3.2.5

Proof

($\Rightarrow$): Suppose $A$ is closed, so then $A$ contains it's limit points. Let $(x_n)\subseteq A$ be a Cauchy Sequence. Then $(x_n)$ converges by being Cauchy. Say it converges to $x$, then since $A$ is closed then $x\in A$ as required since $A$ contains its limit points, which are equivalent to the limits of the convergent sequences of $A$.

($\Leftarrow$): Suppose $\mathrm{\forall }(x_n)\subseteq A$ where $(x_n)\to x$ (ie: the sequence $(x_n)$ is Cauchy because it converges) where $x\in A$. We need to show that $A$ contains it's limit points.

Let $\ell$ be a limit point of $A$. We want to show that $\ell \in A$. Since $\ell$ is a limit point of $A$ then equivalently there is some sequence of points in $A-\{\ell \}$ where it converges to $\ell$. Denote this sequence $(a_n)\in A-\{\ell \}$ where $(a_n)\to \ell$. then since all convergent sequences in $A$ converge to a point in $A$ then $\ell \in A$.

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