# Thread:

1. ## Question involving a collection of closed intervals

Let $F$ be the collection of closed intervals $A_n=[1/n, 1-1/n]$ for $n=3,4,5,...$. What do you notice about $\bigcup F$? Is it closed, open, both, or neither?

I'm afraid the wording and available material has got me stumped on this, and being sick at the moment doesn't help.

2. Notice that $\left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1$.
Can either 1 or 0 be included?

3. Originally Posted by Plato
Notice that $\left( {\frac{1}{n}} \right) \to 0\;\& \;\left( {1 - \frac{1}{n}} \right) \to 1$.
Can either 1 or 0 be included?
I should note, the original question said that $n=1,2,3,...$, but our Prof. told us to change it to $n=3,4,5,...$ when he realized that when $n=1, A_n=[1,0]$ and when $n=2, A_n=[1/2,1/2]$ (or at least that's what I heard him say). I wrote the question word-for-word (besides the change our Prof. gave us), so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.

4. Originally Posted by Runty
so it doesn't say anything about 0 or 1 being inclusive or not. It just asks us whether the set is open, closed, both or neither.
That is the whole point: neither 0 nor 1 can be included.
So what is the union?

5. Originally Posted by Plato
That is the whole point: neither 0 nor 1 can be included.
So what is the union?
I am presently guessing, but here is my answer thus far.

$\bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]$

My guess so far indicates that the union is both open and closed, though I could easily be wrong.

6. Originally Posted by Runty
I am presently guessing, but here is my answer thus far.

$\bigcup\limits_{n = 1}^\infty {[1/n, 1-1/n]}=(0,1]$
For any $k$ does $1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?$
If not how can the union be $(0,1]?$

7. Originally Posted by Plato
For any $k$ does $1\in\left[ {\frac{1}{k},1 - \frac{1}{k}} \right]?$
If not how can the union be $(0,1]?$
Point taken. I guess it's a completely open interval, then.

EDIT: One more thing, I've looked at some work done on this in another forum, for which I will provide the link.

http://www.mymathforum.com/viewtopic...=15805&start=0

Is the work found here accurate?
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