Some questions about limits in metric spaces.

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Definition: Let $f: X to Y$ be a function between metric spaces. Let $E subseteq X$ and $p$ be a limit point of $E$. Let $q in Y.$



We say that $lim_x to p, x in E f(x) = q$ if



$$forall epsilon > 0: exists delta > 0: forall x in E: (0 <
d_X(x,p)< delta implies d_Y(f(x),q) < epsilon)$$




I am self studying the topic limits of functions and I made some conjectures. Can someone verify my conjectures and their proofs?




(1): Let $E_1 subseteq E_2 subseteq X$, $p$ be a limit point of $E_1$. Suppose $lim_x to p, x in E_2f(x)$ exists.



Then $lim_x to p, x in E_1f(x)$ exists and is equal to $lim_x to p, x in E_2f(x)$.




Proof: Evident from the definition, since $E_1 subseteq E_2$.




(2): The converse is not true. That is, if $lim_x to p, x in E_1f(x)$ exists, then $lim_x to p, x in E_2f(x)$ need not exist.
However, if it exists, it must be equal to $lim_x to p, x in
E_1f(x)$.




Proof: Take $mathbbR$ with the usual metric. Consider the function



$$f: mathbbR to mathbbR: x mapsto begincases 1 quad x > 0 \ -1 quad x leq 0endcases$$



Then $lim_x to 0,x in (0,1)f(x) = 1$ but $lim_x to 0,x in mathbbR f(x)$ does not exist (Take $epsilon = 1/2$, no $delta$ will work).



Suppose now that $lim_x to p, x in E_1f(x) = q_1, lim_x to p, x in E_2 f(x)= q_2$



Let $epsilon > 0.$ Choose $delta_1$ s.t. $forall x in E_1: 0 < d_X(x,p) < delta_1 implies d_Y(f(x),q_1) < epsilon$. Choose $delta_2$ s.t. $forall x in E_2: 0 < d_X(x,p) < delta_2 implies d_Y(f(x),q_2) < epsilon$.



Suppose $x in E_1$ with $0 < d_X(x,p) < mindelta_1, delta_2$. By the triangle inequality, $d_Y(q_1,q_2) < 2 epsilon$.



Because $epsilon$ was arbitrary, $d_Y(q_1,q_2) = 0$ and $q_1 = q_2$, as desired.




real-analysis limits metric-spaces




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    To me, everything is O.K.
    – user539887
    Jul 24 at 19:19














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Definition: Let $f: X to Y$ be a function between metric spaces. Let $E subseteq X$ and $p$ be a limit point of $E$. Let $q in Y.$



We say that $lim_x to p, x in E f(x) = q$ if



$$forall epsilon > 0: exists delta > 0: forall x in E: (0 <
d_X(x,p)< delta implies d_Y(f(x),q) < epsilon)$$




I am self studying the topic limits of functions and I made some conjectures. Can someone verify my conjectures and their proofs?




(1): Let $E_1 subseteq E_2 subseteq X$, $p$ be a limit point of $E_1$. Suppose $lim_x to p, x in E_2f(x)$ exists.



Then $lim_x to p, x in E_1f(x)$ exists and is equal to $lim_x to p, x in E_2f(x)$.




Proof: Evident from the definition, since $E_1 subseteq E_2$.




(2): The converse is not true. That is, if $lim_x to p, x in E_1f(x)$ exists, then $lim_x to p, x in E_2f(x)$ need not exist.
However, if it exists, it must be equal to $lim_x to p, x in
E_1f(x)$.




Proof: Take $mathbbR$ with the usual metric. Consider the function



$$f: mathbbR to mathbbR: x mapsto begincases 1 quad x > 0 \ -1 quad x leq 0endcases$$



Then $lim_x to 0,x in (0,1)f(x) = 1$ but $lim_x to 0,x in mathbbR f(x)$ does not exist (Take $epsilon = 1/2$, no $delta$ will work).



Suppose now that $lim_x to p, x in E_1f(x) = q_1, lim_x to p, x in E_2 f(x)= q_2$



Let $epsilon > 0.$ Choose $delta_1$ s.t. $forall x in E_1: 0 < d_X(x,p) < delta_1 implies d_Y(f(x),q_1) < epsilon$. Choose $delta_2$ s.t. $forall x in E_2: 0 < d_X(x,p) < delta_2 implies d_Y(f(x),q_2) < epsilon$.



Suppose $x in E_1$ with $0 < d_X(x,p) < mindelta_1, delta_2$. By the triangle inequality, $d_Y(q_1,q_2) < 2 epsilon$.



Because $epsilon$ was arbitrary, $d_Y(q_1,q_2) = 0$ and $q_1 = q_2$, as desired.




real-analysis limits metric-spaces




share|cite|improve this question















  • 2




    To me, everything is O.K.
    – user539887
    Jul 24 at 19:19












up vote
1
down vote

favorite









up vote
1
down vote

favorite












Definition: Let $f: X to Y$ be a function between metric spaces. Let $E subseteq X$ and $p$ be a limit point of $E$. Let $q in Y.$



We say that $lim_x to p, x in E f(x) = q$ if



$$forall epsilon > 0: exists delta > 0: forall x in E: (0 <
d_X(x,p)< delta implies d_Y(f(x),q) < epsilon)$$




I am self studying the topic limits of functions and I made some conjectures. Can someone verify my conjectures and their proofs?




(1): Let $E_1 subseteq E_2 subseteq X$, $p$ be a limit point of $E_1$. Suppose $lim_x to p, x in E_2f(x)$ exists.



Then $lim_x to p, x in E_1f(x)$ exists and is equal to $lim_x to p, x in E_2f(x)$.




Proof: Evident from the definition, since $E_1 subseteq E_2$.




(2): The converse is not true. That is, if $lim_x to p, x in E_1f(x)$ exists, then $lim_x to p, x in E_2f(x)$ need not exist.
However, if it exists, it must be equal to $lim_x to p, x in
E_1f(x)$.




Proof: Take $mathbbR$ with the usual metric. Consider the function



$$f: mathbbR to mathbbR: x mapsto begincases 1 quad x > 0 \ -1 quad x leq 0endcases$$



Then $lim_x to 0,x in (0,1)f(x) = 1$ but $lim_x to 0,x in mathbbR f(x)$ does not exist (Take $epsilon = 1/2$, no $delta$ will work).



Suppose now that $lim_x to p, x in E_1f(x) = q_1, lim_x to p, x in E_2 f(x)= q_2$



Let $epsilon > 0.$ Choose $delta_1$ s.t. $forall x in E_1: 0 < d_X(x,p) < delta_1 implies d_Y(f(x),q_1) < epsilon$. Choose $delta_2$ s.t. $forall x in E_2: 0 < d_X(x,p) < delta_2 implies d_Y(f(x),q_2) < epsilon$.



Suppose $x in E_1$ with $0 < d_X(x,p) < mindelta_1, delta_2$. By the triangle inequality, $d_Y(q_1,q_2) < 2 epsilon$.



Because $epsilon$ was arbitrary, $d_Y(q_1,q_2) = 0$ and $q_1 = q_2$, as desired.




real-analysis limits metric-spaces




share|cite|improve this question












Definition: Let $f: X to Y$ be a function between metric spaces. Let $E subseteq X$ and $p$ be a limit point of $E$. Let $q in Y.$



We say that $lim_x to p, x in E f(x) = q$ if



$$forall epsilon > 0: exists delta > 0: forall x in E: (0 <
d_X(x,p)< delta implies d_Y(f(x),q) < epsilon)$$




I am self studying the topic limits of functions and I made some conjectures. Can someone verify my conjectures and their proofs?




(1): Let $E_1 subseteq E_2 subseteq X$, $p$ be a limit point of $E_1$. Suppose $lim_x to p, x in E_2f(x)$ exists.



Then $lim_x to p, x in E_1f(x)$ exists and is equal to $lim_x to p, x in E_2f(x)$.




Proof: Evident from the definition, since $E_1 subseteq E_2$.




(2): The converse is not true. That is, if $lim_x to p, x in E_1f(x)$ exists, then $lim_x to p, x in E_2f(x)$ need not exist.
However, if it exists, it must be equal to $lim_x to p, x in
E_1f(x)$.




Proof: Take $mathbbR$ with the usual metric. Consider the function



$$f: mathbbR to mathbbR: x mapsto begincases 1 quad x > 0 \ -1 quad x leq 0endcases$$



Then $lim_x to 0,x in (0,1)f(x) = 1$ but $lim_x to 0,x in mathbbR f(x)$ does not exist (Take $epsilon = 1/2$, no $delta$ will work).



Suppose now that $lim_x to p, x in E_1f(x) = q_1, lim_x to p, x in E_2 f(x)= q_2$



Let $epsilon > 0.$ Choose $delta_1$ s.t. $forall x in E_1: 0 < d_X(x,p) < delta_1 implies d_Y(f(x),q_1) < epsilon$. Choose $delta_2$ s.t. $forall x in E_2: 0 < d_X(x,p) < delta_2 implies d_Y(f(x),q_2) < epsilon$.



Suppose $x in E_1$ with $0 < d_X(x,p) < mindelta_1, delta_2$. By the triangle inequality, $d_Y(q_1,q_2) < 2 epsilon$.



Because $epsilon$ was arbitrary, $d_Y(q_1,q_2) = 0$ and $q_1 = q_2$, as desired.





real-analysis limits metric-spaces





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asked Jul 24 at 19:03









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  • 2




    To me, everything is O.K.
    – user539887
    Jul 24 at 19:19












  • 2




    To me, everything is O.K.
    – user539887
    Jul 24 at 19:19







2




2




To me, everything is O.K.
– user539887
Jul 24 at 19:19




To me, everything is O.K.
– user539887
Jul 24 at 19:19















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