Visualization / sketch for this basic proof about subspace topologyIn a metric space, closure is closedChain...

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Visualization / sketch for this basic proof about subspace topology


In a metric space, closure is closedChain of compact subsets and their intersectionOpen set with respect to french railway metricFor compact $K$ and open $U supseteq K$, there exists $varepsilon>0$ such that $B(K,varepsilon) subseteq U$Open or closed set in $mathbb{R}$Proof regarding nested compact subsets of a metric spaceProve that closed and bounded subsets of metric spaces are compactThe fact that subspace topology coincides with induced metric topology leads to a contradictionA curiosity on covers of compactIntersection of neighborhoods of closed subsets in a metric space













0












$begingroup$



Let $(X,d)$ be a metric space and $Asubset X$ a subset equipped with the induced metric $d_{A}$.
Then the open subsets of $(A,d_{A})$ are exactly the intersections of open subsets of of $(X,d)$ with $A$: $B subset A$ is open in $(A,d_{A})$ iff there exists an open subset $Y subset X$, so that $B = A cap Y$.




My proof of "$implies$" is pretty straight forward:



Let $Bsubset A$ be an open subset.
Define
begin{equation*}
U_{varepsilon}^{Z}(x)
:= { y in Z: d_A(x,y) = d(x,y) < varepsilon}
end{equation*}
for any subset $Z in X$.
Because $B$ is open, for every $x in B$ there exists an $varepsilon_{x} > 0$ so that
begin{equation*}
Bsupset U^{A}_{varepsilon_{x}}(x)
= Acap U^{X}_{varepsilon_{x}}(x).
end{equation*}
The set
begin{equation*}
Y := bigcup_{x in B}U^{X}_{varepsilon_{x}}(x)
end{equation*}
is open in $X$ because it's the union of open sets in $X$.
Furthermore we have $B = Acap Y$.



But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $U^{A}_{varepsilon_{x}}(x) neq U^{X}_{varepsilon_{x}}(x)$, since when I draw $X$ to be a box and $A$ a circle inside it and $B$ a smaller circle in $B$, we alway have $U^{A}_{varepsilon_{x}}(x) = U^{X}_{varepsilon_{x}}(x)$.



Any help is greatly appreciated :)






general-topology metric-spaces visualization







share|cite|improve this question









$endgroup$

















    0












    $begingroup$



    Let $(X,d)$ be a metric space and $Asubset X$ a subset equipped with the induced metric $d_{A}$.
    Then the open subsets of $(A,d_{A})$ are exactly the intersections of open subsets of of $(X,d)$ with $A$: $B subset A$ is open in $(A,d_{A})$ iff there exists an open subset $Y subset X$, so that $B = A cap Y$.




    My proof of "$implies$" is pretty straight forward:



    Let $Bsubset A$ be an open subset.
    Define
    begin{equation*}
    U_{varepsilon}^{Z}(x)
    := { y in Z: d_A(x,y) = d(x,y) < varepsilon}
    end{equation*}
    for any subset $Z in X$.
    Because $B$ is open, for every $x in B$ there exists an $varepsilon_{x} > 0$ so that
    begin{equation*}
    Bsupset U^{A}_{varepsilon_{x}}(x)
    = Acap U^{X}_{varepsilon_{x}}(x).
    end{equation*}
    The set
    begin{equation*}
    Y := bigcup_{x in B}U^{X}_{varepsilon_{x}}(x)
    end{equation*}
    is open in $X$ because it's the union of open sets in $X$.
    Furthermore we have $B = Acap Y$.



    But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $U^{A}_{varepsilon_{x}}(x) neq U^{X}_{varepsilon_{x}}(x)$, since when I draw $X$ to be a box and $A$ a circle inside it and $B$ a smaller circle in $B$, we alway have $U^{A}_{varepsilon_{x}}(x) = U^{X}_{varepsilon_{x}}(x)$.



    Any help is greatly appreciated :)






    general-topology metric-spaces visualization







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$



      Let $(X,d)$ be a metric space and $Asubset X$ a subset equipped with the induced metric $d_{A}$.
      Then the open subsets of $(A,d_{A})$ are exactly the intersections of open subsets of of $(X,d)$ with $A$: $B subset A$ is open in $(A,d_{A})$ iff there exists an open subset $Y subset X$, so that $B = A cap Y$.




      My proof of "$implies$" is pretty straight forward:



      Let $Bsubset A$ be an open subset.
      Define
      begin{equation*}
      U_{varepsilon}^{Z}(x)
      := { y in Z: d_A(x,y) = d(x,y) < varepsilon}
      end{equation*}
      for any subset $Z in X$.
      Because $B$ is open, for every $x in B$ there exists an $varepsilon_{x} > 0$ so that
      begin{equation*}
      Bsupset U^{A}_{varepsilon_{x}}(x)
      = Acap U^{X}_{varepsilon_{x}}(x).
      end{equation*}
      The set
      begin{equation*}
      Y := bigcup_{x in B}U^{X}_{varepsilon_{x}}(x)
      end{equation*}
      is open in $X$ because it's the union of open sets in $X$.
      Furthermore we have $B = Acap Y$.



      But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $U^{A}_{varepsilon_{x}}(x) neq U^{X}_{varepsilon_{x}}(x)$, since when I draw $X$ to be a box and $A$ a circle inside it and $B$ a smaller circle in $B$, we alway have $U^{A}_{varepsilon_{x}}(x) = U^{X}_{varepsilon_{x}}(x)$.



