Mathematics, 02.10.2019 19:00, joel3410

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}

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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 $\cup_{i=1}^{n}A_i$ .

Step-by-step explanation:

answered: Guest

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answered: Guest

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