Need Help Understanding Why These Conditions Imply Closedness I would like to know why these conditions imply that the set $A \subseteq \mathbb{R}^{n}$ is closed:


*

*for every Cauchy sequence $(x_n)_{n\in \mathbf{N}}$ in $A$, its limit  $x\in A$.

*for every sequence $(x_n)_{n\in \mathbf{N}}$ there exists a subsequence $(x_{n_{k}})_{k\in \mathbf{N}} \rightarrow x$ such that it limit $x \in A$.
From what I know the first argument is the definition of a closed set, but I don't really understand why. I don't know why the second one guarantees a closed set.
Any insight will be highly appreciated.
 A: *

*Let $(x_n)$ be a convergent sequence in $A$ with limit $x_0$. Then $(x_n)$ is Cauchy. By the first condition: $x_0 \in A$. This shows that $A$ is closed.

*If for every sequence $(x_n)$ in A there exists a convergent subsequence with limit in $A$, then $A$ is compact, hence closed.
A: I'll assume that we have the following definition. 
Definition: A set $A\subset \mathbf{R}^k$ is said to be closed if it contains its limit points. That is, given any convergent sequence $(x_n)_{n\in \mathbf{N}}\to x$ with $x_n\in A$ for all $n\in \mathbf{N}$, we have $x\in A$.
Now, $\mathbf{R}^k$ is a complete metric space, so all Cauchy sequences converge and conversely. Thus, if $(x_n)_{n\in \mathbf{N}}$ is a convergent sequence to a limit $x$, it is also a Cauchy sequence. As such, $x\in A$. So, $A$ is closed.
If every sequence $(x_n)_{n\in \mathbf{N}}$ in $A$ admits a convergent subsequence, which converges in $A$ then we know that $A$ is compact, by the sequential compactness definition. By the Heine-Borel Theorem, this in turn implies that $A$ is closed and bounded, in particular – closed.
