Countable Union/Intersection of Open/Closed sets

[kathrynag]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

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.

 

[jbunniii]

Homework Statement

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.

Homework Equations

The Attempt at a Solution

a) Let [a,b] be a closed interval.
Well, I know a set is closed if it contains all its limit pointsI'm kinda confused on this whole idea.

For (a), try constructing a sequence of open intervals I_n with the following properties:

1. [a,b] \subset I_n for every n

2. I_{n+1} \subset I_n for every n

3. L(I_n) \searrow b - a as n \rightarrow \infty, where L(I_n) is the length of I_n

You can do something similar with (b).

(c) is really, really easy.

 

[kathrynag]

Ok I get the idea behind a, but I don't see how that is countable.

b)Let(a,b] be contained in In for every n
I_n+1 is contained in In for every nc)Q is the set of rationals. We know a rational can be written as p/q

 

[jbunniii]

Ok I get the idea behind a, but I don't see how that is countable.

A sequence is countable, almost by definition.

If you define a collection of sets I_1, I_2, I_3, \ldots then the fact that the collection can be indexed by integers means precisely that the collection is countable.

The individual SETS are not countable (any interval with nonzero length contains uncountably many points), but that is not what is required. You just need to have a countable NUMBER of sets.

For (c), you don't need to express the rationals in the form p/q. All you need to know are two things:

1. If you have a set containing a single rational \{x\}, is that a closed set?
2. How many rationals are there?

 

[micromass]

The following proves that {0} is a G-set.

\bigcap_{n\in \mathbb{N}}{]-1/n,1/n[}=\{0\}

Can you generalize this idea?

 

[kathrynag]

c) p/q
Yes, a single rational will be closed and all rationals are countable. So we can write them as a countable union.
For irrationals, I cannot be written as p/q
Not sure where to go with this idea

 

[micromass]

So Q is an F-set since

\mathbb{Q}=\bigcup_{q\in \mathbb{Q}}{\{q\}}

To prove the irrationals are a G-set, try taking complements and apply De Morgans law.

 

[kathrynag]

So try the complement of Q
Q^C=x is not an element of Q.
If I take the union of Q=intersection Q^C by Demorgans laws.
Ok, In get that, but I'm not sure how to tell we have intersection of open sets. Is it because a set O is open iff O^C is closed?

 

[micromass]

Yes! A set is open if and only if it's complement is closed.

What you have proven now can be generalized: the complement of a G-set is an F-set and vice versa.