Mixing
Find $bigcap mathcal A^c$ and $bigcup mathcal A^c$ with the proof.
$begingroup$
Let $A_n ={x in Bbb R :-frac {1}{n} lt x lt frac{1}{n}}$,$n in Bbb N$ and define the indexed family $mathcal A^c = { A_{n}^{c} :n in Bbb N }$.
Find $bigcap mathcal A^c$ and $bigcup mathcal A^c$ with the proof.
I have no idea how to do.
index-notation
asked Jan 6 at 3:45
MaggieMaggie
938
$endgroup$
add a comment |
$begingroup$
Let $A_n ={x in Bbb R :-frac {1}{n} lt x lt frac{1}{n}}$,$n in Bbb N$ and define the indexed family $mathcal A^c = { A_{n}^{c} :n in Bbb N }$.
Find $bigcap mathcal A^c$ and $bigcup mathcal A^c$ with the proof.
I have no idea how to do.
index-notation
asked Jan 6 at 3:45
MaggieMaggie
938
$endgroup$
2
$begingroup$
If $A_{n} = left(-frac{1}{n},frac{1}{n}right)$, then $A_{n}^{c} = left(-infty,-frac{1}{n}right]cupleft[frac{1}{n},inftyright)$. Now, draw the sets on the real line.
$endgroup$
– Lucas Corrêa
Jan 6 at 3:53
$begingroup$
Also, did you mean $bigcap A_{n}^{c}$?
$endgroup$
– Lucas Corrêa
Jan 6 at 3:54
$begingroup$
Yes, I mean $bigcap A_{n}^{c}$
$endgroup$
– Maggie
Jan 6 at 3:59
add a comment |
$begingroup$
Let $A_n ={x in Bbb R :-frac {1}{n} lt x lt frac{1}{n}}$,$n in Bbb N$ and define the indexed family $mathcal A^c = { A_{n}^{c} :n in Bbb N }$.
Find $bigcap mathcal A^c$ and $bigcup mathcal A^c$ with the proof.
I have no idea how to do.
index-notation
asked Jan 6 at 3:45
MaggieMaggie
938
$endgroup$
Let $A_n ={x in Bbb R :-frac {1}{n} lt x lt frac{1}{n}}$,$n in Bbb N$ and define the indexed family $mathcal A^c = { A_{n}^{c} :n in Bbb N }$.
Find $bigcap mathcal A^c$ and $bigcup mathcal A^c$ with the proof.
I have no idea how to do.
index-notation
index-notation
asked Jan 6 at 3:45
MaggieMaggie
938
asked Jan 6 at 3:45
MaggieMaggie
938
asked Jan 6 at 3:45
MaggieMaggie
938
asked Jan 6 at 3:45
MaggieMaggie
938
asked Jan 6 at 3:45
MaggieMaggie
938
938
2
$begingroup$
If $A_{n} = left(-frac{1}{n},frac{1}{n}right)$, then $A_{n}^{c} = left(-infty,-frac{1}{n}right]cupleft[frac{1}{n},inftyright)$. Now, draw the sets on the real line.
$endgroup$
– Lucas Corrêa
Jan 6 at 3:53
$begingroup$
Also, did you mean $bigcap A_{n}^{c}$?
$endgroup$
– Lucas Corrêa
Jan 6 at 3:54
$begingroup$
Yes, I mean $bigcap A_{n}^{c}$
$endgroup$
– Maggie
Jan 6 at 3:59
add a comment |
2
$begingroup$
If $A_{n} = left(-frac{1}{n},frac{1}{n}right)$, then $A_{n}^{c} = left(-infty,-frac{1}{n}right]cupleft[frac{1}{n},inftyright)$. Now, draw the sets on the real line.
$endgroup$
– Lucas Corrêa
Jan 6 at 3:53
$begingroup$
Also, did you mean $bigcap A_{n}^{c}$?
$endgroup$
– Lucas Corrêa
Jan 6 at 3:54
$begingroup$
Yes, I mean $bigcap A_{n}^{c}$
$endgroup$
– Maggie
Jan 6 at 3:59
2
2
$begingroup$
If $A_{n} = left(-frac{1}{n},frac{1}{n}right)$, then $A_{n}^{c} = left(-infty,-frac{1}{n}right]cupleft[frac{1}{n},inftyright)$. Now, draw the sets on the real line.
