Subsets and Supersets when evaluating null sets
$begingroup$
In tryin to evaluate whether the following is true am I expanding these sets correctly:
{{$emptyset$}} = { $emptyset$, {$emptyset$}} therefore {$emptyset$} $in$ {{$emptyset$}} $equiv$ T
But I'm not sure if the following is equivalent due to the nature of the empty set..
{{$emptyset$}, {$emptyset$}} = {$emptyset$ ,{$emptyset$}} ?
elementary-set-theory
edited Jan 24 at 5:12
Jack Pfaffinger
374112
asked Jan 24 at 3:00
ElliottElliott
596
$endgroup$
add a comment |
$begingroup$
In tryin to evaluate whether the following is true am I expanding these sets correctly:
{{$emptyset$}} = { $emptyset$, {$emptyset$}} therefore {$emptyset$} $in$ {{$emptyset$}} $equiv$ T
But I'm not sure if the following is equivalent due to the nature of the empty set..
{{$emptyset$}, {$emptyset$}} = {$emptyset$ ,{$emptyset$}} ?
elementary-set-theory
edited Jan 24 at 5:12
Jack Pfaffinger
374112
asked Jan 24 at 3:00
ElliottElliott
596
$endgroup$
add a comment |
$begingroup$
In tryin to evaluate whether the following is true am I expanding these sets correctly:
{{$emptyset$}} = { $emptyset$, {$emptyset$}} therefore {$emptyset$} $in$ {{$emptyset$}} $equiv$ T
But I'm not sure if the following is equivalent due to the nature of the empty set..
{{$emptyset$}, {$emptyset$}} = {$emptyset$ ,{$emptyset$}} ?
elementary-set-theory
edited Jan 24 at 5:12
Jack Pfaffinger
374112
asked Jan 24 at 3:00
ElliottElliott
596
$endgroup$
In tryin to evaluate whether the following is true am I expanding these sets correctly:
{{$emptyset$}} = { $emptyset$, {$emptyset$}} therefore {$emptyset$} $in$ {{$emptyset$}} $equiv$ T
But I'm not sure if the following is equivalent due to the nature of the empty set..
{{$emptyset$}, {$emptyset$}} = {$emptyset$ ,{$emptyset$}} ?
elementary-set-theory
elementary-set-theory
edited Jan 24 at 5:12
Jack Pfaffinger
374112
asked Jan 24 at 3:00
ElliottElliott
596
edited Jan 24 at 5:12
Jack Pfaffinger
374112
asked Jan 24 at 3:00
ElliottElliott
596
edited Jan 24 at 5:12
Jack Pfaffinger
374112
edited Jan 24 at 5:12
Jack Pfaffinger
374112
edited Jan 24 at 5:12
Jack Pfaffinger
374112
374112
asked Jan 24 at 3:00
ElliottElliott
596
asked Jan 24 at 3:00
ElliottElliott
596
asked Jan 24 at 3:00
ElliottElliott
596
596
add a comment |
add a comment |
1 Answer
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$begingroup$
The equation: ${{emptyset}} = {emptyset,{emptyset}} $ is false. You have just added another element (the empty set) arbitrarily. The equation ${emptyset}in{{emptyset}}$ is true, however it is irrelevant to the equation written before it. Just look at the one term in the set, it is exactly ${emptyset}$.
The final equation is again false, because ${emptyset} neq emptyset$. Also the first set there has a repeated term which is not allowed.
answered Jan 24 at 4:22
Jack PfaffingerJack Pfaffinger
374112
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
The equation: ${{emptyset}} = {emptyset,{emptyset}} $ is false. You have just added another element (the empty set) arbitrarily. The equation ${emptyset}in{{emptyset}}$ is true, however it is irrelevant to the equation written before it. Just look at the one term in the set, it is exactly ${emptyset}$.
The final equation is again false, because ${emptyset} neq emptyset$. Also the first set there has a repeated term which is not allowed.
answered Jan 24 at 4:22
Jack PfaffingerJack Pfaffinger
374112
$endgroup$
add a comment |
$begingroup$
The equation: ${{emptyset}} = {emptyset,{emptyset}} $ is false. You have just added another element (the empty set) arbitrarily. The equation ${emptyset}in{{emptyset}}$ is true, however it is irrelevant to the equation written before it. Just look at the one term in the set, it is exactly ${emptyset}$.
The final equation is again false, because ${emptyset} neq emptyset$. Also the first set there has a repeated term which is not allowed.
answered Jan 24 at 4:22
Jack PfaffingerJack Pfaffinger
374112
$endgroup$
add a comment |
$begingroup$
The equation: ${{emptyset}} = {emptyset,{emptyset}} $ is false. You have just added another element (the empty set) arbitrarily. The equation ${emptyset}in{{emptyset}}$ is true, however it is irrelevant to the equation written before it. Just look at the one term in the set, it is exactly ${emptyset}$.
The final equation is again false, because ${emptyset} neq emptyset$. Also the first set there has a repeated term which is not allowed.
answered Jan 24 at 4:22
Jack PfaffingerJack Pfaffinger
374112
$endgroup$
The equation: ${{emptyset}} = {emptyset,{emptyset}} $ is false. You have just added another element (the empty set) arbitrarily. The equation ${emptyset}in{{emptyset}}$ is true, however it is irrelevant to the equation written before it. Just look at the one term in the set, it is exactly ${emptyset}$.
The final equation is again false, because ${emptyset} neq emptyset$. Also the first set there has a repeated term which is not allowed.
answered Jan 24 at 4:22
Jack PfaffingerJack Pfaffinger
374112
answered Jan 24 at 4:22
Jack PfaffingerJack Pfaffinger
374112
answered Jan 24 at 4:22
Jack PfaffingerJack Pfaffinger
374112
answered Jan 24 at 4:22
Jack PfaffingerJack Pfaffinger
374112
374112
add a comment |
add a comment |
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