Closed sets in induced metric space











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Let $(X,d)$ be a metric space, let $Y$ be a subset of $X$ and let $E$ be a subset of $Y$, then



$(i)$ $E$ is relatively open with respect to $Y$ if and only if $E=V cap Y$ for some set $V subseteq X$ which is open in $X$.



$(ii)$ $E$ is relatively closed with respect to $Y$ if and only if $E=K cap Y$ for some set $k subseteq X$ which is closed in $X$.



I need to prove $(ii)$. I looked at this solution for a similar problem in a topological space, but it uses $(i)$. I tried to write a proof that doesn't depend on (i), but I don't know if it works. I am particularly suspicious about the bold text:



$Proof.$ Let $E$ be closed in $(Y,d)$ (with the metric $d$ restricted to $Ytimes Y$) and let $K = bar E$, where $bar E$ is the closure of $E$ with respect to $(X,d)$. Then $K$ is closed in $(X,d)$ and since $K$ is the union of adherent points to $E$ in $(Y^c,d)$ and adherent points to $E$ in $(Y,d)$, then $K cap Y =$ adherent points to $E$ in $(Y,d)$. Since $E$ is closed, it contains all its adherent points, so $K cap Y = E$.
Similarly, let $K$ be a subset of $X$ closed in $(X,d)$ such that $K cap Y = E$ , then $E$ = all adherent points of $K$ in $Y$. Thus $E$ is closed. $q.e.d.$










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    Let $(X,d)$ be a metric space, let $Y$ be a subset of $X$ and let $E$ be a subset of $Y$, then



    $(i)$ $E$ is relatively open with respect to $Y$ if and only if $E=V cap Y$ for some set $V subseteq X$ which is open in $X$.



    $(ii)$ $E$ is relatively closed with respect to $Y$ if and only if $E=K cap Y$ for some set $k subseteq X$ which is closed in $X$.



    I need to prove $(ii)$. I looked at this solution for a similar problem in a topological space, but it uses $(i)$. I tried to write a proof that doesn't depend on (i), but I don't know if it works. I am particularly suspicious about the bold text:



    $Proof.$ Let $E$ be closed in $(Y,d)$ (with the metric $d$ restricted to $Ytimes Y$) and let $K = bar E$, where $bar E$ is the closure of $E$ with respect to $(X,d)$. Then $K$ is closed in $(X,d)$ and since $K$ is the union of adherent points to $E$ in $(Y^c,d)$ and adherent points to $E$ in $(Y,d)$, then $K cap Y =$ adherent points to $E$ in $(Y,d)$. Since $E$ is closed, it contains all its adherent points, so $K cap Y = E$.
    Similarly, let $K$ be a subset of $X$ closed in $(X,d)$ such that $K cap Y = E$ , then $E$ = all adherent points of $K$ in $Y$. Thus $E$ is closed. $q.e.d.$










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      Let $(X,d)$ be a metric space, let $Y$ be a subset of $X$ and let $E$ be a subset of $Y$, then



      $(i)$ $E$ is relatively open with respect to $Y$ if and only if $E=V cap Y$ for some set $V subseteq X$ which is open in $X$.



      $(ii)$ $E$ is relatively closed with respect to $Y$ if and only if $E=K cap Y$ for some set $k subseteq X$ which is closed in $X$.



      I need to prove $(ii)$. I looked at this solution for a similar problem in a topological space, but it uses $(i)$. I tried to write a proof that doesn't depend on (i), but I don't know if it works. I am particularly suspicious about the bold text:



      $Proof.$ Let $E$ be closed in $(Y,d)$ (with the metric $d$ restricted to $Ytimes Y$) and let $K = bar E$, where $bar E$ is the closure of $E$ with respect to $(X,d)$. Then $K$ is closed in $(X,d)$ and since $K$ is the union of adherent points to $E$ in $(Y^c,d)$ and adherent points to $E$ in $(Y,d)$, then $K cap Y =$ adherent points to $E$ in $(Y,d)$. Since $E$ is closed, it contains all its adherent points, so $K cap Y = E$.
      Similarly, let $K$ be a subset of $X$ closed in $(X,d)$ such that $K cap Y = E$ , then $E$ = all adherent points of $K$ in $Y$. Thus $E$ is closed. $q.e.d.$










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      Let $(X,d)$ be a metric space, let $Y$ be a subset of $X$ and let $E$ be a subset of $Y$, then



      $(i)$ $E$ is relatively open with respect to $Y$ if and only if $E=V cap Y$ for some set $V subseteq X$ which is open in $X$.



      $(ii)$ $E$ is relatively closed with respect to $Y$ if and only if $E=K cap Y$ for some set $k subseteq X$ which is closed in $X$.



      I need to prove $(ii)$. I looked at this solution for a similar problem in a topological space, but it uses $(i)$. I tried to write a proof that doesn't depend on (i), but I don't know if it works. I am particularly suspicious about the bold text:



      $Proof.$ Let $E$ be closed in $(Y,d)$ (with the metric $d$ restricted to $Ytimes Y$) and let $K = bar E$, where $bar E$ is the closure of $E$ with respect to $(X,d)$. Then $K$ is closed in $(X,d)$ and since $K$ is the union of adherent points to $E$ in $(Y^c,d)$ and adherent points to $E$ in $(Y,d)$, then $K cap Y =$ adherent points to $E$ in $(Y,d)$. Since $E$ is closed, it contains all its adherent points, so $K cap Y = E$.
      Similarly, let $K$ be a subset of $X$ closed in $(X,d)$ such that $K cap Y = E$ , then $E$ = all adherent points of $K$ in $Y$. Thus $E$ is closed. $q.e.d.$







      proof-verification metric-spaces






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      asked Nov 14 at 15:22









      user258607

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      This question has an open bounty worth +50
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          Your proof is correct!





