- #1

- 211

- 0

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

Compute [tex]\bigcup[/tex] (i [tex]\in[/tex] I) Ui and [tex]\bigcap[/tex] (i [tex]\in[/tex] I) Ui if

a. I=

i.

[tex]\bigcap[/tex] (i [tex]\in[/tex] 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.

[tex]\bigcup[/tex] (i [tex]\in[/tex] 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=

i.

[tex]\bigcap[/tex] (i [tex]\in[/tex] I) Ui = (0,0)

For the double containment argument do I write what if x is not equal to 0?

ii.

[tex]\bigcup[/tex] (i [tex]\in[/tex] I) Ui = (0, 0.5)

Compute [tex]\bigcup[/tex] (i [tex]\in[/tex] I) Ui and [tex]\bigcap[/tex] (i [tex]\in[/tex] I) Ui if

a. I=

**N**, the set of all positive natural numbers and U_n = (-n,n) [tex]\subseteq[/tex]**R**(U_n is the open interval between -n and n)i.

[tex]\bigcap[/tex] (i [tex]\in[/tex] 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.

[tex]\bigcup[/tex] (i [tex]\in[/tex] 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)) [tex]\subseteq[/tex]**R**(U_r is the open interval between 0 and 1/(1+r^2))i.

[tex]\bigcap[/tex] (i [tex]\in[/tex] I) Ui = (0,0)

For the double containment argument do I write what if x is not equal to 0?

ii.

[tex]\bigcup[/tex] (i [tex]\in[/tex] I) Ui = (0, 0.5)

Last edited: