If \{A_1, A_2,\dots,A_n\} is a collection of closed subsets of LaTeX: \mathbb{R} , then prove that \cup_{i=1}^{n}A_i is closed in LaTeX: \mathbb{R}

Answer:The proof of this statement depends on the definition of closed set we are using. So, in the following explanation I will use that a subset [tex]A\subset\mathbb{R}[/tex] is closed if and only if it contains all its limit points.The first step is to prove the statement for [tex]n=2[/tex]. So, let us prove that if [tex]A_1,A_2\subset\mathbb{R}[/tex] are closed, then [tex]A_1\cup A_2[/tex] is closed.Fix [tex]x\in A_1\cup A_2[/tex] a limit point then, there exists a sequence [tex]\{x_n\}_{n=1}^\infty\subset A_1\cup A_2[/tex] such that [tex]x_n\rightarrow x[/tex]. Here we have three possibilities:First: All the elements of [tex]\{x_n\}_{n=1}^\infty[/tex] are in [tex]A_1[/tex], except for a finite number of elements.Second: All the elements of [tex]\{x_n\}_{n=1}^\infty[/tex] are in [tex]A_2[/tex], except for a finite number of elements.Third: There are infinite elements of [tex]\{x_n\}_{n=1}^\infty[/tex] in [tex]A_1[/tex] and in [tex]A_2[/tex].In the first case we have that all the sequence, but a finite number of terms, is contained in [tex]A_1[/tex], which is closed. So, the limit of [tex]\{x_n\}_{n=1}^\infty[/tex] is in [tex]A_1[/tex], hence [tex]x\in A_1\cup A_2[/tex].The second case has the same proof, just changing the indices 1 by 2.The third case is less simpler. Let us call [tex]y_n[/tex] the part of [tex]\{x_n\}_{n=1}^\infty[/tex] that is contained in [tex]A_1[/tex], and [tex]z_n[/tex] the part of [tex]\{x_n\}_{n=1}^\infty[/tex] that is contained in [tex]A_2[/tex]. As every subsequence of a convergent sequence is also convergent, and to the same limit. So, [tex]z_n\rightarrow x[/tex] and [tex]y_n\rightarrow x[/tex].Now, as [tex]\{y_n\}\subset A_1[/tex] and [tex]\{z_n\}\subset A_2[/tex], and both sets are closed, we conclude that [tex]x\in A_1[/tex] and [tex]x\in A_2[/tex], therefore [tex]x\in A_1\cup A_2[/tex].As the three options give us that [tex]x\in A_1\cup A_2[/tex], we deduce that [tex]A_1\cup A_2[/tex] is closed.Finally, with a little inductive step we conclude that [tex] \cup_{i=1}^{n}A_i[/tex].Step-by-step explanation: