How can I prove that the infinite union of certain closed sets is not closed?

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The discussion centers on the properties of closed sets and their infinite unions in topology. The sets defined as $A_n = [-\frac{1}{n}, \frac{1}{n}]$ are closed, but their infinite union results in the closed interval $[-1, 1]$, which is closed. In contrast, the sets $B_n = [-1+\frac{1}{n}, 1-\frac{1}{n}]$ yield an infinite union that approaches the open interval $(-1, 1)$, demonstrating that this union is not closed. Participants clarify the definitions and provide proofs regarding the behavior of these unions as $n$ approaches infinity.

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Hello everyone!

I'm trying to find a set a closed set ${A_n}$ whose inifinte union is not closed. Now, I can picture the following:

If I let $A_n=[-\frac{1}{n}, \frac{1}{n}]$, then $A_n$ is closed, but their union is not, simply because the point $x=0$ seems to be a limit point at """""infinity"""", but it is not in any of the $\{A_n\}$, so the union is therefore not closed.

Now, of course, I'm having trouble showing that, because, I can't always find an $r>0$ s.t. the neighborhood of center $0$ and radius $r$ intersects any of the $\{A_n\}$s.

Similarly, I'm having trouble proving that, if $B_n = [-1+\frac{1}{n}, 1-\frac{1}{n}]$, then their infinite union is not closed.
I would appreciate any help in this... :)
 
Last edited:
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OhMyMarkov said:
Hello everyone!
I'm trying to find a set a closed set ${A_n}$ whose inifinte union is not closed. Now, I can picture the following:
If I let $A_n=[-\frac{1}{n}, \frac{1}{n}]$, then $A_n$ is closed, but their union is not, simply because the point $x=0$ seems to be a limit point at """""infinity"""", but it is not in any of the $\{A_n\}$, so the union is therefore not closed.

Now, of course, I'm having trouble showing that, because, I can't always find an $r>0$ s.t. the neighborhood of center $0$ and radius $r$ intersects any of the $\{A_n\}$s.

Similarly, I'm having trouble proving that, if $B_n = [-1+\frac{1}{n}, 1-\frac{1}{n}]$, then their infinite union is not closed.
You are confused on notation.
$\bigcup\limits_n {\left[ { - \frac{1}{n},\frac{1}{n}} \right]} = \left[ { - 1,1} \right]$ which is a closed set.

$\bigcup\limits_n {\left[ { - 1 + \frac{1}{n},1 - \frac{1}{n}} \right]} = \left( { - 1,1} \right)$ which is an open set.
Note that $\left( { - 1 + \frac{1}{n}} \right) \to - 1$ is a decreasing sequence.
Note that $\left( {1 - \frac{1}{n}} \right) \to 1$ is an increasing sequence.
What is your question?
 
Hi Plato!

Thanks for your reply.

My question is, how can I prove that the union of $[-1+\frac{1}{n}, 1-\frac{1}{n}]$ tends to $(-1,1)$ as $n$ tends to infinity?
 
OhMyMarkov said:
Hi Plato!

Thanks for your reply.

My question is, how can I prove that the union of $[-1+\frac{1}{n}, 1-\frac{1}{n}]$ tends to $(-1,1)$ as $n$ tends to infinity?

I will assume that you have no trouble showing that \(x \in (-1,1)\) is in at least one of the sets: \([-1+\frac{1}{n}, 1-\frac{1}{n}],\ \ n \in \mathbb{N}_+\) and hence in their union . Also that you have no trouble showing that \(\pm 1\) are in none of \([-1+\frac{1}{n}, 1-\frac{1}{n}], \ \ n\in \mathbb{N}_+\) and hence not in their union.

(I am also assuming you have no problem for x >1 or x<-1 either)

Then you are done.

CB
 
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