          Here is another way to write the same argument; perhaps looking at the statements from a different angle will be more convincing.



          First, observe that if $x$ is a limit point of a set $A$ in $Y subset X$, then $x$ is also a limit point of $A$ in $X$.



          Now, let $K subset X$ be the closure of $E$ in $X$, as you have defined it. We want to show that $K cap Y = E$; that is, if $x$ is a limit point of $E$ in $X$ that does not lie in $E$, then in fact $x$ does not lie in $Y$. In other words, we want to prove the following:




          Let $xin X$ be a limit point of $E subset X$. Then, $x notin E implies x notin Y$.




          But now, the contrapositive of this statement makes things clear:




          Let $xin X$ be a limit point of $E subset X$. Then, $x in Y implies x in E$.




          But, this is true simply because, as we observed, if $x in X$ is a limit point of $E$ and $x$ also belongs to $Y$, then $x$ is a limit point of $E$ in $Y$. Since $E$ is closed in $Y$, this implies that $x in E$. $tag{$blacksquare$}$






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            up vote
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            Your proof is correct!





            Here is another way to write the same argument; perhaps looking at the statements from a different angle will be more convincing.



            First, observe that if $x$ is a limit point of a set $A$ in $Y subset X$, then $x$ is also a limit point of $A$ in $X$.



            Now, let $K subset X$ be the closure of $E$ in $X$, as you have defined it. We want to show that $K cap Y = E$; that is, if $x$ is a limit point of $E$ in $X$ that does not lie in $E$, then in fact $x$ does not lie in $Y$. In other words, we want to prove the following:




            Let $xin X$ be a limit point of $E subset X$. Then, $x notin E implies x notin Y$.




            But now, the contrapositive of this statement makes things clear:




            Let $xin X$ be a limit point of $E subset X$. Then, $x in Y implies x in E$.




            But, this is true simply because, as we observed, if $x in X$ is a limit point of $E$ and $x$ also belongs to $Y$, then $x$ is a limit point of $E$ in $Y$. Since $E$ is closed in $Y$, this implies that $x in E$. $tag{$blacksquare$}$






            share|cite|improve this answer

























              up vote
              0
              down vote













              Your proof is correct!





              Here is another way to write the same argument; perhaps looking at the statements from a different angle will be more convincing.



              First, observe that if $x$ is a limit point of a set $A$ in $Y subset X$, then $x$ is also a limit point of $A$ in $X$.



              Now, let $K subset X$ be the closure of $E$ in $X$, as you have defined it. We want to show that $K cap Y = E$; that is, if $x$ is a limit point of $E$ in $X$ that does not lie in $E$, then in fact $x$ does not lie in $Y$. In other words, we want to prove the following:




              Let $xin X$ be a limit point of $E subset X$. Then, $x notin E implies x notin Y$.




              But now, the contrapositive of this statement makes things clear:




              Let $xin X$ be a limit point of $E subset X$. Then, $x in Y implies x in E$.




              But, this is true simply because, as we observed, if $x in X$ is a limit point of $E$ and $x$ also belongs to $Y$, then $x$ is a limit point of $E$ in $Y$. Since $E$ is closed in $Y$, this implies that $x in E$. $tag{$blacksquare$}$






              share|cite|improve this answer























                up vote
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                down vote










                up vote
                0
                down vote









                Your proof is correct!





                Here is another way to write the same argument; perhaps looking at the statements from a different angle will be more convincing.



                First, observe that if $x$ is a limit point of a set $A$ in $Y subset X$, then $x$ is also a limit point of $A$ in $X$.



                Now, let $K subset X$ be the closure of $E$ in $X$, as you have defined it. We want to show that $K cap Y = E$; that is, if $x$ is a limit point of $E$ in $X$ that does not lie in $E$, then in fact $x$ does not lie in $Y$. In other words, we want to prove the following:




                Let $xin X$ be a limit point of $E subset X$. Then, $x notin E implies x notin Y$.




                But now, the contrapositive of this statement makes things clear:




                Let $xin X$ be a limit point of $E subset X$. Then, $x in Y implies x in E$.




                But, this is true simply because, as we observed, if $x in X$ is a limit point of $E$ and $x$ also belongs to $Y$, then $x$ is a limit point of $E$ in $Y$. Since $E$ is closed in $Y$, this implies that $x in E$. $tag{$blacksquare$}$






                share|cite|improve this answer












                Your proof is correct!





                Here is another way to write the same argument; perhaps looking at the statements from a different angle will be more convincing.



                First, observe that if $x$ is a limit point of a set $A$ in $Y subset X$, then $x$ is also a limit point of $A$ in $X$.



                Now, let $K subset X$ be the closure of $E$ in $X$, as you have defined it. We want to show that $K cap Y = E$; that is, if $x$ is a limit point of $E$ in $X$ that does not lie in $E$, then in fact $x$ does not lie in $Y$. In other words, we want to prove the following:




                Let $xin X$ be a limit point of $E subset X$. Then, $x notin E implies x notin Y$.




                But now, the contrapositive of this statement makes things clear:




                Let $xin X$ be a limit point of $E subset X$. Then, $x in Y implies x in E$.




                But, this is true simply because, as we observed, if $x in X$ is a limit point of $E$ and $x$ also belongs to $Y$, then $x$ is a limit point of $E$ in $Y$. Since $E$ is closed in $Y$, this implies that $x in E$. $tag{$blacksquare$}$







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                answered Nov 19 at 7:17









                Brahadeesh

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