      Any help is greatly appreciated :)






      general-topology metric-spaces visualization







      share|cite|improve this question









      $endgroup$





      Let $(X,d)$ be a metric space and $Asubset X$ a subset equipped with the induced metric $d_{A}$.
      Then the open subsets of $(A,d_{A})$ are exactly the intersections of open subsets of of $(X,d)$ with $A$: $B subset A$ is open in $(A,d_{A})$ iff there exists an open subset $Y subset X$, so that $B = A cap Y$.




      My proof of "$implies$" is pretty straight forward:



      Let $Bsubset A$ be an open subset.
      Define
      begin{equation*}
      U_{varepsilon}^{Z}(x)
      := { y in Z: d_A(x,y) = d(x,y) < varepsilon}
      end{equation*}
      for any subset $Z in X$.
      Because $B$ is open, for every $x in B$ there exists an $varepsilon_{x} > 0$ so that
      begin{equation*}
      Bsupset U^{A}_{varepsilon_{x}}(x)
      = Acap U^{X}_{varepsilon_{x}}(x).
      end{equation*}
      The set
      begin{equation*}
      Y := bigcup_{x in B}U^{X}_{varepsilon_{x}}(x)
      end{equation*}
      is open in $X$ because it's the union of open sets in $X$.
      Furthermore we have $B = Acap Y$.



      But I am having trouble visualizing this: I can't think of a way to sketch the scenario so that $U^{A}_{varepsilon_{x}}(x) neq U^{X}_{varepsilon_{x}}(x)$, since when I draw $X$ to be a box and $A$ a circle inside it and $B$ a smaller circle in $B$, we alway have $U^{A}_{varepsilon_{x}}(x) = U^{X}_{varepsilon_{x}}(x)$.



      Any help is greatly appreciated :)






      general-topology metric-spaces visualization




      general-topology metric-spaces visualization






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 2 days ago









      Viktor GlombikViktor Glombik

      1,0121528




      1,0121528






















          1 Answer
          1






          active

          oldest

          votes


















          1












          $begingroup$

          Draw $X$ to be the plane and $A$ to be the $x$-axis.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
            $endgroup$
            – Viktor Glombik
            2 days ago










          • $begingroup$
            Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
            $endgroup$
            – Lee Mosher
            2 days ago










          • $begingroup$
            Sure, but when I want to draw a sketch, an interval is more convenient, right?
            $endgroup$
            – Viktor Glombik
            2 days ago












          • $begingroup$
            Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
            $endgroup$
            – Lee Mosher
            2 days ago











          Your Answer





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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Draw $X$ to be the plane and $A$ to be the $x$-axis.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
            $endgroup$
            – Viktor Glombik
            2 days ago










          • $begingroup$
            Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
            $endgroup$
            – Lee Mosher
            2 days ago










          • $begingroup$
            Sure, but when I want to draw a sketch, an interval is more convenient, right?
            $endgroup$
            – Viktor Glombik
            2 days ago












          • $begingroup$
            Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
            $endgroup$
            – Lee Mosher
            2 days ago
















          1












          $begingroup$

          Draw $X$ to be the plane and $A$ to be the $x$-axis.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
            $endgroup$
            – Viktor Glombik
            2 days ago










          • $begingroup$
            Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
            $endgroup$
            – Lee Mosher
            2 days ago










          • $begingroup$
            Sure, but when I want to draw a sketch, an interval is more convenient, right?
            $endgroup$
            – Viktor Glombik
            2 days ago












          • $begingroup$
            Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
            $endgroup$
            – Lee Mosher
            2 days ago














          1












          1








          1





          $begingroup$

          Draw $X$ to be the plane and $A$ to be the $x$-axis.






          share|cite|improve this answer









          $endgroup$



          Draw $X$ to be the plane and $A$ to be the $x$-axis.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Lee MosherLee Mosher

          50.3k33787




          50.3k33787












          • $begingroup$
            This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
            $endgroup$
            – Viktor Glombik
            2 days ago










          • $begingroup$
            Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
            $endgroup$
            – Lee Mosher
            2 days ago










          • $begingroup$
            Sure, but when I want to draw a sketch, an interval is more convenient, right?
            $endgroup$
            – Viktor Glombik
            2 days ago












          • $begingroup$
            Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
            $endgroup$
            – Lee Mosher
            2 days ago


















          • $begingroup$
            This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
            $endgroup$
            – Viktor Glombik
            2 days ago










          • $begingroup$
            Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
            $endgroup$
            – Lee Mosher
            2 days ago










          • $begingroup$
            Sure, but when I want to draw a sketch, an interval is more convenient, right?
            $endgroup$
            – Viktor Glombik
            2 days ago












          • $begingroup$
            Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
            $endgroup$
            – Lee Mosher
            2 days ago
















          $begingroup$
          This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
          $endgroup$
          – Viktor Glombik
          2 days ago




          $begingroup$
          This is in two dimensions, right?, So $B$ would be an intervall $(a,b)$ on the $x$-axis?
          $endgroup$
          – Viktor Glombik
          2 days ago












          $begingroup$
          Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
          $endgroup$
          – Lee Mosher
          2 days ago




          $begingroup$
          Sort of, but more properly speaking $B$ would be a union of open intervals on the $x$-axis, since that is the usual description of open subsets of the $x$-axis.
          $endgroup$
          – Lee Mosher
          2 days ago












          $begingroup$
          Sure, but when I want to draw a sketch, an interval is more convenient, right?
          $endgroup$
          – Viktor Glombik
          2 days ago






          $begingroup$
          Sure, but when I want to draw a sketch, an interval is more convenient, right?
          $endgroup$
          – Viktor Glombik
          2 days ago














          $begingroup$
          Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
          $endgroup$
          – Lee Mosher
          2 days ago




          $begingroup$
          Sure, for purposes of a sketch that's fine, since the open intervals form a basis for the topology on $mathbb R$.
          $endgroup$
          – Lee Mosher
          2 days ago


















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