Is ∅ ⊈ { ∅, 1, 2 } False?












0












$begingroup$


Is this ∅ ⊈ { ∅, 1, 2 } true or false ?



Also, I am confuse since this { ∅, 1, 2 } has already contain a ∅, does it still contain another ∅ meaning like : { ∅, ∅, 1, 2 } ?



Is ∅ ∈ { ∅, 1, 2 } true ? & {∅} ∈ { ∅, 1, 2 } false ?






elementary-set-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
    $endgroup$
    – Le Anh Dung
    Jan 21 at 5:47






  • 2




    $begingroup$
    Don't confuse $in$ with $subseteq$.
    $endgroup$
    – Cheerful Parsnip
    Jan 21 at 5:55










  • $begingroup$
    $emptyset subset A $ is always true no matter what set $A $ is.
    $endgroup$
    – fleablood
    Jan 21 at 7:21
















0












$begingroup$


Is this ∅ ⊈ { ∅, 1, 2 } true or false ?



Also, I am confuse since this { ∅, 1, 2 } has already contain a ∅, does it still contain another ∅ meaning like : { ∅, ∅, 1, 2 } ?



Is ∅ ∈ { ∅, 1, 2 } true ? & {∅} ∈ { ∅, 1, 2 } false ?






elementary-set-theory







share|cite|improve this question









$endgroup$












  • $begingroup$
    It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
    $endgroup$
    – Le Anh Dung
    Jan 21 at 5:47






  • 2




    $begingroup$
    Don't confuse $in$ with $subseteq$.
    $endgroup$
    – Cheerful Parsnip
    Jan 21 at 5:55










  • $begingroup$
    $emptyset subset A $ is always true no matter what set $A $ is.
    $endgroup$
    – fleablood
    Jan 21 at 7:21














0












0








0





$begingroup$


Is this ∅ ⊈ { ∅, 1, 2 } true or false ?



Also, I am confuse since this { ∅, 1, 2 } has already contain a ∅, does it still contain another ∅ meaning like : { ∅, ∅, 1, 2 } ?



Is ∅ ∈ { ∅, 1, 2 } true ? & {∅} ∈ { ∅, 1, 2 } false ?






elementary-set-theory







share|cite|improve this question









$endgroup$




Is this ∅ ⊈ { ∅, 1, 2 } true or false ?



Also, I am confuse since this { ∅, 1, 2 } has already contain a ∅, does it still contain another ∅ meaning like : { ∅, ∅, 1, 2 } ?



Is ∅ ∈ { ∅, 1, 2 } true ? & {∅} ∈ { ∅, 1, 2 } false ?






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










asked Jan 21 at 5:44









J.SJ.S

555




555












  • $begingroup$
    It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
    $endgroup$
    – Le Anh Dung
    Jan 21 at 5:47






  • 2




    $begingroup$
    Don't confuse $in$ with $subseteq$.
    $endgroup$
    – Cheerful Parsnip
    Jan 21 at 5:55










  • $begingroup$
    $emptyset subset A $ is always true no matter what set $A $ is.
    $endgroup$
    – fleablood
    Jan 21 at 7:21


















  • $begingroup$
    It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
    $endgroup$
    – Le Anh Dung
    Jan 21 at 5:47






  • 2




    $begingroup$
    Don't confuse $in$ with $subseteq$.
    $endgroup$
    – Cheerful Parsnip
    Jan 21 at 5:55










  • $begingroup$
    $emptyset subset A $ is always true no matter what set $A $ is.
    $endgroup$
    – fleablood
    Jan 21 at 7:21
















$begingroup$
It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
$endgroup$
– Le Anh Dung
Jan 21 at 5:47




$begingroup$
It is trivially false since empty set is the subset of any set. This comes from the fact that empty set does not contain any element.
$endgroup$
– Le Anh Dung
Jan 21 at 5:47




2




2




$begingroup$
Don't confuse $in$ with $subseteq$.
$endgroup$
– Cheerful Parsnip
Jan 21 at 5:55




$begingroup$
Don't confuse $in$ with $subseteq$.
$endgroup$
– Cheerful Parsnip
Jan 21 at 5:55












$begingroup$
$emptyset subset A $ is always true no matter what set $A $ is.
$endgroup$
– fleablood
Jan 21 at 7:21




$begingroup$
$emptyset subset A $ is always true no matter what set $A $ is.
$endgroup$
– fleablood
Jan 21 at 7:21










3 Answers
3






active

oldest

votes


















2












$begingroup$

$$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
Also, $${oslash}subset{oslash,1,2}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry, why is {⊘} ⊂ {⊘,1,2} ?
    $endgroup$
    – J.S
    Jan 21 at 5:53










  • $begingroup$
    Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
    $endgroup$
    – Anik Bhowmick
    Jan 21 at 5:54












  • $begingroup$
    @J.S Because $oslashin{oslash,1,2}.$
    $endgroup$
    – Michael Rozenberg
    Jan 21 at 5:54



















0












$begingroup$

No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    $neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.



