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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!
real-analysis proof-verification proof-writing cantor-set
edited 23 hours ago
Martin Sleziak
44.3k7115266
asked Jan 4 '15 at 13:11
Kolmin
1,84311332
add a comment |
up vote
6
down vote
favorite
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
edited 23 hours ago
Martin Sleziak
44.3k7115266
asked Jan 4 '15 at 13:11
Kolmin
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
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
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
edited 23 hours ago
Martin Sleziak
44.3k7115266
asked Jan 4 '15 at 13:11
Kolmin
1,84311332
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
real-analysis proof-verification proof-writing cantor-set
edited 23 hours ago
Martin Sleziak
44.3k7115266
asked Jan 4 '15 at 13:11
Kolmin
1,84311332
edited 23 hours ago
Martin Sleziak
44.3k7115266
asked Jan 4 '15 at 13:11
Kolmin
1,84311332
edited 23 hours ago
Martin Sleziak
44.3k7115266
edited 23 hours ago
Martin Sleziak
44.3k7115266
edited 23 hours ago
Martin Sleziak
44.3k7115266
44.3k7115266
asked Jan 4 '15 at 13:11
Kolmin
1,84311332
asked Jan 4 '15 at 13:11
Kolmin
1,84311332
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
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 $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.
answered Jan 4 '15 at 22:46
Brian M. Scott
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
add a comment |
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.
answered yesterday
Unknown x
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.
answered Jan 4 '15 at 22:46
Brian M. Scott
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
add a comment |
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 $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.
answered Jan 4 '15 at 22:46
Brian M. Scott
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
add a comment |
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 $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.
answered Jan 4 '15 at 22:46
Brian M. Scott
453k38503901
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 $b$, $Deltacapbig(B_epsilon(x)setminus{x}big)nevarnothing$.
answered Jan 4 '15 at 22:46
Brian M. Scott
453k38503901
answered Jan 4 '15 at 22:46
Brian M. Scott
453k38503901
answered Jan 4 '15 at 22:46
Brian M. Scott
453k38503901
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
add a comment |
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
add a comment |
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.
answered yesterday
Unknown x
2,42711025
add a comment |
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.
answered yesterday
Unknown x
2,42711025
add a comment |
up vote
0
down vote
up vote
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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.
answered yesterday
Unknown x
2,42711025
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.
answered yesterday
Unknown x
2,42711025
answered yesterday
Unknown x
2,42711025
answered yesterday
Unknown x
2,42711025
answered yesterday
Unknown x
2,42711025
2,42711025
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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