given a set closed under finite complementation and union; disprove closeness under countable union and...
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 yesterday
Za Ira
147113
add a comment |
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 yesterday
Za Ira
147113
add a comment |
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 yesterday
Za Ira
147113
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 yesterday
Za Ira
147113
asked yesterday
Za Ira
147113
edited 12 hours ago
asked yesterday
Za Ira
147113
asked yesterday
Za Ira
147113
asked yesterday
Za Ira
147113
147113
add a comment |
add a comment |
1 Answer
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${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathcal A$ for each $n$ but the intersection of these is not in$mathcal A$. For unions just take complements of these sets.
answered yesterday
Kavi Rama Murthy
51.2k31855
add a comment |
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${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathcal A$ for each $n$ but the intersection of these is not in$mathcal A$. For unions just take complements of these sets.
answered yesterday
Kavi Rama Murthy
51.2k31855
add a comment |
${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathcal A$ for each $n$ but the intersection of these is not in$mathcal A$. For unions just take complements of these sets.
answered yesterday
Kavi Rama Murthy
51.2k31855
add a comment |
${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathcal A$ for each $n$ but the intersection of these is not in$mathcal A$. For unions just take complements of these sets.
answered yesterday
Kavi Rama Murthy
51.2k31855
${0,0,cdots,0} times Omega$ (where there are $n$ zeros) is in $mathcal A$ for each $n$ but the intersection of these is not in$mathcal A$. For unions just take complements of these sets.
answered yesterday
Kavi Rama Murthy
51.2k31855
answered yesterday
Kavi Rama Murthy
51.2k31855
answered yesterday
Kavi Rama Murthy
51.2k31855
answered yesterday
Kavi Rama Murthy
51.2k31855
51.2k31855
add a comment |
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