A proof that the Cantor set is Perfect











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I found in a book a proof that the Cantor Set $Delta$ is perfect, however I would like to know if "my proof" does the job in the same way.




Theorem: The Cantor Set $Delta$ is perfect.



Proof: Let $x in Delta$ and fix $epsilon > 0$. Then, we can take a $n_0 = n$ sufficiently large to have $epsilon > 1/3^{n_0}$.
Thus, the interval $[a, b]$ where $x$ lies is a subset of $B_epsilon
> (x)$. Hence, by iterating the construction of the Cantor set for $N >
n_0$, we have intervals of length $1/3^N$ all included in $B_epsilon
(x)$, but with only one of those intervals such that $x$ lies within.




The intution behind the proof was that we should prove that for every $x$, if $x in Delta$, then for every $epsilon >0$, $B_epsilon (x) setminus {x} cap Delta neq varnothing$.



Now, I do not particularly like my reference to the $[a, b]$ interval that is not mentioned before. Moreover, here – by choosing a closed interval – I am trying to address all at once the case in which $x$ is an endpoint of one of the closed intervals that form $Delta$. Finally, I did not close the proof with a statement like "Thus, there are infinitely many points that differ from $x$ and that lie within $B_epsilon (x)$.



In the end, I am not completely sure if this can be considered a proof or not. The intuition is correct (I am kind of positive about it), but I am not sure if I was actually able to write down my intuition in a good way.



As always any feedback is more than welcome.

Thank you!










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  • 1




    Do you know that the Cantor set is closed (hence compact)? Now, does the Cantor set have an isolated point?
    – Somabha Mukherjee
    Jan 4 '15 at 13:22










  • @SomabhaMukherjee: Thanks! Yes, I see your point. Still, the point of the question is if my argumentation (and the way in which it is written!) works, not if there is another – much quicker way – to get the result. Hence, the tags on proof-verification and proof-writing. :)
    – Kolmin
    Jan 4 '15 at 14:24















up vote
6
down vote

favorite
2












I found in a book a proof that the Cantor Set $Delta$ is perfect, however I would like to know if "my proof" does the job in the same way.




Theorem: The Cantor Set $Delta$ is perfect.



Proof: Let $x in Delta$ and fix $epsilon > 0$. Then, we can take a $n_0 = n$ sufficiently large to have $epsilon > 1/3^{n_0}$.
Thus, the interval $[a, b]$ where $x$ lies is a subset of $B_epsilon
> (x)$. Hence, by iterating the construction of the Cantor set for $N >
n_0$, we have intervals of length $1/3^N$ all included in $B_epsilon
(x)$, but with only one of those intervals such that $x$ lies within.




The intution behind the proof was that we should prove that for every $x$, if $x in Delta$, then for every $epsilon >0$, $B_epsilon (x) setminus {x} cap Delta neq varnothing$.



Now, I do not particularly like my reference to the $[a, b]$ interval that is not mentioned before. Moreover, here – by choosing a closed interval – I am trying to address all at once the case in which $x$ is an endpoint of one of the closed intervals that form $Delta$. Finally, I did not close the proof with a statement like "Thus, there are infinitely many points that differ from $x$ and that lie within $B_epsilon (x)$.



In the end, I am not completely sure if this can be considered a proof or not. The intuition is correct (I am kind of positive about it), but I am not sure if I was actually able to write down my intuition in a good way.



As always any feedback is more than welcome.

Thank you!










share|cite|improve this question




















  • 1




    Do you know that the Cantor set is closed (hence compact)? Now, does the Cantor set have an isolated point?
    – Somabha Mukherjee
    Jan 4 '15 at 13:22










  • @SomabhaMukherjee: Thanks! Yes, I see your point. Still, the point of the question is if my argumentation (and the way in which it is written!) works, not if there is another – much quicker way – to get the result. Hence, the tags on proof-verification and proof-writing. :)
    – Kolmin
    Jan 4 '15 at 14:24













up vote
6
down vote

favorite
2









up vote
6
down vote

favorite
2






2





I found in a book a proof that the Cantor Set $Delta$ is perfect, however I would like to know if "my proof" does the job in the same way.