    Consider the case $A=phi.$



    Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.






    share|cite|improve this answer









    $endgroup$













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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2












      $begingroup$

      $$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
      Also, $${oslash}subset{oslash,1,2}.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Sorry, why is {⊘} ⊂ {⊘,1,2} ?
        $endgroup$
        – J.S
        Jan 21 at 5:53










      • $begingroup$
        Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
        $endgroup$
        – Anik Bhowmick
        Jan 21 at 5:54












      • $begingroup$
        @J.S Because $oslashin{oslash,1,2}.$
        $endgroup$
        – Michael Rozenberg
        Jan 21 at 5:54
















      2












      $begingroup$

      $$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
      Also, $${oslash}subset{oslash,1,2}.$$






      share|cite|improve this answer









      $endgroup$













      • $begingroup$
        Sorry, why is {⊘} ⊂ {⊘,1,2} ?
        $endgroup$
        – J.S
        Jan 21 at 5:53










      • $begingroup$
        Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
        $endgroup$
        – Anik Bhowmick
        Jan 21 at 5:54












      • $begingroup$
        @J.S Because $oslashin{oslash,1,2}.$
        $endgroup$
        – Michael Rozenberg
        Jan 21 at 5:54














      2












      2








      2





      $begingroup$

      $$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
      Also, $${oslash}subset{oslash,1,2}.$$






      share|cite|improve this answer









      $endgroup$



      $$oslashsubset{oslash,1,2}$$ and $$oslashin{oslash,1,2}.$$
      Also, $${oslash}subset{oslash,1,2}.$$







      share|cite|improve this answer












      share|cite|improve this answer



      share|cite|improve this answer










      answered Jan 21 at 5:47









      Michael RozenbergMichael Rozenberg

      107k1894198




      107k1894198












      • $begingroup$
        Sorry, why is {⊘} ⊂ {⊘,1,2} ?
        $endgroup$
        – J.S
        Jan 21 at 5:53










      • $begingroup$
        Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
        $endgroup$
        – Anik Bhowmick
        Jan 21 at 5:54












      • $begingroup$
        @J.S Because $oslashin{oslash,1,2}.$
        $endgroup$
        – Michael Rozenberg
        Jan 21 at 5:54


















      • $begingroup$
        Sorry, why is {⊘} ⊂ {⊘,1,2} ?
        $endgroup$
        – J.S
        Jan 21 at 5:53










      • $begingroup$
        Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
        $endgroup$
        – Anik Bhowmick
        Jan 21 at 5:54












      • $begingroup$
        @J.S Because $oslashin{oslash,1,2}.$
        $endgroup$
        – Michael Rozenberg
        Jan 21 at 5:54
















      $begingroup$
      Sorry, why is {⊘} ⊂ {⊘,1,2} ?
      $endgroup$
      – J.S
      Jan 21 at 5:53




      $begingroup$
      Sorry, why is {⊘} ⊂ {⊘,1,2} ?
      $endgroup$
      – J.S
      Jan 21 at 5:53












      $begingroup$
      Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
      $endgroup$
      – Anik Bhowmick
      Jan 21 at 5:54






      $begingroup$
      Because it's an element of the set. Don't you write $$ {1} subset {1, 2, 3} $$??
      $endgroup$
      – Anik Bhowmick
      Jan 21 at 5:54














      $begingroup$
      @J.S Because $oslashin{oslash,1,2}.$
      $endgroup$
      – Michael Rozenberg
      Jan 21 at 5:54




      $begingroup$
      @J.S Because $oslashin{oslash,1,2}.$
      $endgroup$
      – Michael Rozenberg
      Jan 21 at 5:54











      0












      $begingroup$

      No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.






      share|cite|improve this answer









      $endgroup$


















        0












        $begingroup$

        No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.






        share|cite|improve this answer









        $endgroup$
















          0












          0








          0





          $begingroup$

          No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.






          share|cite|improve this answer









          $endgroup$



          No, not like that. Since null set is subset of all sets, the question in your title is valid. At least I think so.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 21 at 5:47









          Anik BhowmickAnik Bhowmick

          583417




          583417























              0












              $begingroup$

              $neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.



              Consider the case $A=phi.$



              Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.



                Consider the case $A=phi.$



                Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.



                  Consider the case $A=phi.$



                  Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.






                  share|cite|improve this answer









                  $endgroup$



                  $neg (Asubset B)$ iff there exists a member of $A$ that is not a member of $B$.



                  Consider the case $A=phi.$



                  Since $phi$ has no members, there cannot exist a member of $phi$ that fails to belong to $B$. Therefore $neg (neg (phisubset B))$, equivalently $phisubset B.$ For $any $ $B$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 21 at 6:59









                  DanielWainfleetDanielWainfleet

                  35.3k31648




                  35.3k31648






























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