$endgroup$
– Lucas Corrêa
Jan 6 at 3:53
$begingroup$
If $A_{n} = left(-frac{1}{n},frac{1}{n}right)$, then $A_{n}^{c} = left(-infty,-frac{1}{n}right]cupleft[frac{1}{n},inftyright)$. Now, draw the sets on the real line.
$endgroup$
– Lucas Corrêa
Jan 6 at 3:53
$begingroup$
Also, did you mean $bigcap A_{n}^{c}$?
$endgroup$
– Lucas Corrêa
Jan 6 at 3:54
$begingroup$
Also, did you mean $bigcap A_{n}^{c}$?
$endgroup$
– Lucas Corrêa
Jan 6 at 3:54
$begingroup$
Yes, I mean $bigcap A_{n}^{c}$
$endgroup$
– Maggie
Jan 6 at 3:59
$begingroup$
Yes, I mean $bigcap A_{n}^{c}$
$endgroup$
– Maggie
Jan 6 at 3:59
add a comment |
1 Answer
1
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$begingroup$
$cap_n mathcal A_n =cap_n (-frac 1 n,frac 1 n)^{c}=(cup_n (-1,1))^{c}=(-1,1)^{c}=mathbb R setminus (-1,1)$. I leave it to you to show that $cup_n mathcal A_n =mathbb R setminus {0}$.
answered Jan 6 at 4:59
Kavi Rama MurthyKavi Rama Murthy
55.6k42057
$endgroup$
add a comment |
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1 Answer
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1 Answer
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votes
$begingroup$
$cap_n mathcal A_n =cap_n (-frac 1 n,frac 1 n)^{c}=(cup_n (-1,1))^{c}=(-1,1)^{c}=mathbb R setminus (-1,1)$. I leave it to you to show that $cup_n mathcal A_n =mathbb R setminus {0}$.
answered Jan 6 at 4:59
Kavi Rama MurthyKavi Rama Murthy
55.6k42057
$endgroup$
add a comment |
$begingroup$
$cap_n mathcal A_n =cap_n (-frac 1 n,frac 1 n)^{c}=(cup_n (-1,1))^{c}=(-1,1)^{c}=mathbb R setminus (-1,1)$. I leave it to you to show that $cup_n mathcal A_n =mathbb R setminus {0}$.
answered Jan 6 at 4:59
Kavi Rama MurthyKavi Rama Murthy
55.6k42057
$endgroup$
add a comment |
$begingroup$
$cap_n mathcal A_n =cap_n (-frac 1 n,frac 1 n)^{c}=(cup_n (-1,1))^{c}=(-1,1)^{c}=mathbb R setminus (-1,1)$. I leave it to you to show that $cup_n mathcal A_n =mathbb R setminus {0}$.
answered Jan 6 at 4:59
Kavi Rama MurthyKavi Rama Murthy
55.6k42057
$endgroup$
$cap_n mathcal A_n =cap_n (-frac 1 n,frac 1 n)^{c}=(cup_n (-1,1))^{c}=(-1,1)^{c}=mathbb R setminus (-1,1)$. I leave it to you to show that $cup_n mathcal A_n =mathbb R setminus {0}$.
answered Jan 6 at 4:59
Kavi Rama MurthyKavi Rama Murthy
55.6k42057
answered Jan 6 at 4:59
Kavi Rama MurthyKavi Rama Murthy
55.6k42057
answered Jan 6 at 4:59
Kavi Rama MurthyKavi Rama Murthy
55.6k42057
answered Jan 6 at 4:59
Kavi Rama MurthyKavi Rama Murthy
55.6k42057
55.6k42057
add a comment |
add a comment |
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$begingroup$
If $A_{n} = left(-frac{1}{n},frac{1}{n}right)$, then $A_{n}^{c} = left(-infty,-frac{1}{n}right]cupleft[frac{1}{n},inftyright)$. Now, draw the sets on the real line.
$endgroup$
– Lucas Corrêa
Jan 6 at 3:53
$begingroup$
Also, did you mean $bigcap A_{n}^{c}$?
$endgroup$
– Lucas Corrêa
Jan 6 at 3:54
$begingroup$
Yes, I mean $bigcap A_{n}^{c}$
$endgroup$
– Maggie
Jan 6 at 3:59