Subspace of metric space with finitely many points is complete Show  that: 
If a  subspace  $Y$  of  a metric space consists  of finitely many 
points,  then  $Y$  is  complete.
 A: Given any finite subspace $Y$ of a metric space $X$, the Cauchy sequences $(x_n)$ are precisely those sequences which are eventually constant. To see this, observe that $\exists \ N\in \mathbb{N}$ such that $\forall \ m,n\geq N$ we have $d(x_m,x_n)<\epsilon$, $\forall \epsilon>0$, but if $x_n\neq x_m$ then the distance won't be zero. Since any Cauchy sequence is eventually constant hence it converges in $Y$ itself,i.e., $Y$ is complete.
A: Any Cauchy sequence in such a space must be eventually constant.
A: Let $\left\{ x_{n}\right\} $ be a Cauchy sequence in $Y$. That is, for all $\epsilon>0$, there exists $N$ s.t. for all $n,m\geq N$, $d\left(x_{n},x_{m}\right)<\epsilon$. Let $\epsilon_{0}$ be the minimum distance between pairs of distinct points in $Y$. $\epsilon_0$ is well-defined since it is a minimum of $\left|Y\right|\left(\left|Y\right|-1\right)$ nonnegative real numbers. From the  hypothesis, there exists an $N_{0}$ s.t. for all $n,m\geq N_{0}$, $d\left(x_{n},x_{m}\right)<\epsilon_{0}$. But, this only occurs if the sequence is constant for all $n\geq N_{0}$!
(hence, every Cauchy sequence converges to some point in the space, and the space is complete)
