What is {empty set} x something with the product topology?












-1














The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies.
From what I understand, then ${emptyset}x]a,b[$ is an open set of $IR^2$ with the product topology, right? But what is this geometrically? I can't see what this is.






general-topology product-space







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  • 3




    Empty cross anything is empty.
    – Randall
    2 days ago










  • The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
    – leibnewtz
    2 days ago












  • Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
    – leibnewtz
    2 days ago


















-1














The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies.
From what I understand, then ${emptyset}x]a,b[$ is an open set of $IR^2$ with the product topology, right? But what is this geometrically? I can't see what this is.






general-topology product-space







share|cite|improve this question




















  • 3




    Empty cross anything is empty.
    – Randall
    2 days ago










  • The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
    – leibnewtz
    2 days ago












  • Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
    – leibnewtz
    2 days ago
















-1












-1








-1







The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies.
From what I understand, then ${emptyset}x]a,b[$ is an open set of $IR^2$ with the product topology, right? But what is this geometrically? I can't see what this is.






general-topology product-space







share|cite|improve this question















The product topology is defined as the topology induced by the basis of the product of open sets from each of the original topologies.
From what I understand, then ${emptyset}x]a,b[$ is an open set of $IR^2$ with the product topology, right? But what is this geometrically? I can't see what this is.






general-topology product-space




general-topology product-space






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 2 days ago









Lehs

6,95231662




6,95231662










asked 2 days ago









J. DionisioJ. Dionisio

9511




9511








  • 3




    Empty cross anything is empty.
    – Randall
    2 days ago










  • The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
    – leibnewtz
    2 days ago












  • Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
    – leibnewtz
    2 days ago
















  • 3




    Empty cross anything is empty.
    – Randall
    2 days ago










  • The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
    – leibnewtz
    2 days ago












  • Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
    – leibnewtz
    2 days ago










3




3




Empty cross anything is empty.
– Randall
2 days ago




Empty cross anything is empty.
– Randall
2 days ago












The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
– leibnewtz
2 days ago






The set ${ emptyset }$ has at most one topology. For any space $A$ you can show ${emptyset } times A$ is homeomorphic to $A$
– leibnewtz
2 days ago














Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
– leibnewtz
2 days ago






Being "homeomorphic to" is intuitively the same as "looking like," so it just looks like $mathbb{R}$. I should say that you're wrong in assuming ${emptyset }$ is open. And thanks!
– leibnewtz
2 days ago












2 Answers
2






active

oldest

votes


















1














By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$



The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$






share|cite|improve this answer























  • The question is asking about ${emptyset } times A$, not $emptyset times A$
    – leibnewtz
    2 days ago










  • Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
    – J. Dionisio
    2 days ago










  • Oh, dear! I misunderstood. I'll edit accordingly.
    – Cameron Buie
    2 days ago










  • Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
    – J. Dionisio
    2 days ago



















0














Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.



Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.



Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have



$$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$



Thus, $X$ and ${*}times X$ are homeomorphic.



If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:



enter image description here






share|cite|improve this answer























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    2 Answers
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    2 Answers
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    active

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    active

    oldest

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    1














    By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$



    The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$






    share|cite|improve this answer























    • The question is asking about ${emptyset } times A$, not $emptyset times A$
      – leibnewtz
      2 days ago










    • Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
      – J. Dionisio
      2 days ago










    • Oh, dear! I misunderstood. I'll edit accordingly.
      – Cameron Buie
      2 days ago










    • Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
      – J. Dionisio
      2 days ago
















    1














    By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$



    The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$






    share|cite|improve this answer























    • The question is asking about ${emptyset } times A$, not $emptyset times A$
      – leibnewtz
      2 days ago










    • Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
      – J. Dionisio
      2 days ago










    • Oh, dear! I misunderstood. I'll edit accordingly.
      – Cameron Buie
      2 days ago










    • Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
      – J. Dionisio
      2 days ago














