Geometrical interpretation of disjoint union












0














Can anybody make me understand the geometrical interpretation of the concept $Disjoint$ $Union $? Mathematically it's fine but I'm unable to grasp it geometrically.










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  • What makes you think there is one? The disjoint union is an operation on sets.
    – John Douma
    Jan 4 at 4:13






  • 2




    You mean disjoint union of topological spaces?
    – Randall
    Jan 4 at 4:21










  • Are you perhaps looking for a visual representation of disjoint union, e.g. this?
    – Alexis
    Jan 4 at 4:44










  • @Randall yes. Do u have any example?
    – Prince Thomas
    Jan 4 at 5:28
















0














Can anybody make me understand the geometrical interpretation of the concept $Disjoint$ $Union $? Mathematically it's fine but I'm unable to grasp it geometrically.










share|cite|improve this question






















  • What makes you think there is one? The disjoint union is an operation on sets.
    – John Douma
    Jan 4 at 4:13






  • 2




    You mean disjoint union of topological spaces?
    – Randall
    Jan 4 at 4:21










  • Are you perhaps looking for a visual representation of disjoint union, e.g. this?
    – Alexis
    Jan 4 at 4:44










  • @Randall yes. Do u have any example?
    – Prince Thomas
    Jan 4 at 5:28














0












0








0







Can anybody make me understand the geometrical interpretation of the concept $Disjoint$ $Union $? Mathematically it's fine but I'm unable to grasp it geometrically.










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Can anybody make me understand the geometrical interpretation of the concept $Disjoint$ $Union $? Mathematically it's fine but I'm unable to grasp it geometrically.







algebraic-topology






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asked Jan 4 at 3:40









Prince ThomasPrince Thomas

601210




601210












  • What makes you think there is one? The disjoint union is an operation on sets.
    – John Douma
    Jan 4 at 4:13






  • 2




    You mean disjoint union of topological spaces?
    – Randall
    Jan 4 at 4:21










  • Are you perhaps looking for a visual representation of disjoint union, e.g. this?
    – Alexis
    Jan 4 at 4:44










  • @Randall yes. Do u have any example?
    – Prince Thomas
    Jan 4 at 5:28


















  • What makes you think there is one? The disjoint union is an operation on sets.
    – John Douma
    Jan 4 at 4:13






  • 2




    You mean disjoint union of topological spaces?
    – Randall
    Jan 4 at 4:21










  • Are you perhaps looking for a visual representation of disjoint union, e.g. this?
    – Alexis
    Jan 4 at 4:44










  • @Randall yes. Do u have any example?
    – Prince Thomas
    Jan 4 at 5:28
















What makes you think there is one? The disjoint union is an operation on sets.
– John Douma
Jan 4 at 4:13




What makes you think there is one? The disjoint union is an operation on sets.
– John Douma
Jan 4 at 4:13




2




2




You mean disjoint union of topological spaces?
– Randall
Jan 4 at 4:21




You mean disjoint union of topological spaces?
– Randall
Jan 4 at 4:21












Are you perhaps looking for a visual representation of disjoint union, e.g. this?
– Alexis
Jan 4 at 4:44




Are you perhaps looking for a visual representation of disjoint union, e.g. this?
– Alexis
Jan 4 at 4:44












@Randall yes. Do u have any example?
– Prince Thomas
Jan 4 at 5:28




@Randall yes. Do u have any example?
– Prince Thomas
Jan 4 at 5:28










1 Answer
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Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.



So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.



I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.






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    1 Answer
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    1 Answer
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    Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.



    So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.



    I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.






    share|cite|improve this answer


























      0














      Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.



      So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.



      I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.






      share|cite|improve this answer
























        0












        0








        0






        Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.



        So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.



        I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.






        share|cite|improve this answer












        Let us take the case of a topological space $X$ with two nonempty subspaces $A,B$ whose union is $X$. In general the topologies of $A,B$ do not determine that of $X$. However if $A cap B$ is empty and a set $U$ is open in $X$ if and only if $U$ intersects $A,B$ in sets open in $X$, then intuitively $X$ is "in two bits" and we write $X =A sqcup B$, the disjoint union of $A,B$. This can also be usefully described by saying that the function $f: X to {0,1}$ taking $A$ to $0$ and $B$ to $1$, where ${0,1}$ has the discrete topology, is continuous.



        So we say $X$ is connected if any continuous function from $X$ to the discrete space ${0,1}$ is constant.This characterisation is often quite useful.



        I leave you to generalise to $n$ subsets, or any "disjoint union", and to draw examples.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 18:26









        Ronnie BrownRonnie Brown

        11.9k12938




        11.9k12938






























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