Mixing
Classification of subspace topologies of countably infinite subsets of $mathbb{R}$
$begingroup$
For example, we know that $mathbb{Z}$ is discrete, while $mathbb{Q}$ is not. The former has no limit point, while for the latter, every point is a limit point. An intermediate object is $X_l={l}cup{l+1/n}_{ninmathbb{N}}$, which has only one limit point. I conjecture that for the desired classification, it suffices to see which point is a limit point, and that there are $2^{mathbb{Z}}$ different subspace topologies. The operations such as $X_lcupmathbb{Z}$ increase a the number of limit points by one, and they may explicitly construct an arbitrary subspace topology from $mathbb{Z}$.
Questions
- Are my conclusion and argument correct? If not, please answer the correct conclusion regarding the classification and justify it.
- For this classification, is there an explicit way to construct any such topology from the known objects such as $mathbb{Z}$ and $mathbb{Q}$?
Edit
Qiaochu showed that limit points are not sufficient for classification. What tools would be needed?
general-topology
asked May 3 '17 at 19:49
Math.StackExchangeMath.StackExchange
2,402921
$endgroup$
add a comment |
$begingroup$
For example, we know that $mathbb{Z}$ is discrete, while $mathbb{Q}$ is not. The former has no limit point, while for the latter, every point is a limit point. An intermediate object is $X_l={l}cup{l+1/n}_{ninmathbb{N}}$, which has only one limit point. I conjecture that for the desired classification, it suffices to see which point is a limit point, and that there are $2^{mathbb{Z}}$ different subspace topologies. The operations such as $X_lcupmathbb{Z}$ increase a the number of limit points by one, and they may explicitly construct an arbitrary subspace topology from $mathbb{Z}$.
Questions
- Are my conclusion and argument correct? If not, please answer the correct conclusion regarding the classification and justify it.
- For this classification, is there an explicit way to construct any such topology from the known objects such as $mathbb{Z}$ and $mathbb{Q}$?
Edit
Qiaochu showed that limit points are not sufficient for classification. What tools would be needed?
general-topology
asked May 3 '17 at 19:49
Math.StackExchangeMath.StackExchange
2,402921
$endgroup$
$begingroup$
See math.stackexchange.com/a/416672 for a very useful answer for the compact case.
$endgroup$
– Henno Brandsma
May 3 '17 at 22:01
$begingroup$
Very informative! Thanks so much.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:12
add a comment |
$begingroup$
For example, we know that $mathbb{Z}$ is discrete, while $mathbb{Q}$ is not. The former has no limit point, while for the latter, every point is a limit point. An intermediate object is $X_l={l}cup{l+1/n}_{ninmathbb{N}}$, which has only one limit point. I conjecture that for the desired classification, it suffices to see which point is a limit point, and that there are $2^{mathbb{Z}}$ different subspace topologies. The operations such as $X_lcupmathbb{Z}$ increase a the number of limit points by one, and they may explicitly construct an arbitrary subspace topology from $mathbb{Z}$.
Questions
- Are my conclusion and argument correct? If not, please answer the correct conclusion regarding the classification and justify it.
- For this classification, is there an explicit way to construct any such topology from the known objects such as $mathbb{Z}$ and $mathbb{Q}$?
Edit
Qiaochu showed that limit points are not sufficient for classification. What tools would be needed?
general-topology
asked May 3 '17 at 19:49
Math.StackExchangeMath.StackExchange
2,402921
$endgroup$
For example, we know that $mathbb{Z}$ is discrete, while $mathbb{Q}$ is not. The former has no limit point, while for the latter, every point is a limit point. An intermediate object is $X_l={l}cup{l+1/n}_{ninmathbb{N}}$, which has only one limit point. I conjecture that for the desired classification, it suffices to see which point is a limit point, and that there are $2^{mathbb{Z}}$ different subspace topologies. The operations such as $X_lcupmathbb{Z}$ increase a the number of limit points by one, and they may explicitly construct an arbitrary subspace topology from $mathbb{Z}$.
Questions
- Are my conclusion and argument correct? If not, please answer the correct conclusion regarding the classification and justify it.
- For this classification, is there an explicit way to construct any such topology from the known objects such as $mathbb{Z}$ and $mathbb{Q}$?
Edit
Qiaochu showed that limit points are not sufficient for classification. What tools would be needed?
general-topology
general-topology
asked May 3 '17 at 19:49
Math.StackExchangeMath.StackExchange
2,402921
asked May 3 '17 at 19:49
Math.StackExchangeMath.StackExchange
2,402921
edited May 3 '17 at 20:38
Math.StackExchange
asked May 3 '17 at 19:49
Math.StackExchangeMath.StackExchange
2,402921
asked May 3 '17 at 19:49
Math.StackExchangeMath.StackExchange
2,402921
asked May 3 '17 at 19:49
Math.StackExchangeMath.StackExchange
2,402921
2,402921
$begingroup$
See math.stackexchange.com/a/416672 for a very useful answer for the compact case.
$endgroup$
– Henno Brandsma
May 3 '17 at 22:01
$begingroup$
Very informative! Thanks so much.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:12
add a comment |
$begingroup$
See math.stackexchange.com/a/416672 for a very useful answer for the compact case.
$endgroup$
– Henno Brandsma
May 3 '17 at 22:01
$begingroup$
Very informative! Thanks so much.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:12
$begingroup$
See math.stackexchange.com/a/416672 for a very useful answer for the compact case.
$endgroup$
– Henno Brandsma
May 3 '17 at 22:01
$begingroup$
See math.stackexchange.com/a/416672 for a very useful answer for the compact case.
