given a set closed under finite complementation and union; disprove closeness under countable union and...
$begingroup$
The collection $mathscr{A}$ of subsets of $Omega={0,1}^{infty}$ is given by
$$mathscr{A}={StimesOmega:Ssubseteq{0,1}^{l},lge1}$$
Then, $mathscr{A}$ is closed under complementation, finite union and hence, finite intersection.
If $A_1,A_2...inmathscr{A}$, is
$(i)$ ${underset{n=1}{stackrel{infty}{bigcup}}}A_ninmathscr{A}$ ?
$(ii)$ ${underset{n=1}{stackrel{infty}{bigcap}}}A_ninmathscr{A}$ ?
Somehow, intuitively the answer is no because here, we consider countable union and intersection rather than finite union and intersection. But, I can't think of a theorem supporting this claim.
probability-theory elementary-set-theory random-walk
asked Jan 5 at 4:52
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
The collection $mathscr{A}$ of subsets of $Omega={0,1}^{infty}$ is given by
$$mathscr{A}={StimesOmega:Ssubseteq{0,1}^{l},lge1}$$
Then, $mathscr{A}$ is closed under complementation, finite union and hence, finite intersection.
If $A_1,A_2...inmathscr{A}$, is
$(i)$ ${underset{n=1}{stackrel{infty}{bigcup}}}A_ninmathscr{A}$ ?
$(ii)$ ${underset{n=1}{stackrel{infty}{bigcap}}}A_ninmathscr{A}$ ?
Somehow, intuitively the answer is no because here, we consider countable union and intersection rather than finite union and intersection. But, I can't think of a theorem supporting this claim.
probability-theory elementary-set-theory random-walk
asked Jan 5 at 4:52
Za IraZa Ira
161115
$endgroup$
add a comment |
$begingroup$
The collection $mathscr{A}$ of subsets of $Omega={0,1}^{infty}$ is given by
$$mathscr{A}={StimesOmega:Ssubseteq{0,1}^{l},lge1}$$
Then, $mathscr{A}$ is closed under complementation, finite union and hence, finite intersection.
If $A_1,A_2...inmathscr{A}$, is
$(i)$ ${underset{n=1}{stackrel{infty}{bigcup}}}A_ninmathscr{A}$ ?
$(ii)$ ${underset{n=1}{stackrel{infty}{bigcap}}}A_ninmathscr{A}$ ?
Somehow, intuitively the answer is no because here, we consider countable union and intersection rather than finite union and intersection. But, I can't think of a theorem supporting this claim.
probability-theory elementary-set-theory random-walk
asked Jan 5 at 4:52
Za IraZa Ira
161115
$endgroup$
The collection $mathscr{A}$ of subsets of $Omega={0,1}^{infty}$ is given by
$$mathscr{A}={StimesOmega:Ssubseteq{0,1}^{l},lge1}$$
Then, $mathscr{A}$ is closed under complementation, finite union and hence, finite intersection.
If $A_1,A_2...inmathscr{A}$, is
$(i)$ ${underset{n=1}{stackrel{infty}{bigcup}}}A_ninmathscr{A}$ ?
$(ii)$ ${underset{n=1}{stackrel{infty}{bigcap}}}A_ninmathscr{A}$ ?
Somehow, intuitively the answer is no because here, we consider countable union and intersection rather than finite union and intersection. But, I can't think of a theorem supporting this claim.
probability-theory elementary-set-theory random-walk
probability-theory elementary-set-theory random-walk
asked Jan 5 at 4:52
Za IraZa Ira
161115
asked Jan 5 at 4:52
Za IraZa Ira
161115
edited Jan 6 at 13:41
Za Ira
asked Jan 5 at 4:52
Za IraZa Ira
161115
asked Jan 5 at 4:52
Za IraZa Ira
161115
asked Jan 5 at 4:52
Za IraZa Ira
161115
161115
add a comment |
add a comment |
1 Answer
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$begingroup$
${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathscr{A}$ for each $n$ but the intersection of these is not in$mathscr{A}$. For unions just take complements of these sets.
edited Jan 16 at 5:20
Za Ira
161115
answered Jan 5 at 4:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42160
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add a comment |
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1 Answer
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$begingroup$
${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathscr{A}$ for each $n$ but the intersection of these is not in$mathscr{A}$. For unions just take complements of these sets.
edited Jan 16 at 5:20
Za Ira
161115
answered Jan 5 at 4:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42160
$endgroup$
add a comment |
$begingroup$
${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathscr{A}$ for each $n$ but the intersection of these is not in$mathscr{A}$. For unions just take complements of these sets.
edited Jan 16 at 5:20
Za Ira
161115
answered Jan 5 at 4:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42160
$endgroup$
add a comment |
$begingroup$
${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathscr{A}$ for each $n$ but the intersection of these is not in$mathscr{A}$. For unions just take complements of these sets.
edited Jan 16 at 5:20
Za Ira
161115
answered Jan 5 at 4:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42160
$endgroup$
${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathscr{A}$ for each $n$ but the intersection of these is not in$mathscr{A}$. For unions just take complements of these sets.
edited Jan 16 at 5:20
Za Ira
161115
answered Jan 5 at 4:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42160
edited Jan 16 at 5:20
Za Ira
161115
edited Jan 16 at 5:20
Za Ira
161115
edited Jan 16 at 5:20
Za Ira
161115
161115
answered Jan 5 at 4:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42160
answered Jan 5 at 4:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42160
answered Jan 5 at 4:59
Kavi Rama MurthyKavi Rama Murthy
58.2k42160
58.2k42160
add a comment |
add a comment |
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