Proving $A in B land B subseteq C to A in C$.












1












$begingroup$


We are given $A in B land B subseteq C to A in C$,
how to proove that it is true for any $A,B,C$?



I am having trouble proving it correctly...
This is what i am thinking:





  • $A in B$ means $B{ A$, other elements from $B}$


  • $B subseteq C$ means $C{ A$, other elements from $B$, other elements from $C}$


  • $A$ belongs to $C$, so it is true.


Is this correct? and how to write every elements excluding $x$?






elementary-set-theory proof-writing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It is okay but it would be better if you could make it more formal. Here are some ideas: if $a in B$ and $B subseteq C$, you can use the definition of $B subseteq C$ to conclude that $a in C$. What's the definition of $B subseteq C$?
    $endgroup$
    – Ashish K
    Dec 31 '18 at 12:31


















1












$begingroup$


We are given $A in B land B subseteq C to A in C$,
how to proove that it is true for any $A,B,C$?



I am having trouble proving it correctly...
This is what i am thinking:





  • $A in B$ means $B{ A$, other elements from $B}$


  • $B subseteq C$ means $C{ A$, other elements from $B$, other elements from $C}$


  • $A$ belongs to $C$, so it is true.


Is this correct? and how to write every elements excluding $x$?






elementary-set-theory proof-writing







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    It is okay but it would be better if you could make it more formal. Here are some ideas: if $a in B$ and $B subseteq C$, you can use the definition of $B subseteq C$ to conclude that $a in C$. What's the definition of $B subseteq C$?
    $endgroup$
    – Ashish K
    Dec 31 '18 at 12:31
















1












1








1





$begingroup$


We are given $A in B land B subseteq C to A in C$,
how to proove that it is true for any $A,B,C$?



I am having trouble proving it correctly...
This is what i am thinking:





  • $A in B$ means $B{ A$, other elements from $B}$


  • $B subseteq C$ means $C{ A$, other elements from $B$, other elements from $C}$


  • $A$ belongs to $C$, so it is true.


Is this correct? and how to write every elements excluding $x$?






elementary-set-theory proof-writing







share|cite|improve this question











$endgroup$




We are given $A in B land B subseteq C to A in C$,
how to proove that it is true for any $A,B,C$?



I am having trouble proving it correctly...
This is what i am thinking:





  • $A in B$ means $B{ A$, other elements from $B}$


  • $B subseteq C$ means $C{ A$, other elements from $B$, other elements from $C}$


  • $A$ belongs to $C$, so it is true.


Is this correct? and how to write every elements excluding $x$?






elementary-set-theory proof-writing




elementary-set-theory proof-writing






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 1 at 14:31







Edvards Zakovskis

















asked Dec 31 '18 at 12:27









Edvards ZakovskisEdvards Zakovskis

354




354








  • 2




    $begingroup$
    It is okay but it would be better if you could make it more formal. Here are some ideas: if $a in B$ and $B subseteq C$, you can use the definition of $B subseteq C$ to conclude that $a in C$. What's the definition of $B subseteq C$?
    $endgroup$
    – Ashish K
    Dec 31 '18 at 12:31
















  • 2




    $begingroup$
    It is okay but it would be better if you could make it more formal. Here are some ideas: if $a in B$ and $B subseteq C$, you can use the definition of $B subseteq C$ to conclude that $a in C$. What's the definition of $B subseteq C$?
    $endgroup$
    – Ashish K
    Dec 31 '18 at 12:31










2




2




$begingroup$
It is okay but it would be better if you could make it more formal. Here are some ideas: if $a in B$ and $B subseteq C$, you can use the definition of $B subseteq C$ to conclude that $a in C$. What's the definition of $B subseteq C$?
$endgroup$
– Ashish K
Dec 31 '18 at 12:31






$begingroup$
It is okay but it would be better if you could make it more formal. Here are some ideas: if $a in B$ and $B subseteq C$, you can use the definition of $B subseteq C$ to conclude that $a in C$. What's the definition of $B subseteq C$?
$endgroup$
– Ashish K
Dec 31 '18 at 12:31












2 Answers
2






active

oldest

votes


















3












$begingroup$

Your attempt at proving it is not clear to me, so I will not comment much on its accuracy. It seems like you have a rough understanding of the underlying notions though.