Theorem: The Cantor Set $Delta$ is perfect.



Proof: Let $x in Delta$ and fix $epsilon > 0$. Then, we can take a $n_0 = n$ sufficiently large to have $epsilon > 1/3^{n_0}$.
Thus, the interval $[a, b]$ where $x$ lies is a subset of $B_epsilon
> (x)$. Hence, by iterating the construction of the Cantor set for $N >
n_0$, we have intervals of length $1/3^N$ all included in $B_epsilon
(x)$, but with only one of those intervals such that $x$ lies within.




The intution behind the proof was that we should prove that for every $x$, if $x in Delta$, then for every $epsilon >0$, $B_epsilon (x) setminus {x} cap Delta neq varnothing$.



Now, I do not particularly like my reference to the $[a, b]$ interval that is not mentioned before. Moreover, here – by choosing a closed interval – I am trying to address all at once the case in which $x$ is an endpoint of one of the closed intervals that form $Delta$. Finally, I did not close the proof with a statement like "Thus, there are infinitely many points that differ from $x$ and that lie within $B_epsilon (x)$.



In the end, I am not completely sure if this can be considered a proof or not. The intuition is correct (I am kind of positive about it), but I am not sure if I was actually able to write down my intuition in a good way.



As always any feedback is more than welcome.

Thank you!










share|cite|improve this question















I found in a book a proof that the Cantor Set $Delta$ is perfect, however I would like to know if "my proof" does the job in the same way.




Theorem: The Cantor Set $Delta$ is perfect.



Proof: Let $x in Delta$ and fix $epsilon > 0$. Then, we can take a $n_0 = n$ sufficiently large to have $epsilon > 1/3^{n_0}$.
Thus, the interval $[a, b]$ where $x$ lies is a subset of $B_epsilon
> (x)$. Hence, by iterating the construction of the Cantor set for $N >
n_0$, we have intervals of length $1/3^N$ all included in $B_epsilon
(x)$, but with only one of those intervals such that $x$ lies within.




The intution behind the proof was that we should prove that for every $x$, if $x in Delta$, then for every $epsilon >0$, $B_epsilon (x) setminus {x} cap Delta neq varnothing$.



Now, I do not particularly like my reference to the $[a, b]$ interval that is not mentioned before. Moreover, here – by choosing a closed interval – I am trying to address all at once the case in which $x$ is an endpoint of one of the closed intervals that form $Delta$. Finally, I did not close the proof with a statement like "Thus, there are infinitely many points that differ from $x$ and that lie within $B_epsilon (x)$.



In the end, I am not completely sure if this can be considered a proof or not. The intuition is correct (I am kind of positive about it), but I am not sure if I was actually able to write down my intuition in a good way.



As always any feedback is more than welcome.

Thank you!







real-analysis proof-verification proof-writing cantor-set






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edited 23 hours ago









Martin Sleziak

44.3k7115266




44.3k7115266










asked Jan 4 '15 at 13:11









Kolmin

1,84311332




1,84311332








  • 1




    Do you know that the Cantor set is closed (hence compact)? Now, does the Cantor set have an isolated point?
    – Somabha Mukherjee
    Jan 4 '15 at 13:22










  • @SomabhaMukherjee: Thanks! Yes, I see your point. Still, the point of the question is if my argumentation (and the way in which it is written!) works, not if there is another – much quicker way – to get the result. Hence, the tags on proof-verification and proof-writing. :)
    – Kolmin
    Jan 4 '15 at 14:24














  • 1




    Do you know that the Cantor set is closed (hence compact)? Now, does the Cantor set have an isolated point?
    – Somabha Mukherjee
    Jan 4 '15 at 13:22










