(ps: I'm teaching myself analysis the first time as a sophomore in high school using Understanding Analysis, which means that I do not have so much experience in dealing with serious mathematics, so please be patient with me. Thanks!)
I'm trying to prove that a finite union of closed sets is also closed using the definition of limit point (that I summarized) as
$$
x\textrm{ is a limit point of } F \Leftrightarrow \forall \epsilon > 0, \exists a \in F, \textrm{such that } a \in V_{\epsilon}(x) \cap F \textrm{ and } a\neq x
$$
where $V_{\epsilon}(x)$ is the $\epsilon$-neighborhood of $x$.
Here is my attempt:
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Goal: prove that if $F_{i}$ closed $\forall 1\le i \le n$, then $\displaystyle \bigcup^{n}_{i=1}F_{i}$ is also closed.
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My proof: if $x$ is a limit point of $\bigcup^{n}_{i=1}F_{i}$, then $\forall \epsilon > 0, \exists a \in \displaystyle \bigcup^{n}_{i=1}F_{i},$ such that $a \in V_{\epsilon}(x) \cap \left(\displaystyle \bigcup^{n}_{i=1}F_{i}\right)$ and $a\neq x$. Now since $a \in V_{\epsilon}(x) \cap \left(\displaystyle \bigcup^{n}_{i=1}F_{i}\right)$, $a$ must be in $\bigcup^{n}_{i=1}F_{i}$, and therefore at least one $F_{i}$. This means that $\exists 1 \le i \le n, \forall \epsilon > 0, \exists a \in F_{i}$, such that $ a \in V_{\epsilon}(x) \cap F_{i} \textrm{ and } a\neq x$, hence $x$ is a limit point of $F_{i}$ for some $i$. Since each $F_{i}$ is closed (this is our assumption), $x \in F_{i}$ for some $i$, which guarantee that $x \in \bigcup^{n}_{i=1}F_{i}$. We picked an arbitrary $x$ in the beginning as a limit point of $\bigcup^{n}_{i=1}F_{i}$ and have just proven that $x$ is in $\bigcup^{n}_{i=1}F_{i}$, indicating that the union is a closed set.
Now here's my real question: I have (seemingly) proven the theorem I claimed I would without employing the fact that the union is finite, which means that (it seems like) my proof can be generalized to prove any arbitrary union of closed sets is also closed, which is certainly false. So where have I made a mistake in my proof? Is there a way to fix it without changing the climate of the entire proof?
(ps: I have seen people using the complement of open sets to prove this theorem, or other proofs using contradictions. I wholeheartedly accept both of those ways, but I just wanted to explore proving it directly.)
Best Answer
It's better to assume that $x$, the limit point of $\bigcup_{i=1}^n F_i$ is not a limit point of any $F_i,i=1,2,\ldots,n$. That gives you $n$ many $\varepsilon_i>0$ that witness the "non-limit pointness" so $V_{\varepsilon_i}(a) \cap F_i \subseteq \{a\}$. (The intersection is empty or just $\{a\}$).
Then $\varepsilon = \min_{i=1}^n \varepsilon_i >0$ has $$V_\varepsilon(a) \cap \left( \bigcup_{i=1}^n F_i \right) \subseteq \{a\}$$ contradicting that $a$ is a limit point of the union.
Going from the contrapositive frees us from the dependence of $i$ on $\varepsilon$ that your proof attempt suffers from. You show $$\forall \varepsilon>0: \exists i: V_\varepsilon(a)\cap F_i\setminus \{a\} \neq \emptyset$$ while you must show
$$\exists i: \forall \varepsilon>0: V_\varepsilon(a)\cap F_i\setminus \{a\} \neq \emptyset$$
(to show it is a limit point of some $F_i$) Order of quantifiers is important...