Suppose $Ain B$ and $Bsubseteq C$. By definition of $Bsubseteq C$, for all $bin B$, we have $bin C$. Let $b=A$. Thus $Ain C$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    In set-theory $in$ is a primitive notion that has no definition.



    It is just some kind of relation that two sets have or have not: $$Bin Ctext{ or }Bnotin C$$



    Based on $in$ a new relation $subseteq$ is defined by stating that: $$Bsubseteq Ciff xin Ctext{ for every }xtext{ that satisfies: }xin B $$



    So - by definition - if $Bsubseteq C$ then $Ain C$ is a true statement whenever $Ain B$.



    This comes to the same as saying that: $$Ain Bwedge Bsubseteq Cimplies Ain C$$is a true statement.



    So here a proof is actually nothing more than the application of a definition.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      I didn't know $in$ was primitive. As such, this comment might not be helpful, but the naïve view of $in$ is that it says the set on its left side is "inside" the set on its right; maybe that's worth editing into your answer (and I'll leave that up to you). Although of heard of, say, geometry being taught without any diagrams or talk of shapes, it is nonetheless instructive to "know" what lines & points are, at least by example, even though they, too, are primitive.
      $endgroup$
      – Shaun
      Dec 31 '18 at 14:15








    • 1




      $begingroup$
      @Shaun I think that your comment on its own is enough on this matter, so will not add something. Actually it is only a shift/translation from $in$ to the primitive notion "inside" (what is that in mathematics?) and gives at most a way to visualize (very useful indeed). Somewhere definitions must stop. You could say that the relation $atext{ defined by means of }b$ is well founded, i.e. every non-empty subset has a least (i.e. primitive) element.
      $endgroup$
      – drhab
      Dec 31 '18 at 15:12













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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3












    $begingroup$

    Your attempt at proving it is not clear to me, so I will not comment much on its accuracy. It seems like you have a rough understanding of the underlying notions though.




    Suppose $Ain B$ and $Bsubseteq C$. By definition of $Bsubseteq C$, for all $bin B$, we have $bin C$. Let $b=A$. Thus $Ain C$.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Your attempt at proving it is not clear to me, so I will not comment much on its accuracy. It seems like you have a rough understanding of the underlying notions though.




      Suppose $Ain B$ and $Bsubseteq C$. By definition of $Bsubseteq C$, for all $bin B$, we have $bin C$. Let $b=A$. Thus $Ain C$.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Your attempt at proving it is not clear to me, so I will not comment much on its accuracy. It seems like you have a rough understanding of the underlying notions though.




        Suppose $Ain B$ and $Bsubseteq C$. By definition of $Bsubseteq C$, for all $bin B$, we have $bin C$. Let $b=A$. Thus $Ain C$.






        share|cite|improve this answer











        $endgroup$



        Your attempt at proving it is not clear to me, so I will not comment much on its accuracy. It seems like you have a rough understanding of the underlying notions though.




        Suppose $Ain B$ and $Bsubseteq C$. By definition of $Bsubseteq C$, for all $bin B$, we have $bin C$. Let $b=A$. Thus $Ain C$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 31 '18 at 14:04

























        answered Dec 31 '18 at 12:35









        ShaunShaun

        8,898113681




        8,898113681























            2












            $begingroup$

            In set-theory $in$ is a primitive notion that has no definition.



            It is just some kind of relation that two sets have or have not: $$Bin Ctext{ or }Bnotin C$$



            Based on $in$ a new relation $subseteq$ is defined by stating that: $$Bsubseteq Ciff xin Ctext{ for every }xtext{ that satisfies: }xin B $$



            So - by definition - if $Bsubseteq C$ then $Ain C$ is a true statement whenever $Ain B$.



