Continuous function on a metric space/topology












1












$begingroup$


X a metric space. Let $gamma$1, $gamma$2 : [0,1] → X be two continuous paths such that $gamma$1 (1) = $gamma$2 (0). Consider the map$gamma$ : [0,1] → X defined as $gamma$(t) = $gamma$1 (2t), if t∈ [0,$frac{1}{2}$)
or $gamma$(t) = $gamma$2 (2t−1), if t ∈ [$frac{1}{2}$,1] Show that $gamma$ is a continuous path in X.



How should I show if this is continuous?
Can I use the "usual" definitions, as the epsilon-delta or the limit definition? Or do I have to use something else since X is a metric space?



Any guidence or help would be greatly appreciated!






functions continuity







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  • $begingroup$
    Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
    $endgroup$
    – Yadati Kiran
    Jan 5 at 14:07


















1












$begingroup$


X a metric space. Let $gamma$1, $gamma$2 : [0,1] → X be two continuous paths such that $gamma$1 (1) = $gamma$2 (0). Consider the map$gamma$ : [0,1] → X defined as $gamma$(t) = $gamma$1 (2t), if t∈ [0,$frac{1}{2}$)
or $gamma$(t) = $gamma$2 (2t−1), if t ∈ [$frac{1}{2}$,1] Show that $gamma$ is a continuous path in X.



How should I show if this is continuous?
Can I use the "usual" definitions, as the epsilon-delta or the limit definition? Or do I have to use something else since X is a metric space?



Any guidence or help would be greatly appreciated!






functions continuity







share|cite|improve this question









$endgroup$












  • $begingroup$
    Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
    $endgroup$
    – Yadati Kiran
    Jan 5 at 14:07
















1












1








1





$begingroup$


X a metric space. Let $gamma$1, $gamma$2 : [0,1] → X be two continuous paths such that $gamma$1 (1) = $gamma$2 (0). Consider the map$gamma$ : [0,1] → X defined as $gamma$(t) = $gamma$1 (2t), if t∈ [0,$frac{1}{2}$)
or $gamma$(t) = $gamma$2 (2t−1), if t ∈ [$frac{1}{2}$,1] Show that $gamma$ is a continuous path in X.



How should I show if this is continuous?
Can I use the "usual" definitions, as the epsilon-delta or the limit definition? Or do I have to use something else since X is a metric space?



Any guidence or help would be greatly appreciated!






functions continuity







share|cite|improve this question









$endgroup$




X a metric space. Let $gamma$1, $gamma$2 : [0,1] → X be two continuous paths such that $gamma$1 (1) = $gamma$2 (0). Consider the map$gamma$ : [0,1] → X defined as $gamma$(t) = $gamma$1 (2t), if t∈ [0,$frac{1}{2}$)
or $gamma$(t) = $gamma$2 (2t−1), if t ∈ [$frac{1}{2}$,1] Show that $gamma$ is a continuous path in X.



How should I show if this is continuous?
Can I use the "usual" definitions, as the epsilon-delta or the limit definition? Or do I have to use something else since X is a metric space?



Any guidence or help would be greatly appreciated!






functions continuity




functions continuity






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asked Jan 5 at 13:42









quisquis

83




83












  • $begingroup$
    Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
    $endgroup$
    – Yadati Kiran
    Jan 5 at 14:07




















  • $begingroup$
    Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
    $endgroup$
    – Yadati Kiran
    Jan 5 at 14:07


















$begingroup$
Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
$endgroup$
– Yadati Kiran
Jan 5 at 14:07






$begingroup$
Since $gamma_1$ and $gamma_2$ are continuous over $[0,1]$, $gamma$ behaves as $gamma_1$ over $[0,frac12)$ and as $gamma_2$ over $[frac12,1]$. So it is sufficient to check the continuity at the join of the two paths i.e. at $x=frac12$ which can be checked by using the left and right hand limits.
$endgroup$
– Yadati Kiran
Jan 5 at 14:07












1 Answer
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$begingroup$

It is also true more generally if $X$ is a topological space not necessarily metrizable.




In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.



To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.



This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.



Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.



But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.



We conclude that $f^{-1}(F)$ is closed.



Similarly we find that $g^{-1}(F)$ is closed and we are ready.




You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.






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    $begingroup$

    It is also true more generally if $X$ is a topological space not necessarily metrizable.




    In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.



    To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.



    This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.



    Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.



    But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.



    We conclude that $f^{-1}(F)$ is closed.



    Similarly we find that $g^{-1}(F)$ is closed and we are ready.




    You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      It is also true more generally if $X$ is a topological space not necessarily metrizable.




      In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.



      To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.



      This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.



      Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.



      But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.



      We conclude that $f^{-1}(F)$ is closed.



      Similarly we find that $g^{-1}(F)$ is closed and we are ready.




      You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        It is also true more generally if $X$ is a topological space not necessarily metrizable.




        In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.



        To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.



        This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.



        Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.



        But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.



        We conclude that $f^{-1}(F)$ is closed.



        Similarly we find that $g^{-1}(F)$ is closed and we are ready.




        You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.






        share|cite|improve this answer









        $endgroup$



        It is also true more generally if $X$ is a topological space not necessarily metrizable.




        In general if $A,B$ are closed subsets of topological space $Y$ and we have function $f:Ato X$ continuous on $A$, function $g:Bto X$ continuous on $B$ such that $f$ and $g$ coincide on $Acap B$ then the function $h=fcup g:Acup Bto X$ is continuous as well.



        To prove that $h$ is continuous it is enough to show that the preimage of a closed set $Fsubseteq X$ is a closed subset of $Y$.



        This preimage is the set $f^{-1}(F)cup g^{-1}(F)$, so we are ready if we can prove that the sets $f^{-1}(F)$ and $g^{-1}(F)$ are closed.



        Since $f$ is continuous the set $f^{-1}(F)$ is a closed set in space $A$ which means that it can be written as $Pcap A$ where $P$ is a closed set in $Y$.



        But that means that $f^{-1}(F)=Pcap A$ is an intersection of two closed sets: $A$ and $P$.



        We conclude that $f^{-1}(F)$ is closed.



        Similarly we find that $g^{-1}(F)$ is closed and we are ready.




        You can apply that here on $A=[0,0.5]$ and $B=[0.5,1]$ wich are closed subsets of $Y:=[0,1]$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 5 at 14:13









        drhabdrhab

        103k545136




        103k545136






























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