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



The Next CEO of Stack OverflowThe Matrix in ZFC: A Set-Theoretic Foundation of Matrices?Ordered Pairs (Ordering multiple elements)Problems with this Cartesian Product definitionOrdered pairs and set -builder as function classAn issue with the Kuratowski ordered pair definitionBijection between cartesian products of sets to a given set of same total cardinality“nonempty” in the definition of the cartesian productOrdered sets - can there be two of the same element? (set theory)Ordered pair - why so complicated?Constructing lists from 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







share|cite|improve this question











$endgroup$








  • 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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 17 at 6:46









drhab

104k545136




104k545136










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














  • 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










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