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Baby rudin subset with 3 limit points question
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Construct a bounded closed subset of $mathbbR$ with exactly three
limit points.
This might be a dumb question but why can't you just choose say $S=1,2,3$? It's bounded, and any sequence in $S$ that converges must eventually become constant. So any sequence must then converge to $1,2$ or $3$ which means $S$ is closed. Using the same reasoning, the limit points are $1,2$ and $3$.
I am using the sequential definition of "closed", and the following definition for the set of limit points of a given subset $S$ inside a metric space $M$:
$$lim S:=p in M mid exists (x_n) in S text such that x_n to p .$$
general-topology
edited Aug 15 at 3:52
Aloizio Macedo♦
22.7k23383
asked Aug 15 at 3:18
RayOfHope
605
add a comment |Â
up vote
3
down vote
favorite
Construct a bounded closed subset of $mathbbR$ with exactly three
limit points.
This might be a dumb question but why can't you just choose say $S=1,2,3$? It's bounded, and any sequence in $S$ that converges must eventually become constant. So any sequence must then converge to $1,2$ or $3$ which means $S$ is closed. Using the same reasoning, the limit points are $1,2$ and $3$.
I am using the sequential definition of "closed", and the following definition for the set of limit points of a given subset $S$ inside a metric space $M$:
$$lim S:=p in M mid exists (x_n) in S text such that x_n to p .$$
general-topology
edited Aug 15 at 3:52
Aloizio Macedo♦
22.7k23383
asked Aug 15 at 3:18
RayOfHope
605
I know it's already been answered, but I think Rudin gives a good hint in the exercise right before when he asks to prove that $K = 0 cup frac1n : n in mathbbN$ is compact. It's clear that $0$ is the only limit point. So take $K_2 = 1 cup 1 + frac1n$ and $K_3$ similarly. The union of these 3 sets has 3 unique limit points.
– Good Morning Captain
Aug 15 at 4:36
What you define is just the closure of $S$, not exactly the same.
– Carsten S
Aug 15 at 8:51
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Construct a bounded closed subset of $mathbbR$ with exactly three
limit points.
This might be a dumb question but why can't you just choose say $S=1,2,3$? It's bounded, and any sequence in $S$ that converges must eventually become constant. So any sequence must then converge to $1,2$ or $3$ which means $S$ is closed. Using the same reasoning, the limit points are $1,2$ and $3$.
I am using the sequential definition of "closed", and the following definition for the set of limit points of a given subset $S$ inside a metric space $M$:
$$lim S:=p in M mid exists (x_n) in S text such that x_n to p .$$
general-topology
edited Aug 15 at 3:52
Aloizio Macedo♦
22.7k23383
asked Aug 15 at 3:18
RayOfHope
605
Construct a bounded closed subset of $mathbbR$ with exactly three
limit points.
This might be a dumb question but why can't you just choose say $S=1,2,3$? It's bounded, and any sequence in $S$ that converges must eventually become constant. So any sequence must then converge to $1,2$ or $3$ which means $S$ is closed. Using the same reasoning, the limit points are $1,2$ and $3$.
I am using the sequential definition of "closed", and the following definition for the set of limit points of a given subset $S$ inside a metric space $M$:
$$lim S:=p in M mid exists (x_n) in S text such that x_n to p .$$
general-topology
edited Aug 15 at 3:52
Aloizio Macedo♦
22.7k23383
asked Aug 15 at 3:18
RayOfHope
605
edited Aug 15 at 3:52
Aloizio Macedo♦
22.7k23383
edited Aug 15 at 3:52
Aloizio Macedo♦
22.7k23383
edited Aug 15 at 3:52
Aloizio Macedo♦
22.7k23383
22.7k23383
asked Aug 15 at 3:18
RayOfHope
605
asked Aug 15 at 3:18
RayOfHope
605
asked Aug 15 at 3:18
RayOfHope
605
605
I know it's already been answered, but I think Rudin gives a good hint in the exercise right before when he asks to prove that $K = 0 cup frac1n : n in mathbbN$ is compact. It's clear that $0$ is the only limit point. So take $K_2 = 1 cup 1 + frac1n$ and $K_3$ similarly. The union of these 3 sets has 3 unique limit points.
– Good Morning Captain
Aug 15 at 4:36
What you define is just the closure of $S$, not exactly the same.
– Carsten S
Aug 15 at 8:51
add a comment |Â
I know it's already been answered, but I think Rudin gives a good hint in the exercise right before when he asks to prove that $K = 0 cup frac1n : n in mathbbN$ is compact. It's clear that $0$ is the only limit point. So take $K_2 = 1 cup 1 + frac1n$ and $K_3$ similarly. The union of these 3 sets has 3 unique limit points.
– Good Morning Captain
Aug 15 at 4:36
What you define is just the closure of $S$, not exactly the same.
– Carsten S
Aug 15 at 8:51
I know it's already been answered, but I think Rudin gives a good hint in the exercise right before when he asks to prove that $K = 0 cup frac1n : n in mathbbN$ is compact. It's clear that $0$ is the only limit point. So take $K_2 = 1 cup 1 + frac1n$ and $K_3$ similarly. The union of these 3 sets has 3 unique limit points.
– Good Morning Captain
Aug 15 at 4:36
I know it's already been answered, but I think Rudin gives a good hint in the exercise right before when he asks to prove that $K = 0 cup frac1n : n in mathbbN$ is compact. It's clear that $0$ is the only limit point. So take $K_2 = 1 cup 1 + frac1n$ and $K_3$ similarly. The union of these 3 sets has 3 unique limit points.
– Good Morning Captain
Aug 15 at 4:36
What you define is just the closure of $S$, not exactly the same.
– Carsten S
Aug 15 at 8:51
What you define is just the closure of $S$, not exactly the same.
– Carsten S
Aug 15 at 8:51
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
4
down vote
accepted
With that definition, the set $S$ that you have actually works out, because every convergent sequence must be eventually constant, so must have as a limit one of the elements of $S$.
However, the definition of limit point in some other books is somewhat different :
$x in S$ is a limit point if for all open sets $O$ with $x in O$, there exists $q neq x in S$ such that $q in O$. In other words, $(O setminus x) cap E neq emptyset$ for all open sets $O$ containing $x$.
The way of interpreting this definition, is that a limit point of $S$ is one such that at any arbitrary distance, one can find a point of $S$ closer than that distance which is different from $x$. In other words, there is a sequence, none of whose terms are $x$, which converges to $x$.
For example,Rudin has this definition.
Now, with this definition $S$ does not work, because any eventually constant sequence cannot be used to show that some point is a limit point. We have to, instead, be cleverer.
Indeed, the following set works out now : $frac 12,frac 23,frac 34,frac45,...,1 cup 1frac 12,1frac 34 , 1 frac 45,...,2 cup 2frac 12,2frac 23,2frac 34,...,3$.This set has exactly three limit points, according to the new definition.
You must not be too worried that your definitions are not tallying with other ones. Proceed with your definitions, and soon enough you will comfortable in switching while having conversations with others who have differing definitions.
edited Aug 15 at 3:54
Aloizio Macedo♦
22.7k23383
answered Aug 15 at 3:42
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.2k22665
add a comment |Â
up vote
1
down vote
For any $m$ distinct reals
(or any domain where
the following makes sense)
$(a_k)_k=1^m
$,
the $m$ sequences
$S_k = (a_k+frac1n)_n=1^infty
$
for
$k=1$ to $m$
are bounded
and
$S_k$ converges to
$a_k$.
You can interlace these
to get a single sequence
with $m$ subsets
that each converge to
one of the $a_k$:
$T_n
= (a_1+(n-1 bmod m)
+frac1mlfloor (n+m-1)/m rfloor
$
for
$n=1$ to $infty$.
answered Aug 15 at 3:57
marty cohen
69.7k446122
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
With that definition, the set $S$ that you have actually works out, because every convergent sequence must be eventually constant, so must have as a limit one of the elements of $S$.
However, the definition of limit point in some other books is somewhat different :
$x in S$ is a limit point if for all open sets $O$ with $x in O$, there exists $q neq x in S$ such that $q in O$. In other words, $(O setminus x) cap E neq emptyset$ for all open sets $O$ containing $x$.
The way of interpreting this definition, is that a limit point of $S$ is one such that at any arbitrary distance, one can find a point of $S$ closer than that distance which is different from $x$. In other words, there is a sequence, none of whose terms are $x$, which converges to $x$.
For example,Rudin has this definition.
Now, with this definition $S$ does not work, because any eventually constant sequence cannot be used to show that some point is a limit point. We have to, instead, be cleverer.