            This comes to the same as saying that: $$Ain Bwedge Bsubseteq Cimplies Ain C$$is a true statement.



            So here a proof is actually nothing more than the application of a definition.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I didn't know $in$ was primitive. As such, this comment might not be helpful, but the naïve view of $in$ is that it says the set on its left side is "inside" the set on its right; maybe that's worth editing into your answer (and I'll leave that up to you). Although of heard of, say, geometry being taught without any diagrams or talk of shapes, it is nonetheless instructive to "know" what lines & points are, at least by example, even though they, too, are primitive.
              $endgroup$
              – Shaun
              Dec 31 '18 at 14:15








            • 1




              $begingroup$
              @Shaun I think that your comment on its own is enough on this matter, so will not add something. Actually it is only a shift/translation from $in$ to the primitive notion "inside" (what is that in mathematics?) and gives at most a way to visualize (very useful indeed). Somewhere definitions must stop. You could say that the relation $atext{ defined by means of }b$ is well founded, i.e. every non-empty subset has a least (i.e. primitive) element.
              $endgroup$
              – drhab
              Dec 31 '18 at 15:12


















            2












            $begingroup$

            In set-theory $in$ is a primitive notion that has no definition.



            It is just some kind of relation that two sets have or have not: $$Bin Ctext{ or }Bnotin C$$



            Based on $in$ a new relation $subseteq$ is defined by stating that: $$Bsubseteq Ciff xin Ctext{ for every }xtext{ that satisfies: }xin B $$



            So - by definition - if $Bsubseteq C$ then $Ain C$ is a true statement whenever $Ain B$.



            This comes to the same as saying that: $$Ain Bwedge Bsubseteq Cimplies Ain C$$is a true statement.



            So here a proof is actually nothing more than the application of a definition.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              I didn't know $in$ was primitive. As such, this comment might not be helpful, but the naïve view of $in$ is that it says the set on its left side is "inside" the set on its right; maybe that's worth editing into your answer (and I'll leave that up to you). Although of heard of, say, geometry being taught without any diagrams or talk of shapes, it is nonetheless instructive to "know" what lines & points are, at least by example, even though they, too, are primitive.
              $endgroup$
              – Shaun
              Dec 31 '18 at 14:15








            • 1




              $begingroup$
              @Shaun I think that your comment on its own is enough on this matter, so will not add something. Actually it is only a shift/translation from $in$ to the primitive notion "inside" (what is that in mathematics?) and gives at most a way to visualize (very useful indeed). Somewhere definitions must stop. You could say that the relation $atext{ defined by means of }b$ is well founded, i.e. every non-empty subset has a least (i.e. primitive) element.
              $endgroup$
              – drhab
              Dec 31 '18 at 15:12
















            2












            2








            2





            $begingroup$

            In set-theory $in$ is a primitive notion that has no definition.



            It is just some kind of relation that two sets have or have not: $$Bin Ctext{ or }Bnotin C$$



            Based on $in$ a new relation $subseteq$ is defined by stating that: $$Bsubseteq Ciff xin Ctext{ for every }xtext{ that satisfies: }xin B $$



            So - by definition - if $Bsubseteq C$ then $Ain C$ is a true statement whenever $Ain B$.



            This comes to the same as saying that: $$Ain Bwedge Bsubseteq Cimplies Ain C$$is a true statement.



            So here a proof is actually nothing more than the application of a definition.






            share|cite|improve this answer











            $endgroup$



            In set-theory $in$ is a primitive notion that has no definition.



            It is just some kind of relation that two sets have or have not: $$Bin Ctext{ or }Bnotin C$$



            Based on $in$ a new relation $subseteq$ is defined by stating that: $$Bsubseteq Ciff xin Ctext{ for every }xtext{ that satisfies: }xin B $$



            So - by definition - if $Bsubseteq C$ then $Ain C$ is a true statement whenever $Ain B$.



            This comes to the same as saying that: $$Ain Bwedge Bsubseteq Cimplies Ain C$$is a true statement.



