Prove that a set A is open with respect to metric norm dp if and only if it is open with dq












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Consider two metrics $d_{p}=(sumlimits_{k=1}^n |x_{k}-y_{k}|^p)^{1/p}$ and $d_{q}=(sumlimits_{k=1}^n |x_{k}-y_{k}|^q)^{1/q}$



Prove that a non-empty subset $A subset mathbb R ^n$ is open with respect to $d_{p}$ iff it is open with respect to $d_{p}$



Attempt



So $A$ is open in both those metrics if $d_{p}$ and $d_{q}$ are equivalent.



Then I have to prove that $alpha d_p(x,y) leq d_q(x,y) leq beta d_p(x,y)$ for $alpha, beta$ positive.



How do I start to prove this?






real-analysis general-topology metric-spaces







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    1














    Consider two metrics $d_{p}=(sumlimits_{k=1}^n |x_{k}-y_{k}|^p)^{1/p}$ and $d_{q}=(sumlimits_{k=1}^n |x_{k}-y_{k}|^q)^{1/q}$



    Prove that a non-empty subset $A subset mathbb R ^n$ is open with respect to $d_{p}$ iff it is open with respect to $d_{p}$



    Attempt



    So $A$ is open in both those metrics if $d_{p}$ and $d_{q}$ are equivalent.



    Then I have to prove that $alpha d_p(x,y) leq d_q(x,y) leq beta d_p(x,y)$ for $alpha, beta$ positive.



    How do I start to prove this?






    real-analysis general-topology metric-spaces







    share|cite|improve this question

























      1












      1








      1







      Consider two metrics $d_{p}=(sumlimits_{k=1}^n |x_{k}-y_{k}|^p)^{1/p}$ and $d_{q}=(sumlimits_{k=1}^n |x_{k}-y_{k}|^q)^{1/q}$



      Prove that a non-empty subset $A subset mathbb R ^n$ is open with respect to $d_{p}$ iff it is open with respect to $d_{p}$



      Attempt



      So $A$ is open in both those metrics if $d_{p}$ and $d_{q}$ are equivalent.



      Then I have to prove that $alpha d_p(x,y) leq d_q(x,y) leq beta d_p(x,y)$ for $alpha, beta$ positive.



      How do I start to prove this?






      real-analysis general-topology metric-spaces







      share|cite|improve this question













      Consider two metrics $d_{p}=(sumlimits_{k=1}^n |x_{k}-y_{k}|^p)^{1/p}$ and $d_{q}=(sumlimits_{k=1}^n |x_{k}-y_{k}|^q)^{1/q}$



      Prove that a non-empty subset $A subset mathbb R ^n$ is open with respect to $d_{p}$ iff it is open with respect to $d_{p}$



      Attempt



      So $A$ is open in both those metrics if $d_{p}$ and $d_{q}$ are equivalent.



      Then I have to prove that $alpha d_p(x,y) leq d_q(x,y) leq beta d_p(x,y)$ for $alpha, beta$ positive.



      How do I start to prove this?






      real-analysis general-topology metric-spaces




      real-analysis general-topology metric-spaces






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 at 6:22









      Snop D.

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      285






















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          If $p<q$ define $A_i=frac{|x_i-y_i|^{1/p}}{sum_{j=1}^n |x_j-y_j|^{1/p}}le 1$ so $f(x)=sum_{i=1}^nA_i^x$ has negative derivative and thus $nge f(0)ge f(p/q)ge f(1)=1$ implies $nd_p(x,y)^{p^2}ge d_q(x,y)^{q^2} ge d_p(x,y)^{p^2}$. Now, if $A$ is open under $d_p$, for any $x in A$ there exists $r$ s.t. $d_p(x,y)<rRightarrow yin A$. Choosing $R=r^{p^2/q^2}$ shows that $d_q(x,y)<R$ implies $d_p(x,y)le d_q(x,y)^{q^2/p^2}<r$ so $yin A$ and $A$ is open under $d_q$. The converse follows similarly.






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            If $p<q$ define $A_i=frac{|x_i-y_i|^{1/p}}{sum_{j=1}^n |x_j-y_j|^{1/p}}le 1$ so $f(x)=sum_{i=1}^nA_i^x$ has negative derivative and thus $nge f(0)ge f(p/q)ge f(1)=1$ implies $nd_p(x,y)^{p^2}ge d_q(x,y)^{q^2} ge d_p(x,y)^{p^2}$. Now, if $A$ is open under $d_p$, for any $x in A$ there exists $r$ s.t. $d_p(x,y)<rRightarrow yin A$. Choosing $R=r^{p^2/q^2}$ shows that $d_q(x,y)<R$ implies $d_p(x,y)le d_q(x,y)^{q^2/p^2}<r$ so $yin A$ and $A$ is open under $d_q$. The converse follows similarly.






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              If $p<q$ define $A_i=frac{|x_i-y_i|^{1/p}}{sum_{j=1}^n |x_j-y_j|^{1/p}}le 1$ so $f(x)=sum_{i=1}^nA_i^x$ has negative derivative and thus $nge f(0)ge f(p/q)ge f(1)=1$ implies $nd_p(x,y)^{p^2}ge d_q(x,y)^{q^2} ge d_p(x,y)^{p^2}$. Now, if $A$ is open under $d_p$, for any $x in A$ there exists $r$ s.t. $d_p(x,y)<rRightarrow yin A$. Choosing $R=r^{p^2/q^2}$ shows that $d_q(x,y)<R$ implies $d_p(x,y)le d_q(x,y)^{q^2/p^2}<r$ so $yin A$ and $A$ is open under $d_q$. The converse follows similarly.






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                If $p<q$ define $A_i=frac{|x_i-y_i|^{1/p}}{sum_{j=1}^n |x_j-y_j|^{1/p}}le 1$ so $f(x)=sum_{i=1}^nA_i^x$ has negative derivative and thus $nge f(0)ge f(p/q)ge f(1)=1$ implies $nd_p(x,y)^{p^2}ge d_q(x,y)^{q^2} ge d_p(x,y)^{p^2}$. Now, if $A$ is open under $d_p$, for any $x in A$ there exists $r$ s.t. $d_p(x,y)<rRightarrow yin A$. Choosing $R=r^{p^2/q^2}$ shows that $d_q(x,y)<R$ implies $d_p(x,y)le d_q(x,y)^{q^2/p^2}<r$ so $yin A$ and $A$ is open under $d_q$. The converse follows similarly.






                share|cite|improve this answer












                If $p<q$ define $A_i=frac{|x_i-y_i|^{1/p}}{sum_{j=1}^n |x_j-y_j|^{1/p}}le 1$ so $f(x)=sum_{i=1}^nA_i^x$ has negative derivative and thus $nge f(0)ge f(p/q)ge f(1)=1$ implies $nd_p(x,y)^{p^2}ge d_q(x,y)^{q^2} ge d_p(x,y)^{p^2}$. Now, if $A$ is open under $d_p$, for any $x in A$ there exists $r$ s.t. $d_p(x,y)<rRightarrow yin A$. Choosing $R=r^{p^2/q^2}$ shows that $d_q(x,y)<R$ implies $d_p(x,y)le d_q(x,y)^{q^2/p^2}<r$ so $yin A$ and $A$ is open under $d_q$. The converse follows similarly.







                share|cite|improve this answer












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                answered Nov 25 at 8:59









                Guacho Perez

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