Indeed, the following set works out now : $frac 12,frac 23,frac 34,frac45,...,1 cup 1frac 12,1frac 34 , 1 frac 45,...,2 cup 2frac 12,2frac 23,2frac 34,...,3$.This set has exactly three limit points, according to the new definition.
You must not be too worried that your definitions are not tallying with other ones. Proceed with your definitions, and soon enough you will comfortable in switching while having conversations with others who have differing definitions.
edited Aug 15 at 3:54
Aloizio Macedo♦
22.7k23383
answered Aug 15 at 3:42
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.2k22665
add a comment |Â
up vote
4
down vote
accepted
With that definition, the set $S$ that you have actually works out, because every convergent sequence must be eventually constant, so must have as a limit one of the elements of $S$.
However, the definition of limit point in some other books is somewhat different :
$x in S$ is a limit point if for all open sets $O$ with $x in O$, there exists $q neq x in S$ such that $q in O$. In other words, $(O setminus x) cap E neq emptyset$ for all open sets $O$ containing $x$.
The way of interpreting this definition, is that a limit point of $S$ is one such that at any arbitrary distance, one can find a point of $S$ closer than that distance which is different from $x$. In other words, there is a sequence, none of whose terms are $x$, which converges to $x$.
For example,Rudin has this definition.
Now, with this definition $S$ does not work, because any eventually constant sequence cannot be used to show that some point is a limit point. We have to, instead, be cleverer.
Indeed, the following set works out now : $frac 12,frac 23,frac 34,frac45,...,1 cup 1frac 12,1frac 34 , 1 frac 45,...,2 cup 2frac 12,2frac 23,2frac 34,...,3$.This set has exactly three limit points, according to the new definition.
You must not be too worried that your definitions are not tallying with other ones. Proceed with your definitions, and soon enough you will comfortable in switching while having conversations with others who have differing definitions.
edited Aug 15 at 3:54
Aloizio Macedo♦
22.7k23383
answered Aug 15 at 3:42
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.2k22665
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
With that definition, the set $S$ that you have actually works out, because every convergent sequence must be eventually constant, so must have as a limit one of the elements of $S$.
However, the definition of limit point in some other books is somewhat different :
$x in S$ is a limit point if for all open sets $O$ with $x in O$, there exists $q neq x in S$ such that $q in O$. In other words, $(O setminus x) cap E neq emptyset$ for all open sets $O$ containing $x$.
The way of interpreting this definition, is that a limit point of $S$ is one such that at any arbitrary distance, one can find a point of $S$ closer than that distance which is different from $x$. In other words, there is a sequence, none of whose terms are $x$, which converges to $x$.
For example,Rudin has this definition.
Now, with this definition $S$ does not work, because any eventually constant sequence cannot be used to show that some point is a limit point. We have to, instead, be cleverer.
Indeed, the following set works out now : $frac 12,frac 23,frac 34,frac45,...,1 cup 1frac 12,1frac 34 , 1 frac 45,...,2 cup 2frac 12,2frac 23,2frac 34,...,3$.This set has exactly three limit points, according to the new definition.
You must not be too worried that your definitions are not tallying with other ones. Proceed with your definitions, and soon enough you will comfortable in switching while having conversations with others who have differing definitions.
edited Aug 15 at 3:54
Aloizio Macedo♦
22.7k23383
answered Aug 15 at 3:42
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.2k22665
With that definition, the set $S$ that you have actually works out, because every convergent sequence must be eventually constant, so must have as a limit one of the elements of $S$.
However, the definition of limit point in some other books is somewhat different :
$x in S$ is a limit point if for all open sets $O$ with $x in O$, there exists $q neq x in S$ such that $q in O$. In other words, $(O setminus x) cap E neq emptyset$ for all open sets $O$ containing $x$.
The way of interpreting this definition, is that a limit point of $S$ is one such that at any arbitrary distance, one can find a point of $S$ closer than that distance which is different from $x$. In other words, there is a sequence, none of whose terms are $x$, which converges to $x$.
For example,Rudin has this definition.
Now, with this definition $S$ does not work, because any eventually constant sequence cannot be used to show that some point is a limit point. We have to, instead, be cleverer.
