Determine the power set, $P(A)$ for $A={ x: x=0 text{ or } x in P({0}) }$
$begingroup$
This is my attempted solution for this question but, is there other cases that I should consider?
case 1: $x=0, A={0}, P(A)={emptyset,0}$
case 2:
$xin P({0}), A={emptyset, {0}}, P(A)={emptyset,{emptyset},{0},{emptyset,{0}}}$
elementary-set-theory
edited Jan 8 at 17:08
gt6989b
35.2k22557
asked Jan 8 at 16:41
Harrison LHarrison L
33
$endgroup$
add a comment |
$begingroup$
This is my attempted solution for this question but, is there other cases that I should consider?
case 1: $x=0, A={0}, P(A)={emptyset,0}$
case 2:
$xin P({0}), A={emptyset, {0}}, P(A)={emptyset,{emptyset},{0},{emptyset,{0}}}$
elementary-set-theory
edited Jan 8 at 17:08
gt6989b
35.2k22557
asked Jan 8 at 16:41
Harrison LHarrison L
33
$endgroup$
$begingroup$
I think you are missing $x=0$ in your set, in addition to $emptyset$ and ${0}$ which you already have.
$endgroup$
– EuxhenH
Jan 8 at 16:53
add a comment |
$begingroup$
This is my attempted solution for this question but, is there other cases that I should consider?
case 1: $x=0, A={0}, P(A)={emptyset,0}$
case 2:
$xin P({0}), A={emptyset, {0}}, P(A)={emptyset,{emptyset},{0},{emptyset,{0}}}$
elementary-set-theory
edited Jan 8 at 17:08
gt6989b
35.2k22557
asked Jan 8 at 16:41
Harrison LHarrison L
33
$endgroup$
This is my attempted solution for this question but, is there other cases that I should consider?
case 1: $x=0, A={0}, P(A)={emptyset,0}$
case 2:
$xin P({0}), A={emptyset, {0}}, P(A)={emptyset,{emptyset},{0},{emptyset,{0}}}$
elementary-set-theory
elementary-set-theory
edited Jan 8 at 17:08
gt6989b
35.2k22557
asked Jan 8 at 16:41
Harrison LHarrison L
33
edited Jan 8 at 17:08
gt6989b
35.2k22557
asked Jan 8 at 16:41
Harrison LHarrison L
33
edited Jan 8 at 17:08
gt6989b
35.2k22557
edited Jan 8 at 17:08
gt6989b
35.2k22557
edited Jan 8 at 17:08
gt6989b
35.2k22557
35.2k22557
asked Jan 8 at 16:41
Harrison LHarrison L
33
asked Jan 8 at 16:41
Harrison LHarrison L
33
asked Jan 8 at 16:41
Harrison LHarrison L
33
33
$begingroup$
I think you are missing $x=0$ in your set, in addition to $emptyset$ and ${0}$ which you already have.
$endgroup$
– EuxhenH
Jan 8 at 16:53
add a comment |
$begingroup$
I think you are missing $x=0$ in your set, in addition to $emptyset$ and ${0}$ which you already have.
$endgroup$
– EuxhenH
Jan 8 at 16:53
$begingroup$
I think you are missing $x=0$ in your set, in addition to $emptyset$ and ${0}$ which you already have.
$endgroup$
– EuxhenH
Jan 8 at 16:53
$begingroup$
I think you are missing $x=0$ in your set, in addition to $emptyset$ and ${0}$ which you already have.
$endgroup$
– EuxhenH
Jan 8 at 16:53
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT
Since $P({0}) = {{0}, emptyset}$, you have
$$
A = { x: x=0 text{ or } x in P({0}) }
= {0} cup P({0})
= {0, {0}, emptyset}
$$
and the power set is given by all subsets of this, so should have a grand total of 8 elements.
answered Jan 8 at 17:11
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT
Since $P({0}) = {{0}, emptyset}$, you have
$$
A = { x: x=0 text{ or } x in P({0}) }
= {0} cup P({0})
= {0, {0}, emptyset}
$$
and the power set is given by all subsets of this, so should have a grand total of 8 elements.
answered Jan 8 at 17:11
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
HINT
Since $P({0}) = {{0}, emptyset}$, you have
$$
A = { x: x=0 text{ or } x in P({0}) }
= {0} cup P({0})
= {0, {0}, emptyset}
$$
and the power set is given by all subsets of this, so should have a grand total of 8 elements.
answered Jan 8 at 17:11
gt6989bgt6989b
35.2k22557
$endgroup$
add a comment |
$begingroup$
HINT
Since $P({0}) = {{0}, emptyset}$, you have
$$
A = { x: x=0 text{ or } x in P({0}) }
= {0} cup P({0})
= {0, {0}, emptyset}
$$
and the power set is given by all subsets of this, so should have a grand total of 8 elements.
answered Jan 8 at 17:11
gt6989bgt6989b
35.2k22557
$endgroup$
HINT
Since $P({0}) = {{0}, emptyset}$, you have
$$
A = { x: x=0 text{ or } x in P({0}) }
= {0} cup P({0})
= {0, {0}, emptyset}
$$
and the power set is given by all subsets of this, so should have a grand total of 8 elements.
answered Jan 8 at 17:11
gt6989bgt6989b
35.2k22557
answered Jan 8 at 17:11
gt6989bgt6989b
35.2k22557
answered Jan 8 at 17:11
gt6989bgt6989b
35.2k22557
answered Jan 8 at 17:11
gt6989bgt6989b
35.2k22557
35.2k22557
add a comment |
add a comment |
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$begingroup$
I think you are missing $x=0$ in your set, in addition to $emptyset$ and ${0}$ which you already have.
$endgroup$
– EuxhenH
Jan 8 at 16:53