Completeness in metric spaces.

Question

Let $X$ be a metric space and $Y\subset X$ a subspace of $X.$

I have to prove the statement:

(a) Prove that if $X$ is complete and $Y$ is closed, then $Y$ is complete.

(b) Prove that if $Y$ is complete then $Y$is closed in $X$.

I'm following the definition (Complete metric spaces). A metric space is called complete if every Cauchy sequence in the space converges.

But, I am confused how to start to prove....

Answer 1

For (a), if $\{x_n\}$ is a Cauchy sequence in $Y$, then it is a Cauchy sequence in $X$, thus converging to some point $x\in X$. Since $Y$ is closed, $x\in Y$ and so $Y$ is complete.

For (b), if $\{x_n\}$ is a sequence in $Y$ converging to some point $x$, we need to show that $x\in Y$. Since $\{x_n\}$ is a Cauchy sequence in $Y$, it converges to some point $y\in Y$. Then $x=y$ by the uniqueness of limits. Thus $x\in Y$ and so $Y$ is closed.

Answer 2

For (a), you can prove that $Y$ is complete using the definition of completeness. That is, take a sequence in $Y$, assume it is cauchy, and prove it converges.

Use the following facts:

1. $Y$ is closed if and only if every convergent sequence of elements from $Y$ has a limit in $Y$.
2. Every sequence of elements from $Y$ is also a sequence of elements from $X$.

For (b), you can prove that $Y$ is closed by proving that every convergent sequence in $Y$ has a limit in $Y$.