Let $cl(A)$ be the closure of $A\subseteq\mathbb R^n$, the smallest closed set that contains $A$. I'd like to prove $cl(A)=A\cup A'$ with $A'$ defined as the set of all accumulation points of $A$. An accumulation point of $A$ is a point $x$ whose open balls must contain a point of $A$ distinct from $x$.
To show $cl(A)\supseteq A\cup A'$, we can consider separate cases. If $x\in A$, then $x$ must be in $cl(A)$ because the closure is itself a closed set containing $A$. If $x\notin A$, then $x$ must be an accumulation point of $A$, which enables us to show each closed set containing $A$ must contain $x$. This proves that $x\in cl(A)$.
Things have gone well so far, but I have a hard time showing that $cl(A)\subseteq A\cup A'$. Let us pick some point $x$ in $cl(A)$ that does NOT belong to $A$. How do I prove that $x$ is an accumulation point? Thank you.
Best Answer
Suppose that $x \in cl(A)$ and $x \notin A$. Let $B(x,\varepsilon)$ be a neighborhood of $x$. $B(x, \varepsilon)$ must contain points from $A$. Otherwise, $\mathbb{R}^n - B(x, \varepsilon)$ would be a closed set containing $A$ and hence $cl(A)$. But then $x \in \mathbb{R}^n - B(x, \varepsilon)$, which is a contradiction.