# Countable Union/Intersection of Open/Closed sets

• Nov 3rd 2010, 07:22 PM
kathrynmath
Countable Union/Intersection of Open/Closed sets
A set A is called a F set if it can be written as the countable union of closed sets. A set B is called a G set if it can be written as the countable intersection of open sets.
a) Show that a closed interval [a,b] is a G set
b) Show that the half-open interval (a,b] is both a G and an F set.
c) Show that Q is an F set and the set of irrationals I forms a G set.

a) Let [a,b] be a closed interval.
Well, I know a set is closed if it contains all its limit points

I'm kinda confused on this whole idea.
• Nov 3rd 2010, 07:51 PM
Drexel28
Quote:

Originally Posted by kathrynmath
A set A is called a F set if it can be written as the countable union of closed sets. A set B is called a G set if it can be written as the countable intersection of open sets.
a) Show that a closed interval [a,b] is a G set
b) Show that the half-open interval (a,b] is both a G and an F set.
c) Show that Q is an F set and the set of irrationals I forms a G set.

a) Let [a,b] be a closed interval.
Well, I know a set is closed if it contains all its limit points

I'm kinda confused on this whole idea.

I'll give some hints

a) What if we extended the endpoints of the interval by a classically small amount in either direction?
b) What if we took intervals of the form $[a+\varepslion,b]$ and intervals of the form $[a-\varepsilon,b]$. What should $\varepsilon$ really be?
c) Remember that $\mathbb{Q}$ is itself countable. And the second part follows from the first from DeMorgan's laws.
• Nov 3rd 2010, 07:55 PM
kathrynmath
a) Let [a,b] be contained in a set of interval I_n
Let I_n+1 be contained in I_n

b) I don't follow b, and how can were you able to make that a closed interval when it was (a,b]?

c) Since Q is countable, you have to be able to write Q as a countable union.
• Nov 3rd 2010, 07:59 PM
Drexel28
Quote:

Originally Posted by kathrynmath
a) Let [a,b] be contained in a set of interval I_n
Let I_n+1 be contained in I_n

Ok, actually construct the intervals.

Quote:

b) I don't follow b, and how can were you able to make that a closed interval when it was (a,b]?
What if we took $I_n=[a+\frac{1}{n},b]$?

Quote:

c) Since Q is countable, you have to be able to write Q as a countable union.
a countable union of what? When you get it it will make me SING, for otherwise I would be afraid I was some kind of simpLETON.
• Nov 3rd 2010, 08:10 PM
kathrynmath
I get stuck when I try to construct the intervals.

I still don't get b really. By your defining of In, then if I_n+1 is contained in In, it can't contain both endpoints?

c) a countable union of closed sets
• Nov 4th 2010, 07:29 AM
kathrynmath
a) How bout considering the interval (a-1/n, b+1/n)
b) consider the intervals (a-1/n,b+1/n) and [a-1/n,b+1/n}

Is that the right idea? Still not sure on c
• Nov 4th 2010, 08:16 AM
Plato
For part b), I would prefer this: $F_n = \left[ {a + \frac{{b - a}}{{2n}},b} \right]\;\& \,G_n = \left( {a,b + \frac{1}{n}} \right)$
Is it clear that $\bigcup\limits_n {F_n } = \left(a,b\right] = \bigcap\limits_n {G_n }~?$

For part c): there is listing of the rational numbers, $\mathbb{Q} = \left\{ {r_n :n \in \mathbb{Z}^ + } \right\}$
What is this union $\bigcup\limits_n {\left\{ {r_n } \right\}}~?$.

What is this intersection $\bigcap\limits_n {\left(\mathbb{R}\setminus \left\{ {r_n } \right\}\right)}~?$
• Nov 4th 2010, 01:37 PM
kathrynmath
Not really sure how you got that for b).

For c) I can use complements and Demorgan's Laws
• Nov 4th 2010, 02:11 PM
Plato
If $a=0~\&~b=0.5$ you cannot use $I_n=[a+\frac{1}{n},b]$ because $I_1$ is not a proper interval.
That is the reason I defined the $F_n$ the way I did in part b. Each $F_n$ is well defined an looks like $[\alpha,\beta]$.