Example of quotient map from Munkres book

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I was reading the notion of quotient map, topology and space but ran into the following example.enter image description here



In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.



Would be thankful if anyone will show the rigorous proof.






general-topology







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  • It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
    – Rushabh Mehta
    3 hours ago














up vote
1
down vote

favorite












I was reading the notion of quotient map, topology and space but ran into the following example.enter image description here



In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.



Would be thankful if anyone will show the rigorous proof.






general-topology







share|cite|improve this question





















  • It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
    – Rushabh Mehta
    3 hours ago












up vote
1
down vote

favorite









up vote
1
down vote

favorite











I was reading the notion of quotient map, topology and space but ran into the following example.enter image description here



In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.



Would be thankful if anyone will show the rigorous proof.






general-topology







share|cite|improve this question













I was reading the notion of quotient map, topology and space but ran into the following example.enter image description here



In this example I have understood almost everything except one moment: How to prove rigorously that $p(x)$ is closed map.



Would be thankful if anyone will show the rigorous proof.






general-topology




general-topology






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asked 3 hours ago









RFZ

4,77031337




4,77031337











  • It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
    – Rushabh Mehta
    3 hours ago
















  • It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
    – Rushabh Mehta
    3 hours ago















It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
– Rushabh Mehta
3 hours ago




It's obvious that both subfunctions of $p(x)$ are closed maps. Proving that the whole function is closed is simply a generalization of that.
– Rushabh Mehta
3 hours ago










2 Answers
2






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Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.






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    If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).



    Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that



    $Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,



    $p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$



    $[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.



    Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.






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      2 Answers
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      2 Answers
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      Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.






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        up vote
        3
        down vote













        Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.






        share|cite|improve this answer






















          up vote
          3
          down vote










          up vote
          3
          down vote









          Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.






          share|cite|improve this answer












          Any closed sunset of $[0,1] cup [2,3]$ is compact. Since $p$ is continuous its image is compact, hence closed.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 3 hours ago









          Kavi Rama Murthy

          32.2k31644




          32.2k31644




















              up vote
              1
              down vote













              If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).



              Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that



              $Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,



              $p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$



              $[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.



              Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.






              share|cite|improve this answer
























                up vote
                1
                down vote













                If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).



                Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that



                $Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,



                $p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$



                $[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.



                Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.






                share|cite|improve this answer






















                  up vote
                  1
                  down vote










                  up vote
                  1
                  down vote









                  If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).



                  Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that



                  $Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,



                  $p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$



                  $[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.



                  Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.






                  share|cite|improve this answer












                  If you want to do it from scratch, note that $f:xmapsto x$ and $g:xmapsto x-1$ are closed maps on $mathbb R$ (this is easy to prove).



                  Let $C$ be closed in $X$, so there is a closed set $C'subseteq mathbb R$ such that



                  $Xcap C'=[0,1]cap C'sqcup [2,3]cap C'=C.$ Then,



                  $p(C)=f([0,1]cap C'))sqcup g([2,3]cap C'))=$



                  $[0,1]cap C'sqcup [1,2]cap g(C')=[0,2]cap (C'cup g(C'))$.



                  Since $C'$ is closed in $mathbb R$ by assumption and $g(C')$ is closed in $mathbb R$ also, by the first remark, we conclude that $p(C)=[0,2]cap (C'cup g(C'))$ is closed in $Y$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 2 hours ago









                  Matematleta

                  8,6442918




                  8,6442918



























                       

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