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}

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 is closed if and only if it contains all its limit points.

The first step is to prove the statement for . So, let us prove that if are closed, then is closed.

Fix a limit point then, there exists a sequence such that .

Here we have three possibilities:

First: All the elements of are in , except for a finite number of elements.

Second: All the elements of are in , except for a finite number of elements.

Third: There are infinite elements of in and in .

In the first case we have that all the sequence, but a finite number of terms, is contained in , which is closed. So, the limit of is in , hence .

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

The third case is less simpler. Let us call the part of that is contained in , and the part of that is contained in . As every subsequence of a convergent sequence is also convergent, and to the same limit. So, and .

Now, as and , and both sets are closed, we conclude that and , therefore .

As the three options give us that , we deduce that is closed.

Finally, with a little inductive step we conclude that .

Step-by-step explanation:

answer: 15

step-by-step explanation:

no association

step-by-step explanation:

if you have eyes and you can use them efficiently you can see that it just looks like random dots like do a line of best fit and like they're so spread out it's such a weak association you might as well consider it nonexistent