            So here a proof is actually nothing more than the application of a definition.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 31 '18 at 12:51

























            answered Dec 31 '18 at 12:46









            drhabdrhab

            99.1k544130




            99.1k544130












            • $begingroup$
              I didn't know $in$ was primitive. As such, this comment might not be helpful, but the naïve view of $in$ is that it says the set on its left side is "inside" the set on its right; maybe that's worth editing into your answer (and I'll leave that up to you). Although of heard of, say, geometry being taught without any diagrams or talk of shapes, it is nonetheless instructive to "know" what lines & points are, at least by example, even though they, too, are primitive.
              $endgroup$
              – Shaun
              Dec 31 '18 at 14:15








            • 1




              $begingroup$
              @Shaun I think that your comment on its own is enough on this matter, so will not add something. Actually it is only a shift/translation from $in$ to the primitive notion "inside" (what is that in mathematics?) and gives at most a way to visualize (very useful indeed). Somewhere definitions must stop. You could say that the relation $atext{ defined by means of }b$ is well founded, i.e. every non-empty subset has a least (i.e. primitive) element.
              $endgroup$
              – drhab
              Dec 31 '18 at 15:12




















            • $begingroup$
              I didn't know $in$ was primitive. As such, this comment might not be helpful, but the naïve view of $in$ is that it says the set on its left side is "inside" the set on its right; maybe that's worth editing into your answer (and I'll leave that up to you). Although of heard of, say, geometry being taught without any diagrams or talk of shapes, it is nonetheless instructive to "know" what lines & points are, at least by example, even though they, too, are primitive.
              $endgroup$
              – Shaun
              Dec 31 '18 at 14:15








            • 1




              $begingroup$
              @Shaun I think that your comment on its own is enough on this matter, so will not add something. Actually it is only a shift/translation from $in$ to the primitive notion "inside" (what is that in mathematics?) and gives at most a way to visualize (very useful indeed). Somewhere definitions must stop. You could say that the relation $atext{ defined by means of }b$ is well founded, i.e. every non-empty subset has a least (i.e. primitive) element.
              $endgroup$
              – drhab
              Dec 31 '18 at 15:12


















            $begingroup$
            I didn't know $in$ was primitive. As such, this comment might not be helpful, but the naïve view of $in$ is that it says the set on its left side is "inside" the set on its right; maybe that's worth editing into your answer (and I'll leave that up to you). Although of heard of, say, geometry being taught without any diagrams or talk of shapes, it is nonetheless instructive to "know" what lines & points are, at least by example, even though they, too, are primitive.
            $endgroup$
            – Shaun
            Dec 31 '18 at 14:15






            $begingroup$
            I didn't know $in$ was primitive. As such, this comment might not be helpful, but the naïve view of $in$ is that it says the set on its left side is "inside" the set on its right; maybe that's worth editing into your answer (and I'll leave that up to you). Although of heard of, say, geometry being taught without any diagrams or talk of shapes, it is nonetheless instructive to "know" what lines & points are, at least by example, even though they, too, are primitive.
            $endgroup$
            – Shaun
            Dec 31 '18 at 14:15






            1




            1




            $begingroup$
            @Shaun I think that your comment on its own is enough on this matter, so will not add something. Actually it is only a shift/translation from $in$ to the primitive notion "inside" (what is that in mathematics?) and gives at most a way to visualize (very useful indeed). Somewhere definitions must stop. You could say that the relation $atext{ defined by means of }b$ is well founded, i.e. every non-empty subset has a least (i.e. primitive) element.
            $endgroup$
            – drhab
            Dec 31 '18 at 15:12






            $begingroup$
            @Shaun I think that your comment on its own is enough on this matter, so will not add something. Actually it is only a shift/translation from $in$ to the primitive notion "inside" (what is that in mathematics?) and gives at most a way to visualize (very useful indeed). Somewhere definitions must stop. You could say that the relation $atext{ defined by means of }b$ is well founded, i.e. every non-empty subset has a least (i.e. primitive) element.
            $endgroup$
            – drhab
            Dec 31 '18 at 15:12




















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