Problem of the Day | 13-03-2025 | Topology - Completeness and Closedness in Metric Spaces: Proving the Relationship Between X and $Y \subset X$

Problem: Let $X$ be a metric space and $Y$ be a subset of $X$.

Hints:

Take a Cauchy sequence in $Y$ (note that this will also be Cauchy in $X$). Now use completeness of $X$ and closedness of $Y$ to conclude that $Y$ is complete.
You need to prove $\bar{Y} = Y$. Take a sequence $(y_n)$ in Y such that it converges to $x \in \bar{Y}$. Note that any convergent sequence is Cauchy and then use completeness of $Y$ to conclude $x \in Y$.

Solution: We recall that a metric space $(X,d)$ is complete if every Cauchy sequence is convergent in $X$.

Given that $X$ is complete and $Y$ is closed. We need to prove that $Y$ is complete. Let $\left( y_n \right) $ be a Cauchy sequence in $Y$. Since $Y \subseteq X$ and $\left( y_n \right) $ is Cauchy in $Y$, it must be Cauchy in $X$. As $X$ is complete, $\left( y_n \right) $ will converge in $X$. Thus, $\left( y_n \right) $ is a convergent sequence. Since $Y$ is closed, the limit must be in $Y$ and therefore $\left( y_n \right) $ is convergent in $Y$. Hence, $Y$ is complete.

We will show that $\bar{Y} = Y$. By definition $Y \subseteq \bar{Y} $. For the other side, let $x \in \bar{Y} $. This implies there is a sequence $\left( y_n \right) \subseteq Y$ such that $y_n \rightarrow x$. Since $\left( y_n \right) $ is convergent sequence, and any convergent sequence in a metric space is Cauchy, so $\left( y_n \right) $ must be Cauchy. As $Y$ is complete, $\left( y_n \right) $ must converge in $Y$. In a metric space, the limit of a sequence is unique, so $y_n \rightarrow x$ and hence $x \in Y$. Thus, $\bar{Y} = Y$.

13-03-2025