Why is this relation not reflexive?

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Determine whether the relation R on the set of all Web
pages is reflexive, symmetric, antisymmetric, and/or transitive,
where (a, b) ∈ R if and only if
there is at least one common link onWeb page a and
Web page b.



Why is this relation not reflexive?







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    Note that, perhaps somewhat counterintuitively, the seemingly stronger condition that the pages have all their links in common does yield a reflexive relation (in fact an equivalence relation).
    – joriki
    Jul 31 at 15:24







  • 1




    If a web page has any link at all, then it has a link in common with itself and thus is related to itself. But if a web page has no links at all then it doesnt have any links in common with itself and is no related to itself (or to any page). So either a page is related to itself or it has no links. But having no links is an option. (It's clearly symmetric [and not anti-symmetric] though, and easily seen to not be transitive.)
    – fleablood
    Jul 31 at 15:41














up vote
0
down vote

favorite












Determine whether the relation R on the set of all Web
pages is reflexive, symmetric, antisymmetric, and/or transitive,
where (a, b) ∈ R if and only if
there is at least one common link onWeb page a and
Web page b.



Why is this relation not reflexive?







share|cite|improve this question















  • 1




    Note that, perhaps somewhat counterintuitively, the seemingly stronger condition that the pages have all their links in common does yield a reflexive relation (in fact an equivalence relation).
    – joriki
    Jul 31 at 15:24







  • 1




    If a web page has any link at all, then it has a link in common with itself and thus is related to itself. But if a web page has no links at all then it doesnt have any links in common with itself and is no related to itself (or to any page). So either a page is related to itself or it has no links. But having no links is an option. (It's clearly symmetric [and not anti-symmetric] though, and easily seen to not be transitive.)
    – fleablood
    Jul 31 at 15:41












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Determine whether the relation R on the set of all Web
pages is reflexive, symmetric, antisymmetric, and/or transitive,
where (a, b) ∈ R if and only if
there is at least one common link onWeb page a and
Web page b.



Why is this relation not reflexive?







share|cite|improve this question











Determine whether the relation R on the set of all Web
pages is reflexive, symmetric, antisymmetric, and/or transitive,
where (a, b) ∈ R if and only if
there is at least one common link onWeb page a and
Web page b.



Why is this relation not reflexive?









share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 31 at 15:19









i11_1997

132




132







  • 1




    Note that, perhaps somewhat counterintuitively, the seemingly stronger condition that the pages have all their links in common does yield a reflexive relation (in fact an equivalence relation).
    – joriki
    Jul 31 at 15:24







  • 1




    If a web page has any link at all, then it has a link in common with itself and thus is related to itself. But if a web page has no links at all then it doesnt have any links in common with itself and is no related to itself (or to any page). So either a page is related to itself or it has no links. But having no links is an option. (It's clearly symmetric [and not anti-symmetric] though, and easily seen to not be transitive.)
    – fleablood
    Jul 31 at 15:41












  • 1




    Note that, perhaps somewhat counterintuitively, the seemingly stronger condition that the pages have all their links in common does yield a reflexive relation (in fact an equivalence relation).
    – joriki
    Jul 31 at 15:24







  • 1




    If a web page has any link at all, then it has a link in common with itself and thus is related to itself. But if a web page has no links at all then it doesnt have any links in common with itself and is no related to itself (or to any page). So either a page is related to itself or it has no links. But having no links is an option. (It's clearly symmetric [and not anti-symmetric] though, and easily seen to not be transitive.)
    – fleablood
    Jul 31 at 15:41







1




1




Note that, perhaps somewhat counterintuitively, the seemingly stronger condition that the pages have all their links in common does yield a reflexive relation (in fact an equivalence relation).
– joriki
Jul 31 at 15:24





Note that, perhaps somewhat counterintuitively, the seemingly stronger condition that the pages have all their links in common does yield a reflexive relation (in fact an equivalence relation).
– joriki
Jul 31 at 15:24





1




1




If a web page has any link at all, then it has a link in common with itself and thus is related to itself. But if a web page has no links at all then it doesnt have any links in common with itself and is no related to itself (or to any page). So either a page is related to itself or it has no links. But having no links is an option. (It's clearly symmetric [and not anti-symmetric] though, and easily seen to not be transitive.)
– fleablood
Jul 31 at 15:41




If a web page has any link at all, then it has a link in common with itself and thus is related to itself. But if a web page has no links at all then it doesnt have any links in common with itself and is no related to itself (or to any page). So either a page is related to itself or it has no links. But having no links is an option. (It's clearly symmetric [and not anti-symmetric] though, and easily seen to not be transitive.)
– fleablood
Jul 31 at 15:41










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Because a webpage can have no links.






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    Because a webpage can have no links.






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      Because a webpage can have no links.






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        up vote
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        Because a webpage can have no links.






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        Because a webpage can have no links.







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        answered Jul 31 at 15:19









        Kenny Lau

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