# Thread: Infintie unions of closed sets in R

1. ## Infintie unions of closed sets in R

Hello, I'm hoping you guys can help me unravel this apparent contradiction and show me where my reasoning went wrong. I'm working out of the book "Mathematical Analysis" by Apostol (second addition) and here a few of Apostol's definitions and notation.

B(a;r). The open ball of radius r centered at a.

Interior Point. Let S be a subset of R (reals), and assume that a is an element of S. Then a is called an interior point of S if there is an open ball with center at a, all of whose points belong to S.

Open Set. A set S in R is called open if all its points are interior points.

Closed Set. A set S in R is called close dif its complement R - S is open.

Later in the book its mentioned that the empty set is open (vacuously). In the argument I wrote, I tried to show that a countably infinite union, S, of closed sets is open. But this union should be equal to R and so the complement is the empty set which is open, implying S is closed. Here is a link to what I wrote up, you can click on the picture to magnify it so its readable:

Image of closed sets - Photobucket - Video and Image Hosting

2. Consider another example:
$\displaystyle A_n = \left[ {\frac{1}{{n + 2}},1 - \frac{1}{{n + 1}}} \right]\quad ,\quad \bigcup\limits_n {A_n } = ?$

3. Oh neat, that union should be the open interval (0,1) right Plato? I should of mentioned that a theorem in Apostol's book says that the union of a finite collection of closed sets is closed. So I started to wonder if all infinite unions of closed sets are closed. This is a cool example.

4. Originally Posted by Jacobsen
Oh neat, that union should be the open interval (0,1) right Plato? I should of mentioned that a theorem in Apostol's book says that the union of a finite collection of closed sets is closed. So I started to wonder if all infinite unions of closed sets are closed. This is a cool example.
A similar situation occurs with open sets in case thou art wondering. Any union of open sets is open and so with finite intersections. However, it does not necessarily work with infinite intesections - see if you can find a counterexample.

5. Thanks for the response Hacker. It took me a bit, but how about if we let A_n = (1 - 1/n, 1 + 1/n). Then the intersection of all A_n should be the singleton {1}. This set is closed because the complement is (-infinity, 1) union (1, infinity) which is open.

6. Originally Posted by Jacobsen
Thanks for the response Hacker. It took me a bit, but how about if we let A_n = (1 - 1/n, 1 + 1/n). Then the intersection of all A_n should be the singleton {1}. This set is closed because the complement is (-infinity, 1) union (1, infinity) which is open.
Yes.