How to prove the double containment argument for infinite operations with sets?

[laminatedevildoll]

Hi, I am having problems with proving the double containment argument for the following problems. I would appreciate any feedback. Thanks.

Compute $$\bigcup$$ (i $$\in$$ I) Ui and $$\bigcap$$ (i $$\in$$ I) Ui if

a. I=N, the set of all positive natural numbers and U_n = (-n,n) $$\subseteq$$ R (U_n is the open interval between -n and n)
i.
$$\bigcap$$ (i $$\in$$ I) Ui = (infinity)

How do I prove the double containment argument?
Do I say that if x is not equal to infinity then there exists a number such that x=1?
ii.
$$\bigcup$$ (i $$\in$$ I) Ui = (-1,1)

Do I write is x is a member of (-1,1) then there exists a number such that |x| < 1 or 1< 1/|x|

b. I=N, the set of all real numbers and U_r = (0, 1/(1+r^2)) $$\subseteq$$ R (U_r is the open interval between 0 and 1/(1+r^2))
i.
$$\bigcap$$ (i $$\in$$ I) Ui = (0,0)

For the double containment argument do I write what if x is not equal to 0?
ii.
$$\bigcup$$ (i $$\in$$ I) Ui = (0, 0.5)

 

[mathwonk]

in a) the correct answers are that the union is all of R, and the intersection is (-1,1).

The union part is deep and requires the archimedean axiom, that every real number lies in some interval of form (-n,n).

the intersection part is trivial.

in b the answers are again both wrong, the correct answers being the union is (0,1) and the intersection is empty, unless you meant to say the index set was again all positive integers, instead of all reals, in which case the union is indeed (0,1/2).

 

[laminatedevildoll]

in a) the correct answers are that the union is all of R, and the intersection is (-1,1).

The union part is deep and requires the archimedean axiom, that every real number lies in some interval of form (-n,n).

the intersection part is trivial.

For the union of (-n,n)

Could i write x $$\in$$ (-n,n) for some n $$\in$$ N
x $$\in$$ (-n,n) $$\subseteq$$ for all n.
x $$\in$$ (-n,n) then there is an integer N such that |x|<1
1<1/|x| then (-N,N), x $$\in$$ N.

For the intersection, do I still have to do soem proof to show that it is all reals?

in b the answers are again both wrong, the correct answers being the union is (0,1) and the intersection is empty, unless you meant to say the index set was again all positive integers, instead of all reals, in which case the union is indeed (0,1/2).

so if it's the set of all reals then the set is empty?

Second question:
a. I={2,4,6,8..} the set of all even numbers and i Ui={ni: nx $$\in$$ Z}
For the union, is it all reals?
For the intersection is it just {2i,4i,6i...}

b. I={1,3,5,7,9..} the set of all odd numbers and i Ui={ni: nx $$\in$$ Z}
For the union, is it all reals?
For the intersection is it just {i,3i,5i,7i...}

Thank you