  • @SomabhaMukherjee: Thanks! Yes, I see your point. Still, the point of the question is if my argumentation (and the way in which it is written!) works, not if there is another – much quicker way – to get the result. Hence, the tags on proof-verification and proof-writing. :)
    – Kolmin
    Jan 4 '15 at 14:24








1




1




Do you know that the Cantor set is closed (hence compact)? Now, does the Cantor set have an isolated point?
– Somabha Mukherjee
Jan 4 '15 at 13:22




Do you know that the Cantor set is closed (hence compact)? Now, does the Cantor set have an isolated point?
– Somabha Mukherjee
Jan 4 '15 at 13:22












@SomabhaMukherjee: Thanks! Yes, I see your point. Still, the point of the question is if my argumentation (and the way in which it is written!) works, not if there is another – much quicker way – to get the result. Hence, the tags on proof-verification and proof-writing. :)
– Kolmin
Jan 4 '15 at 14:24




@SomabhaMukherjee: Thanks! Yes, I see your point. Still, the point of the question is if my argumentation (and the way in which it is written!) works, not if there is another – much quicker way – to get the result. Hence, the tags on proof-verification and proof-writing. :)
– Kolmin
Jan 4 '15 at 14:24










2 Answers
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up vote
7
down vote



accepted










Your idea is sound, but you’ve not expressed it clearly enough for you to have a real proof. I’ll write up an argument along the general lines that you have in mind.




Let $xinDelta$ and $epsilon>0$ be arbitrary. Choose $ninBbb N$ large enough so that $3^{-n}<epsilon$. $C_n$, the $n$-th stage in the standard construction of $Delta$, is the union of $2^n$ pairwise disjoint closed intervals, each of length $3^{-n}$; let $I$ be the one of these intervals containing $x$, clearly $Isubseteq B_epsilon(x)$.



Now consider $C_{n+1}$: it’s a disjoint union of $2^{n+1}$ closed intervals, each of length $3^{-(n+1)}$, and exactly two of these intervals, say $I_0$ and $I_1$, are subsets of $I$. Let $I_0$ be the one that contains $x$. $Deltacap I_1$ is non-empty for the same reason that $Delta$ is non-empty (why is that?), so let $yinDeltacap I_1$. Then $yinDeltacapbig(B_epsilon(x)setminus{x}big)$, and since $epsilon>0$ was arbitrary, $x$ is a limit point of $Delta$. Finally, $xinDelta$ was arbitrary, and $Delta$ is closed, so $Delta$ is perfect. $dashv$




It isn’t actually necessary to split $I$ into $I_0$ and $I_1$. Let $I=[a,b]$; then $a,binDeltacap B_epsilon(x)$, and since $x$ cannot be equal to both $a$ and $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.






share|cite|improve this answer





















  • First of all, as always, thanks a lot! Regarding the proof and your last statement, actually this is exactly what made me think if the proof was ok. Indeed, the proof I found in a book referred exactly to what you mentioned in the end, but in my proof I was not referring to it. Beyond this, now I have to admit that the comment to my original question by Somabha Mukherjee is making me wonder: why do we actually need any sort of "long" proof like mine (more precisely... yours) or another that does not use a split of the interval?
    – Kolmin
    Jan 5 '15 at 10:43






  • 1




    @Kolmin: You're very welcome! We need some such argument in order to show that it has no isolated points. That comment seems to take this for granted, but it requires proof.
    – Brian M. Scott
    Jan 5 '15 at 10:57


















up vote
0
down vote













Let $xin Delta$, For any $epsilon>0$, consider $(x-epsilon,x+epsilon)$ Using archemedean propery, $exists Nin mathbb N: frac{1}{3^N}<epsilon$. Let $frac{M}{3^N}=max {frac{m}{3^N}:frac{m}{3^N}<x space text{and} space min mathbb N}$. So $xin [frac{M}{3^N},frac{M+1}{3^N}]subset (x-epsilon,x+epsilon).$ Consider the $N+1$ the stage of construction Removing $(frac{3M+1}{3^{N+1}},frac{3M+2}{3^{N+1}})$ from $[frac{M}{3^N},frac{M+1}{3^N}]$.



