Question

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 1

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
`A\subset\mathbb{R}` is closed if and only if it contains all its limit points.

The first step is to prove the statement for
`n=2`. So, let us prove that if
`A_1,A_2\subset\mathbb{R}` are closed, then
`A_1\cup A_2` is closed.

Fix
`x\in A_1\cup A_2` a limit point then, there exists a sequence
`\{x_n\}_(n=1)^\infty\subset A_1\cup A_2` such that
`x_n\rightarrow x`.

Here we have three possibilities:

First: All the elements of
`\{x_n\}_(n=1)^\infty` are in
`A_1`, except for a finite number of elements.

Second: All the elements of
`\{x_n\}_(n=1)^\infty` are in
`A_2`, except for a finite number of elements.

Third: There are infinite elements of
`\{x_n\}_(n=1)^\infty` in
`A_1` and in
`A_2`.

In the first case we have that all the sequence, but a finite number of terms, is contained in
`A_1`, which is closed. So, the limit of
`\{x_n\}_(n=1)^\infty` is in
`A_1`, hence
`x\in A_1\cup A_2`.

The second case has the same proof, just changing the indices 1 by 2.

The third case is less simpler. Let us call
`y_n` the part of
`\{x_n\}_(n=1)^\infty` that is contained in
`A_1`, and
`z_n` the part of
`\{x_n\}_(n=1)^\infty` that is contained in
`A_2`. As every subsequence of a convergent sequence is also convergent, and to the same limit. So,
`z_n\rightarrow x` and
`y_n\rightarrow x`.

Now, as
`\{y_n\}\subset A_1` and
`\{z_n\}\subset A_2`, and both sets are closed, we conclude that
`x\in A_1` and
`x\in A_2`, therefore
`x\in A_1\cup A_2`.

As the three options give us that
`x\in A_1\cup A_2`, we deduce that
`A_1\cup A_2` is closed.

Finally, with a little inductive step we conclude that
`\cup_(i=1)^(n)A_i`.

Explanation: