I'm trying to prove the following.
For any $A \subset \mathbb{R}^n$, $(\bar{A})' \subset A'$.
where $\bar{A} = A \cup A'$ : closure of $A$.
My attempt
If $(\bar{A})' = \emptyset$ then trivial. Otherwise, $\forall a \in (\bar{A})'$, $\forall \epsilon > 0$ $\exists b \in \bar{A} $ such that $0 < |b-a| < \epsilon/2$.
(Case1) If $b \in A$, since $\epsilon$ was arbitrary, we have $a \in A'$.
(Case2) If $b \in A' \cap A^{c} $, then $\exists c \in A - \{a\}$ such that $0 < |c - b| < \epsilon/2$. By triangle inequality, we have $0 < |c - a| < \epsilon$, since $\epsilon$ was arbitrary, we have $a \in A'$.
Is my proof right? I feel somewhat uncomfortable with the Cases part in my proof. It seems, that my proof is alright, only in the following scenario.
Someone gives me $\epsilon>0$. Then every time I always have such $b \in \bar{A}$, but my $b$ can be in Case 1 or in Case 2, depending on $\epsilon$ So in the situation that the fate of $b$ depends on the choice of $\epsilon$, my reasoning does not seem to be held true. Any ideas?
Just choose $\varepsilon_n =\frac{1}{n}$ then you have a sequence $b_n$ such that either of the cases hold. This sequence is infinite so at least one of the cases must happen for infinitely many $n$ that is when you approach $a$ arbitarily close.