Case 1 $xin [frac{M}{3^N},frac{3M+1}{3^{N+1}}].$ there exists $cin Delta$: $cin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$. Hence $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ So, $x$ is a limit point of $Delta$.



Similarly



Case 2 suppose $xin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$, We have $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ Hence, $x$ is the limit point of $Delta$.
So, $Delta$ is a perfect set.






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    up vote
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    accepted










    Your idea is sound, but you’ve not expressed it clearly enough for you to have a real proof. I’ll write up an argument along the general lines that you have in mind.




    Let $xinDelta$ and $epsilon>0$ be arbitrary. Choose $ninBbb N$ large enough so that $3^{-n}<epsilon$. $C_n$, the $n$-th stage in the standard construction of $Delta$, is the union of $2^n$ pairwise disjoint closed intervals, each of length $3^{-n}$; let $I$ be the one of these intervals containing $x$, clearly $Isubseteq B_epsilon(x)$.



    Now consider $C_{n+1}$: it’s a disjoint union of $2^{n+1}$ closed intervals, each of length $3^{-(n+1)}$, and exactly two of these intervals, say $I_0$ and $I_1$, are subsets of $I$. Let $I_0$ be the one that contains $x$. $Deltacap I_1$ is non-empty for the same reason that $Delta$ is non-empty (why is that?), so let $yinDeltacap I_1$. Then $yinDeltacapbig(B_epsilon(x)setminus{x}big)$, and since $epsilon>0$ was arbitrary, $x$ is a limit point of $Delta$. Finally, $xinDelta$ was arbitrary, and $Delta$ is closed, so $Delta$ is perfect. $dashv$




    It isn’t actually necessary to split $I$ into $I_0$ and $I_1$. Let $I=[a,b]$; then $a,binDeltacap B_epsilon(x)$, and since $x$ cannot be equal to both $a$ and $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.






    share|cite|improve this answer





















    • First of all, as always, thanks a lot! Regarding the proof and your last statement, actually this is exactly what made me think if the proof was ok. Indeed, the proof I found in a book referred exactly to what you mentioned in the end, but in my proof I was not referring to it. Beyond this, now I have to admit that the comment to my original question by Somabha Mukherjee is making me wonder: why do we actually need any sort of "long" proof like mine (more precisely... yours) or another that does not use a split of the interval?
      – Kolmin
      Jan 5 '15 at 10:43






    • 1




      @Kolmin: You're very welcome! We need some such argument in order to show that it has no isolated points. That comment seems to take this for granted, but it requires proof.
      – Brian M. Scott
      Jan 5 '15 at 10:57















    up vote
    7
    down vote



    accepted










    Your idea is sound, but you’ve not expressed it clearly enough for you to have a real proof. I’ll write up an argument along the general lines that you have in mind.




    Let $xinDelta$ and $epsilon>0$ be arbitrary. Choose $ninBbb N$ large enough so that $3^{-n}<epsilon$. $C_n$, the $n$-th stage in the standard construction of $Delta$, is the union of $2^n$ pairwise disjoint closed intervals, each of length $3^{-n}$; let $I$ be the one of these intervals containing $x$, clearly $Isubseteq B_epsilon(x)$.



