Let $E$ and $F$ be subsets of a metric space $X.$ Show that $$(E\cup F)' = E' \cup F'$$
I have seen proofs of this problem but what is bothering me is the contradicting nature of two arguments that I have come up with.
I shall denote $D_r(x)$ to mean the deleted neighborhood of $x$ with radius $r.$
Argument I :
\begin{align} x\in(E\cup F)'\\ \iff D_r(x)\cap(E\cup F)\ne \emptyset,\forall r>0\\ \iff (D_r(x)\cap E)\cup(D_r(x)\cap F)\ne \emptyset,\forall r>0\\ \iff D_r(x)\cap E \ne \emptyset \text{ or }D_r(x)\cap F \ne \emptyset, \forall r>0\\ \iff x\in E' \text{ or }x\in F' \\ \iff x\in E'\cup F' \\\end{align}
It was all well and good until (for the fun of it) I decided to prove the same using contraposition :
Argument II : \begin{align} x\notin(E\cup F)'\\ \iff \exists r>0 \text{ s.t }D_r(x)\cap(E\cup F) = \emptyset,\\ \iff \exists r>0 \text{ s.t }(D_r(x)\cap E)\cup(D_r(x)\cap F)= \emptyset\\ \iff \exists r>0 \text{ s.t }D_r(x)\cap E = \emptyset \text{ and }D_r(x)\cap F = \emptyset\\ \iff x\notin E' \text{ and }x\notin F' \\ \iff x\notin E'\cap F' \\\end{align}
Where am I going wrong in the second argument? I feel like I got my negations all wrong but I am not able to find it.
Your reasoning in Argument II is correct up to the second last line. However, \begin{equation} x\not\in E' \text{ and } x\not\in F' \iff x \not\in E'\cap F' \end{equation} is incorrect. In fact, we have (where $\land$ denotes "and"; $\lor$ denotes "or"; and $\lnot$ denotes "not"): \begin{split} x\not\in E' \text{ and } x\not\in F' &\iff \lnot(x\in E')\land \lnot(x\in F') \\ &\iff \lnot (x\in E' \lor x\in F') \qquad \text{(by De Morgan's law)}\\ &\iff \lnot (x \in E'\cup F') \\ &\iff x \not\in E' \cup F'. \end{split}
Remark. This mistake is not specific to metric spaces, but rather to set theory.