Let $Y$ be a dense subspace of a space $(X, d)$. Prove that if every Cauchy sequence $(y_n)⊂Y$ is convergent on $X$, so $X$ is complete.
Hello,it is difficult for me to use the properties of the metric spaces, I would be very grateful if you could help me to do this proof, thanks.
Let $(x_n)$ be a Cauchy sequence in $X$. To prove that $X$ is complete we need to show that $(x_n)$ converges. Since $Y$ is dense in $X$ each element in the sequence $(x_n)$ can be approximated by an element in $Y$. Then we can find a sequence $(y_n)$ in $Y$ such that $d(x_n,y_n)<\frac{1}{n}$ for each $n \in \mathbb{Z}_{>0}$. Since $(x_n)$ is Cauchy and $$ d(y_m,y_n) \leq d(y_m,x_m)+d(x_m,x_n)+d(x_n,y_m) < \frac{1}{m}+d(x_m,x_n)+\frac{1}{n}, $$
it follows that $(y_n)$ is also Cauchy. Thus, by hypothesis there is an $x \in X$ such that $y_n \to x$ in $X$. Let $\varepsilon >0$ and choose $N$ such that $\frac{1}{n}< \frac{\varepsilon}{2}$ and $d(x, y_n)< \frac{\varepsilon}{2}$ for all $n \geq N$. Then, $$ d(x,x_n) \leq d(x,y_n)+d(y_n,x_n) < \varepsilon $$ for all $n \geq N$. This proves that $x_n \to x$ in $X$ and therefore $X$ is complete.