Rudin's proof that the Cantor set has no segments
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In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(alpha,beta)$ contains a segment of the form $(frac{3k+1}{3^{n}},frac{3k+2}{3^{n}})$ if $3^{-n}ltfrac{beta-alpha}{6}$. I don't know how $frac{beta-alpha}{6}$ is obtained.
real-analysis cantor-set
asked Oct 10 '15 at 19:03
MiladMilad
819
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add a comment |
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In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(alpha,beta)$ contains a segment of the form $(frac{3k+1}{3^{n}},frac{3k+2}{3^{n}})$ if $3^{-n}ltfrac{beta-alpha}{6}$. I don't know how $frac{beta-alpha}{6}$ is obtained.
real-analysis cantor-set
asked Oct 10 '15 at 19:03
MiladMilad
819
$endgroup$
add a comment |
$begingroup$
In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(alpha,beta)$ contains a segment of the form $(frac{3k+1}{3^{n}},frac{3k+2}{3^{n}})$ if $3^{-n}ltfrac{beta-alpha}{6}$. I don't know how $frac{beta-alpha}{6}$ is obtained.
real-analysis cantor-set
asked Oct 10 '15 at 19:03
MiladMilad
819
$endgroup$
In Principles of Mathematical Analysis, on page 42, Rudin proves that the Cantor set has no segments by stating that every segment $(alpha,beta)$ contains a segment of the form $(frac{3k+1}{3^{n}},frac{3k+2}{3^{n}})$ if $3^{-n}ltfrac{beta-alpha}{6}$. I don't know how $frac{beta-alpha}{6}$ is obtained.
real-analysis cantor-set
real-analysis cantor-set
asked Oct 10 '15 at 19:03
MiladMilad
819
asked Oct 10 '15 at 19:03
MiladMilad
819
edited Oct 12 '15 at 18:14
Milad
asked Oct 10 '15 at 19:03
MiladMilad
819
asked Oct 10 '15 at 19:03
MiladMilad
819
asked Oct 10 '15 at 19:03
MiladMilad
819
819
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Your denominators should be $3^n$, not $3^{-n}$.
Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is
$$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$
if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then
$$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$
so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that
$$frac1{3^n}lefrac{beta-alpha}4;,$$
the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.
Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.
answered Oct 10 '15 at 19:17
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:24
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in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
$endgroup$
– Milad
Oct 10 '15 at 19:42
1
$begingroup$
@Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:45
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:59
add a comment |
$begingroup$
We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,
so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.
Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so
$(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.
answered Oct 10 '15 at 19:16
user84413user84413
23k11848
$endgroup$
$begingroup$
Very nice proof!
$endgroup$
– ZFR
Oct 10 '15 at 19:22
$begingroup$
@RFZ Thanks for your comment; I appreciate it!
$endgroup$
– user84413
Oct 10 '15 at 19:24
$begingroup$
You are always welcome!
$endgroup$
– ZFR
Oct 10 '15 at 19:25
$begingroup$
your proof is really smart. thanks @user84413
$endgroup$
– Milad
Oct 10 '15 at 19:35
add a comment |
$begingroup$
If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.
See the above picture.
answered Jan 10 at 1:31
tchappy hatchappy ha
778412
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your denominators should be $3^n$, not $3^{-n}$.
Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is
$$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$
if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then
$$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$
so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that
$$frac1{3^n}lefrac{beta-alpha}4;,$$
the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.
Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.
answered Oct 10 '15 at 19:17
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:24
$begingroup$
in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
$endgroup$
– Milad
Oct 10 '15 at 19:42
1
$begingroup$
@Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:45
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:59
add a comment |
$begingroup$
Your denominators should be $3^n$, not $3^{-n}$.
Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is
$$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$
if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then
$$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$
so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that
$$frac1{3^n}lefrac{beta-alpha}4;,$$
the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.
Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.
answered Oct 10 '15 at 19:17
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:24
$begingroup$
in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
$endgroup$
– Milad
Oct 10 '15 at 19:42
1
$begingroup$
@Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:45
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:59
add a comment |
$begingroup$
Your denominators should be $3^n$, not $3^{-n}$.
Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is
$$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$
if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then
$$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$
so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that
$$frac1{3^n}lefrac{beta-alpha}4;,$$
the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.
Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.
answered Oct 10 '15 at 19:17
Brian M. ScottBrian M. Scott
460k40517918
$endgroup$
Your denominators should be $3^n$, not $3^{-n}$.
Suppose that $alpha$ is just a little bigger than $frac{3k+1}{3^n}$ for some $k$, so that $(alpha,beta)$ is just barely too far to the right to contain $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$ no matter how big $beta-alpha$ is. The next interval of that form to the right is
$$left(frac{3k+4}{3^n},frac{3k+5}{3^n}right);;tag{1}$$
if ensure that $frac{3k+5}{3^n}lebeta$, the interval $(alpha,beta)$ will contain the interval $(1)$. If $alpha=frac{3k+1}{3^n}+epsilon$, then
$$frac{3k+5}{3^n}=alpha-epsilon+frac4{3^n};,$$
so we need to make sure that $betagealpha-epsilon+frac4{3^n}$ no matter how small $epsilon$ is. This will certainly be the case if $betagealpha+frac4{3^n}$, i.e., if $beta-alphagefrac4{3^n}$. In other words, if we choose $n$ big enough so that
$$frac1{3^n}lefrac{beta-alpha}4;,$$
the interval $(alpha,beta)$ is certain to contain an interval of the form $left(frac{3k+1}{3^n},frac{3k+2}{3^n}right)$.
Clearly this will be the case if $frac1{3^n}lefrac{beta-alpha}6$.
answered Oct 10 '15 at 19:17
Brian M. ScottBrian M. Scott
460k40517918
answered Oct 10 '15 at 19:17
Brian M. ScottBrian M. Scott
460k40517918
answered Oct 10 '15 at 19:17
Brian M. ScottBrian M. Scott
460k40517918
answered Oct 10 '15 at 19:17
Brian M. ScottBrian M. Scott
460k40517918
460k40517918
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:24
$begingroup$
in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
$endgroup$
– Milad
Oct 10 '15 at 19:42
1
$begingroup$
@Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:45
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:59
add a comment |
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:24
$begingroup$
in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
$endgroup$
– Milad
Oct 10 '15 at 19:42
1
$begingroup$
@Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:45
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:59
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:24
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:24
$begingroup$
in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
$endgroup$
– Milad
Oct 10 '15 at 19:42
$begingroup$
in fact I had obtained $frac{beta-alpha}{4}$ but thought I had a mistake because Rudin uses $frac{beta-alpha}{6}$ @Brian
$endgroup$
– Milad
Oct 10 '15 at 19:42
1
1
$begingroup$
@Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:45
$begingroup$
@Milad: Using $frac{beta-alpha}6$ makes it just a little easier to see that $(alpha,beta)$ must contain an interval of the desired form. It means that $(alpha,beta)$ is at least $frac6{3^n}$ units long, long enough for two full cycles from $frac{3k}{3^n}$ to $frac{3(k+1)}{3^n}$, and it’s clear without really even doing any arithmetic that no matter where $alpha$ falls, two full cycles have to contain an interval of the desired type.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:45
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:59
$begingroup$
@Milad: You’re welcome.
$endgroup$
– Brian M. Scott
Oct 10 '15 at 19:59
add a comment |
$begingroup$
We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,
so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.
Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so
$(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.
answered Oct 10 '15 at 19:16
user84413user84413
23k11848
$endgroup$
$begingroup$
Very nice proof!
$endgroup$
– ZFR
Oct 10 '15 at 19:22
$begingroup$
@RFZ Thanks for your comment; I appreciate it!
$endgroup$
– user84413
Oct 10 '15 at 19:24
$begingroup$
You are always welcome!
$endgroup$
– ZFR
Oct 10 '15 at 19:25
$begingroup$
your proof is really smart. thanks @user84413
$endgroup$
– Milad
Oct 10 '15 at 19:35
add a comment |
$begingroup$
We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,
so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.
Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so
$(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.
answered Oct 10 '15 at 19:16
user84413user84413
23k11848
$endgroup$
$begingroup$
Very nice proof!
$endgroup$
– ZFR
Oct 10 '15 at 19:22
$begingroup$
@RFZ Thanks for your comment; I appreciate it!
$endgroup$
– user84413
Oct 10 '15 at 19:24
$begingroup$
You are always welcome!
$endgroup$
– ZFR
Oct 10 '15 at 19:25
$begingroup$
your proof is really smart. thanks @user84413
$endgroup$
– Milad
Oct 10 '15 at 19:35
add a comment |
$begingroup$
We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,
so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.
Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so
$(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.
answered Oct 10 '15 at 19:16
user84413user84413
23k11848
$endgroup$
We want to show that there is an integer $k$ such that $alphalefrac{3k+1}{3^n}$ and $frac{3k+2}{3^n}lebeta$,
so we want to have $frac{alpha(3^n)-1}{3}le klefrac{beta(3^n)-2}{3}$.
Such an integer $k$ will exist if $frac{beta(3^n)-2}{3}-frac{alpha(3^n)-1}{3}ge1$, so
$(beta-alpha)3^nge4$ gives $3^{-n}lefrac{beta-alpha}{4}$.
answered Oct 10 '15 at 19:16
user84413user84413
23k11848
edited Oct 10 '15 at 19:22
answered Oct 10 '15 at 19:16
user84413user84413
23k11848
answered Oct 10 '15 at 19:16
user84413user84413
23k11848
answered Oct 10 '15 at 19:16
user84413user84413
23k11848
23k11848
$begingroup$
Very nice proof!
$endgroup$
– ZFR
Oct 10 '15 at 19:22
$begingroup$
@RFZ Thanks for your comment; I appreciate it!
$endgroup$
– user84413
Oct 10 '15 at 19:24
$begingroup$
You are always welcome!
$endgroup$
– ZFR
Oct 10 '15 at 19:25
$begingroup$
your proof is really smart. thanks @user84413
$endgroup$
– Milad
Oct 10 '15 at 19:35
add a comment |
$begingroup$
Very nice proof!
$endgroup$
– ZFR
Oct 10 '15 at 19:22
$begingroup$
@RFZ Thanks for your comment; I appreciate it!
$endgroup$
– user84413
Oct 10 '15 at 19:24
$begingroup$
You are always welcome!
$endgroup$
– ZFR
Oct 10 '15 at 19:25
$begingroup$
your proof is really smart. thanks @user84413
$endgroup$
– Milad
Oct 10 '15 at 19:35
$begingroup$
Very nice proof!
$endgroup$
– ZFR
Oct 10 '15 at 19:22
$begingroup$
Very nice proof!
$endgroup$
– ZFR
Oct 10 '15 at 19:22
$begingroup$
@RFZ Thanks for your comment; I appreciate it!
$endgroup$
– user84413
Oct 10 '15 at 19:24
$begingroup$
@RFZ Thanks for your comment; I appreciate it!
$endgroup$
– user84413
Oct 10 '15 at 19:24
$begingroup$
You are always welcome!
$endgroup$
– ZFR
Oct 10 '15 at 19:25
$begingroup$
You are always welcome!
$endgroup$
– ZFR
Oct 10 '15 at 19:25
$begingroup$
your proof is really smart. thanks @user84413
$endgroup$
– Milad
Oct 10 '15 at 19:35
$begingroup$
your proof is really smart. thanks @user84413
$endgroup$
– Milad
Oct 10 '15 at 19:35
add a comment |
$begingroup$
If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.
See the above picture.
answered Jan 10 at 1:31
tchappy hatchappy ha
778412
$endgroup$
add a comment |
$begingroup$
If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.
See the above picture.
answered Jan 10 at 1:31
tchappy hatchappy ha
778412
$endgroup$
add a comment |
$begingroup$
If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.
See the above picture.
answered Jan 10 at 1:31
tchappy hatchappy ha
778412
$endgroup$
If $alpha$ and $beta$ satisfy $3^{-m} < frac{beta-alpha}{5}$, then every segment $(alpha, beta)$ contains a segment which was removed when the Cantor set was constructed.
See the above picture.
answered Jan 10 at 1:31
tchappy hatchappy ha
778412
answered Jan 10 at 1:31
tchappy hatchappy ha
778412
answered Jan 10 at 1:31
tchappy hatchappy ha
778412
answered Jan 10 at 1:31
tchappy hatchappy ha
778412
778412
add a comment |
add a comment |
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