How Kuratowski representation of ordered paris produce different set than Hausdorff representation of ordered...
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How Kuratowski representation of ordered paris produce different set than Hausdorff representation of ordered pairs
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While defining Cartesian products I came across the kuratowski representation of ordered pair of two elements $(a,b)$ as: $${{a},{a,b}}$$ and a slight modification of Hausdorff definition of ordered pair of two elements $(a,b)$ as: $${{a,varnothing},{b,{varnothing}}}$$
It was stated that they do not form equal sets but not stated how.
All of this is in reference to ZFC set theory.
elementary-set-theory
edited Mar 17 at 6:46
drhab
104k545136
asked Mar 17 at 5:36
Ujjal MajumdarUjjal Majumdar
142
$endgroup$
add a comment |
$begingroup$
While defining Cartesian products I came across the kuratowski representation of ordered pair of two elements $(a,b)$ as: $${{a},{a,b}}$$ and a slight modification of Hausdorff definition of ordered pair of two elements $(a,b)$ as: $${{a,varnothing},{b,{varnothing}}}$$
It was stated that they do not form equal sets but not stated how.
All of this is in reference to ZFC set theory.
elementary-set-theory
edited Mar 17 at 6:46
drhab
104k545136
asked Mar 17 at 5:36
Ujjal MajumdarUjjal Majumdar
142
$endgroup$
1
$begingroup$
Consider the pair $(varnothing,varnothing)$. The Kuratowski set corresponding to it is ${{varnothing},{varnothing,varnothing}}= {{varnothing}}$. The Hausdoff set corresponding to it ${{varnothing,varnothing},{varnothing,{varnothing}}} = {{varnothing},{varnothing,{varnothing}}}$. So the Kuratowski set has one element, the Hausdorff set has two.
$endgroup$
– Arturo Magidin
Mar 17 at 5:39
1
$begingroup$
Well, they certainly don't look equal. It would be more surprising if they were equal than if they weren't.
$endgroup$
– Eric Wofsey
Mar 17 at 5:42
add a comment |
$begingroup$
While defining Cartesian products I came across the kuratowski representation of ordered pair of two elements $(a,b)$ as: $${{a},{a,b}}$$ and a slight modification of Hausdorff definition of ordered pair of two elements $(a,b)$ as: $${{a,varnothing},{b,{varnothing}}}$$
It was stated that they do not form equal sets but not stated how.
All of this is in reference to ZFC set theory.
elementary-set-theory
edited Mar 17 at 6:46
drhab
104k545136
asked Mar 17 at 5:36
Ujjal MajumdarUjjal Majumdar
142
$endgroup$
While defining Cartesian products I came across the kuratowski representation of ordered pair of two elements $(a,b)$ as: $${{a},{a,b}}$$ and a slight modification of Hausdorff definition of ordered pair of two elements $(a,b)$ as: $${{a,varnothing},{b,{varnothing}}}$$
It was stated that they do not form equal sets but not stated how.
All of this is in reference to ZFC set theory.
elementary-set-theory
elementary-set-theory
edited Mar 17 at 6:46
drhab
104k545136
asked Mar 17 at 5:36
Ujjal MajumdarUjjal Majumdar
142
edited Mar 17 at 6:46
drhab
104k545136
asked Mar 17 at 5:36
Ujjal MajumdarUjjal Majumdar
142
edited Mar 17 at 6:46
drhab
104k545136
edited Mar 17 at 6:46
drhab
104k545136
edited Mar 17 at 6:46
drhab
104k545136
104k545136
asked Mar 17 at 5:36
Ujjal MajumdarUjjal Majumdar
142
asked Mar 17 at 5:36
Ujjal MajumdarUjjal Majumdar
142
asked Mar 17 at 5:36
Ujjal MajumdarUjjal Majumdar
142
142
1
$begingroup$
Consider the pair $(varnothing,varnothing)$. The Kuratowski set corresponding to it is ${{varnothing},{varnothing,varnothing}}= {{varnothing}}$. The Hausdoff set corresponding to it ${{varnothing,varnothing},{varnothing,{varnothing}}} = {{varnothing},{varnothing,{varnothing}}}$. So the Kuratowski set has one element, the Hausdorff set has two.
$endgroup$
– Arturo Magidin
Mar 17 at 5:39
1
$begingroup$
Well, they certainly don't look equal. It would be more surprising if they were equal than if they weren't.
$endgroup$
– Eric Wofsey
Mar 17 at 5:42
add a comment |
1
$begingroup$
Consider the pair $(varnothing,varnothing)$. The Kuratowski set corresponding to it is ${{varnothing},{varnothing,varnothing}}= {{varnothing}}$. The Hausdoff set corresponding to it ${{varnothing,varnothing},{varnothing,{varnothing}}} = {{varnothing},{varnothing,{varnothing}}}$. So the Kuratowski set has one element, the Hausdorff set has two.
$endgroup$
– Arturo Magidin
Mar 17 at 5:39
1
$begingroup$
Well, they certainly don't look equal. It would be more surprising if they were equal than if they weren't.
$endgroup$
– Eric Wofsey
Mar 17 at 5:42
1
1
$begingroup$
Consider the pair $(varnothing,varnothing)$. The Kuratowski set corresponding to it is ${{varnothing},{varnothing,varnothing}}= {{varnothing}}$. The Hausdoff set corresponding to it ${{varnothing,varnothing},{varnothing,{varnothing}}} = {{varnothing},{varnothing,{varnothing}}}$. So the Kuratowski set has one element, the Hausdorff set has two.
$endgroup$
– Arturo Magidin
Mar 17 at 5:39
$begingroup$
Consider the pair $(varnothing,varnothing)$. The Kuratowski set corresponding to it is ${{varnothing},{varnothing,varnothing}}= {{varnothing}}$. The Hausdoff set corresponding to it ${{varnothing,varnothing},{varnothing,{varnothing}}} = {{varnothing},{varnothing,{varnothing}}}$. So the Kuratowski set has one element, the Hausdorff set has two.
$endgroup$
– Arturo Magidin
Mar 17 at 5:39
1
1
$begingroup$
Well, they certainly don't look equal. It would be more surprising if they were equal than if they weren't.
$endgroup$
– Eric Wofsey
Mar 17 at 5:42
$begingroup$
Well, they certainly don't look equal. It would be more surprising if they were equal than if they weren't.
$endgroup$
– Eric Wofsey
Mar 17 at 5:42
add a comment |
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$begingroup$
Consider the pair $(varnothing,varnothing)$. The Kuratowski set corresponding to it is ${{varnothing},{varnothing,varnothing}}= {{varnothing}}$. The Hausdoff set corresponding to it ${{varnothing,varnothing},{varnothing,{varnothing}}} = {{varnothing},{varnothing,{varnothing}}}$. So the Kuratowski set has one element, the Hausdorff set has two.
$endgroup$
– Arturo Magidin
Mar 17 at 5:39
1
$begingroup$
Well, they certainly don't look equal. It would be more surprising if they were equal than if they weren't.
$endgroup$
– Eric Wofsey
Mar 17 at 5:42