If a sequence in $M$ is a Cauchy sequence, then it converges in $\mathbb{R}^{2}$, since $M$ is closed, that limit lies in $M$ which means that $M$ is a complete metric space.