    Now consider $C_{n+1}$: it’s a disjoint union of $2^{n+1}$ closed intervals, each of length $3^{-(n+1)}$, and exactly two of these intervals, say $I_0$ and $I_1$, are subsets of $I$. Let $I_0$ be the one that contains $x$. $Deltacap I_1$ is non-empty for the same reason that $Delta$ is non-empty (why is that?), so let $yinDeltacap I_1$. Then $yinDeltacapbig(B_epsilon(x)setminus{x}big)$, and since $epsilon>0$ was arbitrary, $x$ is a limit point of $Delta$. Finally, $xinDelta$ was arbitrary, and $Delta$ is closed, so $Delta$ is perfect. $dashv$




    It isn’t actually necessary to split $I$ into $I_0$ and $I_1$. Let $I=[a,b]$; then $a,binDeltacap B_epsilon(x)$, and since $x$ cannot be equal to both $a$ and $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.






    share|cite|improve this answer





















    • First of all, as always, thanks a lot! Regarding the proof and your last statement, actually this is exactly what made me think if the proof was ok. Indeed, the proof I found in a book referred exactly to what you mentioned in the end, but in my proof I was not referring to it. Beyond this, now I have to admit that the comment to my original question by Somabha Mukherjee is making me wonder: why do we actually need any sort of "long" proof like mine (more precisely... yours) or another that does not use a split of the interval?
      – Kolmin
      Jan 5 '15 at 10:43






    • 1




      @Kolmin: You're very welcome! We need some such argument in order to show that it has no isolated points. That comment seems to take this for granted, but it requires proof.
      – Brian M. Scott
      Jan 5 '15 at 10:57













    up vote
    7
    down vote



    accepted







    up vote
    7
    down vote



    accepted






    Your idea is sound, but you’ve not expressed it clearly enough for you to have a real proof. I’ll write up an argument along the general lines that you have in mind.




    Let $xinDelta$ and $epsilon>0$ be arbitrary. Choose $ninBbb N$ large enough so that $3^{-n}<epsilon$. $C_n$, the $n$-th stage in the standard construction of $Delta$, is the union of $2^n$ pairwise disjoint closed intervals, each of length $3^{-n}$; let $I$ be the one of these intervals containing $x$, clearly $Isubseteq B_epsilon(x)$.



    Now consider $C_{n+1}$: it’s a disjoint union of $2^{n+1}$ closed intervals, each of length $3^{-(n+1)}$, and exactly two of these intervals, say $I_0$ and $I_1$, are subsets of $I$. Let $I_0$ be the one that contains $x$. $Deltacap I_1$ is non-empty for the same reason that $Delta$ is non-empty (why is that?), so let $yinDeltacap I_1$. Then $yinDeltacapbig(B_epsilon(x)setminus{x}big)$, and since $epsilon>0$ was arbitrary, $x$ is a limit point of $Delta$. Finally, $xinDelta$ was arbitrary, and $Delta$ is closed, so $Delta$ is perfect. $dashv$




    It isn’t actually necessary to split $I$ into $I_0$ and $I_1$. Let $I=[a,b]$; then $a,binDeltacap B_epsilon(x)$, and since $x$ cannot be equal to both $a$ and $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.






    share|cite|improve this answer












    Your idea is sound, but you’ve not expressed it clearly enough for you to have a real proof. I’ll write up an argument along the general lines that you have in mind.




    Let $xinDelta$ and $epsilon>0$ be arbitrary. Choose $ninBbb N$ large enough so that $3^{-n}<epsilon$. $C_n$, the $n$-th stage in the standard construction of $Delta$, is the union of $2^n$ pairwise disjoint closed intervals, each of length $3^{-n}$; let $I$ be the one of these intervals containing $x$, clearly $Isubseteq B_epsilon(x)$.



