Prove that a finite union of closed sets is also closed (using limit points) Let $F_i$ be a family of closed sets, then we know that $\bigcup_{i=1}^nF_i$ is closed.
Proving that statement is equivalent to proving:

If $p$ is a limit point of $\bigcup_{i=1}^nF_i$ then $p\in\bigcup_{i=1}^nF_i$

It is easy to prove the contrapositive: if $p\notin\bigcup_{i=1}^nF_i$ then $p$ is not a limit point of $\bigcup_{i=1}^nF_i$
However I tried the following direct proof that I am sure it is wrong because it does not use the countable nature of the union. I want to know where I am making a mistake in the following chain of reasonings:
If $p$ is a limit point of $\bigcup_{i=1}^nF_i$ then in every neighbourhood there is a point $q\neq p$, such that $q\in \bigcup_{i=1}^nF_i$. Since $q\in \bigcup_{i=1}^nF_i$ then $q$ belongs to at least one $F_i$, then (and this is what I suspect is false) $p$ is a limit point of $F_i$. Since $F_i$ is closed, then $p\in F_i$, then $p\in\bigcup_{i=1}^nF_i$.
Therefore: If $p$ is a limit point of $\bigcup_{i=1}^nF_i$ then $p\in\bigcup_{i=1}^nF_i$.
Thank you very much in advance.
 A: I know from your other questions you like to see rigorous answers; so I thought I'd do the same on this one too. I did this a slightly longer way since I do not know which definition of closed you are using (so I used a standard definition of open)
def. $U$ is open $\Leftrightarrow \forall x \in U\; \exists \varepsilon >0 : B(x, \varepsilon) \subset U$
def. $x$ is a limit point of $U \Leftrightarrow \forall \varepsilon >0 \, (B(x, \varepsilon)\setminus \{x\}) \cap U \neq \emptyset$


*

*$\forall i\, F_i$ is closed

*$p$ is a limit point of $\cup_{i=1}^n F_i$

*$\forall \varepsilon >0 \, B(p, \varepsilon)\setminus \{p\} \cup (\cap_{i=1}^n F_i) \neq \emptyset$

*$\forall \varepsilon >0 \, B(p, \varepsilon) \cap (\cup_{i=1}^n F_i) \neq \emptyset$

*Suppose $p \not \in \cup_{i=1}^n F_i$ 

*$p \in (\cup_{i=1}^n F_i)^c$

*$p \in \cap_{i=1}^n F_i^c $

*$\forall i \leq n : p \in F_i^c$

*$p \in F_k^c$

*$F_k^c$ is open

*$\forall x \in F_k^c\; \exists \varepsilon >0 : B(x, \varepsilon)\subset F_k^c$

*$\exists \varepsilon >0 : B(p, \varepsilon)\subset F_k^c$

*$B(p, \varepsilon)\subset F_k^c$

*$\forall i \, B(p, \varepsilon)\subset F_i^c$

*$B(p, \varepsilon) \subset  \cap_{i=1}^n F_i^c $

*$B(p, \varepsilon) \cap (\cap_{i=1}^n F_i^c)^c = \emptyset$

*$B(p, \varepsilon) \cap (\cup_{i=1}^n F_i) = \emptyset$

*$ \exists \varepsilon >0: B(p, \varepsilon) \cap (\cup_{i=1}^n F_i) = \emptyset$

*$\lnot \forall \varepsilon >0 \, B(p, \varepsilon) \cap (\cup_{i=1}^n F_i) \neq \emptyset$

*Contradition

*$p \in \cup_{i=1}^n F_i$


Notice that I did not explicitly use the fact that $n$ is finite. This is a very subtle point in the proof that is used implicitly in steps 6 - 8; As a counter example in the case $n \rightarrow \infty$, let $(X, d)$ be a metric space, and $\lim_{n\rightarrow \infty} \cup_{i=1}^{n} F_i = X$ where $F_i \subset F_{i+1}$. Can you see how this would nullify this proof?
A: Use the infinite pigeon-hole principle: If aninfinite set is presented as the union of finitely many subsets,at least one subset is infinite.So if a sequence p(n)  converges to p, where each p(n) belongs to some F(j), then,for at least one j, there are infinitely many n for which p(n) belongs to F(j), so p belongs to this F(j).