Indeed, the following set works out now : $frac 12,frac 23,frac 34,frac45,...,1 cup 1frac 12,1frac 34 , 1 frac 45,...,2 cup 2frac 12,2frac 23,2frac 34,...,3$.This set has exactly three limit points, according to the new definition.
You must not be too worried that your definitions are not tallying with other ones. Proceed with your definitions, and soon enough you will comfortable in switching while having conversations with others who have differing definitions.
edited Aug 15 at 3:54
Aloizio Macedo♦
22.7k23383
answered Aug 15 at 3:42
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.2k22665
edited Aug 15 at 3:54
Aloizio Macedo♦
22.7k23383
edited Aug 15 at 3:54
Aloizio Macedo♦
22.7k23383
edited Aug 15 at 3:54
Aloizio Macedo♦
22.7k23383
22.7k23383
answered Aug 15 at 3:42
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.2k22665
answered Aug 15 at 3:42
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.2k22665
answered Aug 15 at 3:42
ðÑÂтþý òіûûð þûþф üÑÂûûñÑÂрó
33.2k22665
33.2k22665
add a comment |Â
add a comment |Â
up vote
1
down vote
For any $m$ distinct reals
(or any domain where
the following makes sense)
$(a_k)_k=1^m
$,
the $m$ sequences
$S_k = (a_k+frac1n)_n=1^infty
$
for
$k=1$ to $m$
are bounded
and
$S_k$ converges to
$a_k$.
You can interlace these
to get a single sequence
with $m$ subsets
that each converge to
one of the $a_k$:
$T_n
= (a_1+(n-1 bmod m)
+frac1mlfloor (n+m-1)/m rfloor
$
for
$n=1$ to $infty$.
answered Aug 15 at 3:57
marty cohen
69.7k446122
add a comment |Â
up vote
1
down vote
For any $m$ distinct reals
(or any domain where
the following makes sense)
$(a_k)_k=1^m
$,
the $m$ sequences
$S_k = (a_k+frac1n)_n=1^infty
$
for
$k=1$ to $m$
are bounded
and
$S_k$ converges to
$a_k$.
You can interlace these
to get a single sequence
with $m$ subsets
that each converge to
one of the $a_k$:
$T_n
= (a_1+(n-1 bmod m)
+frac1mlfloor (n+m-1)/m rfloor
$
for
$n=1$ to $infty$.
answered Aug 15 at 3:57
marty cohen
69.7k446122
add a comment |Â
up vote
1
down vote
up vote
1
down vote
For any $m$ distinct reals
(or any domain where
the following makes sense)
$(a_k)_k=1^m
$,
the $m$ sequences
$S_k = (a_k+frac1n)_n=1^infty
$
for
$k=1$ to $m$
are bounded
and
$S_k$ converges to
$a_k$.
You can interlace these
to get a single sequence
with $m$ subsets
that each converge to
one of the $a_k$:
$T_n
= (a_1+(n-1 bmod m)
+frac1mlfloor (n+m-1)/m rfloor
$
for
$n=1$ to $infty$.
answered Aug 15 at 3:57
marty cohen
69.7k446122
For any $m$ distinct reals
(or any domain where
the following makes sense)
$(a_k)_k=1^m
$,
the $m$ sequences
$S_k = (a_k+frac1n)_n=1^infty
$
for
$k=1$ to $m$
are bounded
and
$S_k$ converges to
$a_k$.
You can interlace these
to get a single sequence
with $m$ subsets
that each converge to
one of the $a_k$:
$T_n
= (a_1+(n-1 bmod m)
+frac1mlfloor (n+m-1)/m rfloor
$
for
$n=1$ to $infty$.
answered Aug 15 at 3:57
marty cohen
69.7k446122
answered Aug 15 at 3:57
marty cohen
69.7k446122
answered Aug 15 at 3:57
marty cohen
69.7k446122
answered Aug 15 at 3:57
marty cohen
69.7k446122
69.7k446122
add a comment |Â
add a comment |Â
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I know it's already been answered, but I think Rudin gives a good hint in the exercise right before when he asks to prove that $K = 0 cup frac1n : n in mathbbN$ is compact. It's clear that $0$ is the only limit point. So take $K_2 = 1 cup 1 + frac1n$ and $K_3$ similarly. The union of these 3 sets has 3 unique limit points.
– Good Morning Captain
Aug 15 at 4:36
What you define is just the closure of $S$, not exactly the same.
– Carsten S
Aug 15 at 8:51