    Now consider $C_{n+1}$: it’s a disjoint union of $2^{n+1}$ closed intervals, each of length $3^{-(n+1)}$, and exactly two of these intervals, say $I_0$ and $I_1$, are subsets of $I$. Let $I_0$ be the one that contains $x$. $Deltacap I_1$ is non-empty for the same reason that $Delta$ is non-empty (why is that?), so let $yinDeltacap I_1$. Then $yinDeltacapbig(B_epsilon(x)setminus{x}big)$, and since $epsilon>0$ was arbitrary, $x$ is a limit point of $Delta$. Finally, $xinDelta$ was arbitrary, and $Delta$ is closed, so $Delta$ is perfect. $dashv$




    It isn’t actually necessary to split $I$ into $I_0$ and $I_1$. Let $I=[a,b]$; then $a,binDeltacap B_epsilon(x)$, and since $x$ cannot be equal to both $a$ and $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 4 '15 at 22:46









    Brian M. Scott

    453k38503901




    453k38503901












    • First of all, as always, thanks a lot! Regarding the proof and your last statement, actually this is exactly what made me think if the proof was ok. Indeed, the proof I found in a book referred exactly to what you mentioned in the end, but in my proof I was not referring to it. Beyond this, now I have to admit that the comment to my original question by Somabha Mukherjee is making me wonder: why do we actually need any sort of "long" proof like mine (more precisely... yours) or another that does not use a split of the interval?
      – Kolmin
      Jan 5 '15 at 10:43






    • 1




      @Kolmin: You're very welcome! We need some such argument in order to show that it has no isolated points. That comment seems to take this for granted, but it requires proof.
      – Brian M. Scott
      Jan 5 '15 at 10:57


















    • First of all, as always, thanks a lot! Regarding the proof and your last statement, actually this is exactly what made me think if the proof was ok. Indeed, the proof I found in a book referred exactly to what you mentioned in the end, but in my proof I was not referring to it. Beyond this, now I have to admit that the comment to my original question by Somabha Mukherjee is making me wonder: why do we actually need any sort of "long" proof like mine (more precisely... yours) or another that does not use a split of the interval?
      – Kolmin
      Jan 5 '15 at 10:43






    • 1




      @Kolmin: You're very welcome! We need some such argument in order to show that it has no isolated points. That comment seems to take this for granted, but it requires proof.
      – Brian M. Scott
      Jan 5 '15 at 10:57
















    First of all, as always, thanks a lot! Regarding the proof and your last statement, actually this is exactly what made me think if the proof was ok. Indeed, the proof I found in a book referred exactly to what you mentioned in the end, but in my proof I was not referring to it. Beyond this, now I have to admit that the comment to my original question by Somabha Mukherjee is making me wonder: why do we actually need any sort of "long" proof like mine (more precisely... yours) or another that does not use a split of the interval?
    – Kolmin
    Jan 5 '15 at 10:43




    First of all, as always, thanks a lot! Regarding the proof and your last statement, actually this is exactly what made me think if the proof was ok. Indeed, the proof I found in a book referred exactly to what you mentioned in the end, but in my proof I was not referring to it. Beyond this, now I have to admit that the comment to my original question by Somabha Mukherjee is making me wonder: why do we actually need any sort of "long" proof like mine (more precisely... yours) or another that does not use a split of the interval?
    – Kolmin
    Jan 5 '15 at 10:43




    1




    1




    @Kolmin: You're very welcome! We need some such argument in order to show that it has no isolated points. That comment seems to take this for granted, but it requires proof.
    – Brian M. Scott
    Jan 5 '15 at 10:57




    @Kolmin: You're very welcome! We need some such argument in order to show that it has no isolated points. That comment seems to take this for granted, but it requires proof.
    – Brian M. Scott
    Jan 5 '15 at 10:57










    up vote
    0
    down vote













    Let $xin Delta$, For any $epsilon>0$, consider $(x-epsilon,x+epsilon)$ Using archemedean propery, $exists Nin mathbb N: frac{1}{3^N}<epsilon$. Let $frac{M}{3^N}=max {frac{m}{3^N}:frac{m}{3^N}<x space text{and} space min mathbb N}$. So $xin [frac{M}{3^N},frac{M+1}{3^N}]subset (x-epsilon,x+epsilon).$ Consider the $N+1$ the stage of construction Removing $(frac{3M+1}{3^{N+1}},frac{3M+2}{3^{N+1}})$ from $[frac{M}{3^N},frac{M+1}{3^N}]$.



