# Thread: Open set, closed set or neither

1. ## Open set, closed set or neither

I want to check for this set to be open, closed or neither

I = {x ∈ R3 : 1 ≤ x1 ≤ 3,0 ≤ x2,−1 ≥ x3}

I think it is closed, but I don't know how to show it.
I have already shown that it is not an open set by saying:
you cannot say that there is for any point x in I an open ball with center z and radius r. Because i we take x1 = 3, then we cannot find a r such that ||x-z||<r.
I know that you can prove that a set is closed by proving that the complement of the set is open, but if I take R3\I, how can I show that for every x there is a open ball? Or have I do it another way?

Thank you

2. ## Re: Open set, closed set or neither

both x2 and x3 are infinite intervals

how is I going to be closed?

3. ## Re: Open set, closed set or neither

But how can you prove it that is not open nor closed?

4. ## Re: Open set, closed set or neither

Originally Posted by Cyn1
But how can you prove it that is not open nor closed?
Ignore what I wrote. It's incorrect. Intervals of the sort $[a,\infty)$ are closed.

5. ## Re: Open set, closed set or neither

the complement is

$\left((-\infty,1)\cup (3, \infty)\right) \times (-\infty, 0) \times (-1, \infty) = (-\infty,1) \times (-\infty, 0) \times (-1, \infty) \bigcup (3, \infty) \times (-\infty, 0) \times (-1, \infty)$

intervals of the form $(-\infty, a)$ or $(a, \infty)$ are open as they contain none of their endpoints.

so $I^c$ is the union of two open 3D intervals which is of course open.

Since $I^c$ is open $I$ is closed

What you need to show is the assertion about open semi-infinite intervals being open which isn't much different from showing finite open intervals are open.

6. ## Re: Open set, closed set or neither

Thank you,

But I have still don't understand it totally.
Have the points in this set satisfy all of the conditions? I mean that the point (2,2,-5) is a point in this set, but is (2,2,2) also a point in this set?
And how can I show that the complement isn't open either? The complement is I^c= {x ∈ R3 : 1>x1, x1>3, 0 > x2, −1 < x3}.

Can someone help me?
Thank you

7. ## Re: Open set, closed set or neither

I am sorry, I saw you answer just a few minutes ago. But I have still one question: Have the points in this set satisfy all of the conditions? I mean that the point (2,2,-5) is a point in this set, but is (2,2,2) also a point in this set?
Thank you

8. ## Re: Open set, closed set or neither

Originally Posted by Cyn1
I am sorry, I saw you answer just a few minutes ago. But I have still one question: Have the points in this set satisfy all of the conditions? I mean that the point (2,2,-5) is a point in this set, but is (2,2,2) also a point in this set?
Thank you
$2 > -1$ so $(2,2,2) \not \in I$

I'm not really sure what's important about that.

9. ## Re: Open set, closed set or neither

Originally Posted by Cyn1
I want to check for this set to be open, closed or neither
I = {x ∈ R3 : 1 ≤ x1 ≤ 3,0 ≤ x2,−1 ≥ x3}
I think it is closed, but I don't know how to show it.
I have already shown that it is not an open set by saying:
you cannot say that there is for any point x in I an open ball with center z and radius r. Because i we take x1 = 3, then we cannot find a r such that ||x-z||<r.
I know that you can prove that a set is closed by proving that the complement of the set is open, but if I take R3\I, how can I show that for every x there is a open ball? Or have I do it another way?
If $\mathcal{O}$ is an open set then $\mathcal{O}$ cannot contain any of its own boundary points.
Is it true that $(3,1,-1)\in\mathcal{I}~?$

Any closed set, $\mathcal{M}$ must contain contain all of its limit points.

Those two facts are very useful in answering this sort of question. The fact that the set $\mathcal{I}$ is connected makes this somewhat easier.

10. ## Re: Open set, closed set or neither

Thank you,

But how can you show that a set M contains all of its limit points?

11. ## Re: Open set, closed set or neither

But actually, I have to prove that the complement of I is open and so that I is closed. But how can you prove that it holds for every x?