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

405




405








  • 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














    Your Answer








    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057660%2fproving-a-in-b-land-b-subseteq-c-to-a-in-c%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    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

        11.1k113688




        11.1k113688























            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

            105k545136




            105k545136












            • $begingroup$
              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




















            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3057660%2fproving-a-in-b-land-b-subseteq-c-to-a-in-c%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Bundesstraße 106

            Image
            Dieser Artikel beschreibt die Bundesstraße 106 in Deutschland. Zur gleichnamigen Straße in Österreich siehe Mölltal Straße. Vorlage:Infobox hochrangige Straße/Wartung/DE-B Bundesstraße 106 in Deutschland Karte Basisdaten Betreiber: Deutschland  Bundesrepublik Deutschland Straßenbeginn: Wismar ( 53° 54′  N , 11° 24′  O 53.900536 11.396875 ) Straßenende: Schwerin ( 53° 36′  N , 11° 24′  O 53.592785 11.404892 ) Gesamtlänge: 40,7 km Bundesland: Mecklenburg-Vorpommern Ausbauzustand: zweistreifig [1] Bundesstraße 106 in Schwerin (Umgehungsstraße) Straßenverlauf Land Mecklenburg-Vorpommern Landkreis Nordwestmecklenburg Ortsumgehung Wismar Wallensteingraben über   Dorf Mecklenburg Wallensteingraben Groß Stieten Bad Kleinen OT Niendorf Bahnstrecke Lübeck–Bad Kleinen Zickhusen Lübstorf Klein Trebbow OT Kirch Stück ...
            Read more

            Verónica Boquete

            Image
            Verónica Boquete im Trikot der Nationalmannschaft (2013) Personalia Name Verónica Boquete Giadans Geburtstag 9. April 1987 Geburtsort Santiago de Compostela, Spanien Größe 159 cm Position Mittelfeld / Sturm Juniorinnen Jahre Station 2001–2005 Xuventú Aguiño Frauen Jahre Station Spiele (Tore) 1 2005–2008 Prainsa Zaragoza 83 (40) 2008–2010 Espanyol Barcelona 53 (30) 2010 Buffalo Flash 9 0 (9) 2010 Chicago Red Stars 3 0 (1) 2010–2011 Espanyol Barcelona 26 (41) 2011 Philadelphia Independence 13 0 (5) 2011 Energija Woronesch 5 0 (1) 2012–2014 Tyresö FF 39 (14) 2014 Portland Thorns FC 15 0 (4) 2014–2015 1. FFC Frankfurt 23 0 (7) 2015–2016 FC Bayern München 9 0 (0) 2016–2018 Paris Saint-Germain 31 0 (8) 2018 Peking BG Phoenix 2019– Utah Royals FC 0 0 (0) Nationalmannschaft Jahre Auswahl ...
            Read more

            Ida-Boy-Ed-Garten

            Image
            Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
            Read more