    Case 1 $xin [frac{M}{3^N},frac{3M+1}{3^{N+1}}].$ there exists $cin Delta$: $cin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$. Hence $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ So, $x$ is a limit point of $Delta$.



    Similarly



    Case 2 suppose $xin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$, We have $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ Hence, $x$ is the limit point of $Delta$.
    So, $Delta$ is a perfect set.






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      Let $xin Delta$, For any $epsilon>0$, consider $(x-epsilon,x+epsilon)$ Using archemedean propery, $exists Nin mathbb N: frac{1}{3^N}<epsilon$. Let $frac{M}{3^N}=max {frac{m}{3^N}:frac{m}{3^N}<x space text{and} space min mathbb N}$. So $xin [frac{M}{3^N},frac{M+1}{3^N}]subset (x-epsilon,x+epsilon).$ Consider the $N+1$ the stage of construction Removing $(frac{3M+1}{3^{N+1}},frac{3M+2}{3^{N+1}})$ from $[frac{M}{3^N},frac{M+1}{3^N}]$.



      Case 1 $xin [frac{M}{3^N},frac{3M+1}{3^{N+1}}].$ there exists $cin Delta$: $cin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$. Hence $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ So, $x$ is a limit point of $Delta$.



      Similarly



      Case 2 suppose $xin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$, We have $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ Hence, $x$ is the limit point of $Delta$.
      So, $Delta$ is a perfect set.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Let $xin Delta$, For any $epsilon>0$, consider $(x-epsilon,x+epsilon)$ Using archemedean propery, $exists Nin mathbb N: frac{1}{3^N}<epsilon$. Let $frac{M}{3^N}=max {frac{m}{3^N}:frac{m}{3^N}<x space text{and} space min mathbb N}$. So $xin [frac{M}{3^N},frac{M+1}{3^N}]subset (x-epsilon,x+epsilon).$ Consider the $N+1$ the stage of construction Removing $(frac{3M+1}{3^{N+1}},frac{3M+2}{3^{N+1}})$ from $[frac{M}{3^N},frac{M+1}{3^N}]$.



        Case 1 $xin [frac{M}{3^N},frac{3M+1}{3^{N+1}}].$ there exists $cin Delta$: $cin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$. Hence $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ So, $x$ is a limit point of $Delta$.



        Similarly



        Case 2 suppose $xin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$, We have $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ Hence, $x$ is the limit point of $Delta$.
        So, $Delta$ is a perfect set.






        share|cite|improve this answer












        Let $xin Delta$, For any $epsilon>0$, consider $(x-epsilon,x+epsilon)$ Using archemedean propery, $exists Nin mathbb N: frac{1}{3^N}<epsilon$. Let $frac{M}{3^N}=max {frac{m}{3^N}:frac{m}{3^N}<x space text{and} space min mathbb N}$. So $xin [frac{M}{3^N},frac{M+1}{3^N}]subset (x-epsilon,x+epsilon).$ Consider the $N+1$ the stage of construction Removing $(frac{3M+1}{3^{N+1}},frac{3M+2}{3^{N+1}})$ from $[frac{M}{3^N},frac{M+1}{3^N}]$.



        Case 1 $xin [frac{M}{3^N},frac{3M+1}{3^{N+1}}].$ there exists $cin Delta$: $cin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$. Hence $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ So, $x$ is a limit point of $Delta$.



        Similarly



        Case 2 suppose $xin [frac{3M+2}{3^{N+1}},frac{M+1}{3^{N}}]$, We have $(x-epsilon,x+epsilon)cap Delta setminus {x}neq emptyset.$ Hence, $x$ is the limit point of $Delta$.
        So, $Delta$ is a perfect set.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










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