    1












    1








    1






    By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$



    The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$






    share|cite|improve this answer














    By definition, the elements of $Atimes B$ (for sets $A$ and $B$) are the ordered pairs $langle a,brangle$ such that $ain A$ and $bin B.$ Since $ainemptyset$ is impossible, then $emptysettimes B$ has no elements, meaning $emptysettimes B=emptyset,$ regardless of the set $B.$



    The open sets of $Bbb R^2$ in the product topology can be obtained from the basis ${Atimes B:A,Btext{ are open subsets of }Bbb R}.$ This will include elements of the form $emptysettimes B=emptyset,$ as well as $Atimesemptyset=emptyset.$ However, it will not have any elements of the form ${emptyset}times B,$ since $emptyset$ isn't an element of $Bbb R.$







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited 2 days ago

























    answered 2 days ago









    Cameron BuieCameron Buie

    85.1k771155




    85.1k771155












    • The question is asking about ${emptyset } times A$, not $emptyset times A$
      – leibnewtz
      2 days ago










    • Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
      – J. Dionisio
      2 days ago










    • Oh, dear! I misunderstood. I'll edit accordingly.
      – Cameron Buie
      2 days ago










    • Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
      – J. Dionisio
      2 days ago


















    • The question is asking about ${emptyset } times A$, not $emptyset times A$
      – leibnewtz
      2 days ago










    • Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
      – J. Dionisio
      2 days ago










    • Oh, dear! I misunderstood. I'll edit accordingly.
      – Cameron Buie
      2 days ago










    • Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
      – J. Dionisio
      2 days ago
















    The question is asking about ${emptyset } times A$, not $emptyset times A$
    – leibnewtz
    2 days ago




    The question is asking about ${emptyset } times A$, not $emptyset times A$
    – leibnewtz
    2 days ago












    Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
    – J. Dionisio
    2 days ago




    Maybe I didn't understand your answer, but just to be sure, ∅ x O, where O is an open set of the topology on B is an open set of the product topology on A x B, right? Different from U x ∅, where U is an open set of the topology on B
    – J. Dionisio
    2 days ago












    Oh, dear! I misunderstood. I'll edit accordingly.
    – Cameron Buie
    2 days ago




    Oh, dear! I misunderstood. I'll edit accordingly.
    – Cameron Buie
    2 days ago












    Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
    – J. Dionisio
    2 days ago




    Ohhh okay, now I understand, thank you! I was mistaking ∅ with {∅}. Thanks a lot for the help!
    – J. Dionisio
    2 days ago











    0














    Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.



    Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.



    Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have



    $$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$



    Thus, $X$ and ${*}times X$ are homeomorphic.



    If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:



    enter image description here






    share|cite|improve this answer




























      0














      Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.



      Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.



      Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have



      $$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$



      Thus, $X$ and ${*}times X$ are homeomorphic.



      If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:



      enter image description here






      share|cite|improve this answer


























        0












        0








        0






        Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.



        Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.



        Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have



        $$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$



        Thus, $X$ and ${*}times X$ are homeomorphic.



        If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:



        enter image description here






        share|cite|improve this answer














        Instead of ${emptyset}$, I'll consider ${*}$ just to avoid confusion. It really doesn't matter, you can use anything instead of $*$. What matters is that ${*}$ is a singleton as a set.



        Let $X$ be a topological space. Then, basis for topology of ${*}times X$ is given by ${ {*}times Umid Usubseteq X text{open}}$.



        Define $fcolon Xto {*}times X$ by $f(x) = (*,x)$ and $gcolon {*}times Xto X$ by $g(*,x) = x$. Obviously, these functions are inverses of each other. They are also continuous, since for any $Usubseteq X$ open we have



        $$f^{-1}({*}times U) = U, g^{-1}(U) = {*}times U.$$



        Thus, $X$ and ${*}times X$ are homeomorphic.



        If $X = mathbb R$, instead of $*$ choose some real number $a$. Then ${a}timesmathbb R$ is naturally a line embedded in $mathbb R^2$. Here are examples for some different $a$'s:



        enter image description here







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited 2 days ago

























        answered 2 days ago









        EnnarEnnar

        14.4k32343




        14.4k32343






























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