$endgroup$
– Henno Brandsma
May 3 '17 at 22:01
$begingroup$
Very informative! Thanks so much.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:12
$begingroup$
Very informative! Thanks so much.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:12
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The book by Zemadeni (IIRC) Banach spaces of continuous functions, has a full proof of the standard classification of countable subspaces of the reals (or in fact countable metric spaces). The compact ones are countable ordinal numbers in essence, the others versions of this with $mathbb{Q}$ as a dense in itself kernel.
The idea is to consider the scattering sequence for a countable space $X$. $X^{(1)} = X'$ (the set of limit points of $X$), $X^{(alpha+1)} = (X^{(alpha)})'$, for successor ordinals $alpha +1$, and $X^{(beta)} =bigcap {X^{(alpha)} alpha < beta}$ for limit ordinals $beta$. This process stops on some countable ordinal $gamma$ such that $X^{(gamma)} = X^{(gamma+1)}$. Then $X$ is uniquely determined by this $gamma$ and if we end with the empty set, how many points there were on the stage before. (This is finite for a compact $X$ but can be countable discrete as well), or whether it ended in a countable dense in itself space (which is $mathbb{Q}$ essentially.)
So in that sense limit points are involved and important but we have to iterate the process. In fact, for this very reason ordinals were invented by Cantor. He came from analysis trying to analyse the possible countable sets of discontinuity of certain sums of goniometric series, and he needed ordinals to formulate his results, which led him to develop set theory for the first time.
answered May 3 '17 at 21:54
Henno BrandsmaHenno Brandsma
114k349125
$endgroup$
$begingroup$
Thanks for your answer! Good to know that it motivated Cantor to invent ordinals.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:10
add a comment |
$begingroup$
It does not suffice to record which points are limit points. For example, the subspace given by
$${ -n, n in mathbb{N} } cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
and the subspace given by points of the form
$$left{ -frac{1}{n}, n in mathbb{N} right} cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
both have the same limit point structure (one limit point "in the middle") but are not homeomorphic.
answered May 3 '17 at 20:28
Qiaochu YuanQiaochu Yuan
281k32595940
$endgroup$
$begingroup$
Thanks so much for your answer. So, limit points are irrelevant. But how can we classify?
$endgroup$
– Math.StackExchange
May 3 '17 at 20:31
$begingroup$
Well, they're not irrelevant, they're just not enough. At each limit point you can in addition keep track of whether it is being approached from the left, from the right, or both. I don't know if this is enough or not.
$endgroup$
– Qiaochu Yuan
May 3 '17 at 20:54
$begingroup$
Sorry for my sloppy use of the word. Thanks for input regarding the direction of the limit, which is really important. If this question is truly non-trivial and will not receive a decisive answer, I will migrate it to MO.
$endgroup$
– Math.StackExchange
May 3 '17 at 21:04
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The book by Zemadeni (IIRC) Banach spaces of continuous functions, has a full proof of the standard classification of countable subspaces of the reals (or in fact countable metric spaces). The compact ones are countable ordinal numbers in essence, the others versions of this with $mathbb{Q}$ as a dense in itself kernel.
The idea is to consider the scattering sequence for a countable space $X$. $X^{(1)} = X'$ (the set of limit points of $X$), $X^{(alpha+1)} = (X^{(alpha)})'$, for successor ordinals $alpha +1$, and $X^{(beta)} =bigcap {X^{(alpha)} alpha < beta}$ for limit ordinals $beta$. This process stops on some countable ordinal $gamma$ such that $X^{(gamma)} = X^{(gamma+1)}$. Then $X$ is uniquely determined by this $gamma$ and if we end with the empty set, how many points there were on the stage before. (This is finite for a compact $X$ but can be countable discrete as well), or whether it ended in a countable dense in itself space (which is $mathbb{Q}$ essentially.)
So in that sense limit points are involved and important but we have to iterate the process. In fact, for this very reason ordinals were invented by Cantor. He came from analysis trying to analyse the possible countable sets of discontinuity of certain sums of goniometric series, and he needed ordinals to formulate his results, which led him to develop set theory for the first time.
answered May 3 '17 at 21:54
Henno BrandsmaHenno Brandsma
114k349125
$endgroup$
$begingroup$
Thanks for your answer! Good to know that it motivated Cantor to invent ordinals.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:10
add a comment |
$begingroup$
The book by Zemadeni (IIRC) Banach spaces of continuous functions, has a full proof of the standard classification of countable subspaces of the reals (or in fact countable metric spaces). The compact ones are countable ordinal numbers in essence, the others versions of this with $mathbb{Q}$ as a dense in itself kernel.
The idea is to consider the scattering sequence for a countable space $X$. $X^{(1)} = X'$ (the set of limit points of $X$), $X^{(alpha+1)} = (X^{(alpha)})'$, for successor ordinals $alpha +1$, and $X^{(beta)} =bigcap {X^{(alpha)} alpha < beta}$ for limit ordinals $beta$. This process stops on some countable ordinal $gamma$ such that $X^{(gamma)} = X^{(gamma+1)}$. Then $X$ is uniquely determined by this $gamma$ and if we end with the empty set, how many points there were on the stage before. (This is finite for a compact $X$ but can be countable discrete as well), or whether it ended in a countable dense in itself space (which is $mathbb{Q}$ essentially.)
So in that sense limit points are involved and important but we have to iterate the process. In fact, for this very reason ordinals were invented by Cantor. He came from analysis trying to analyse the possible countable sets of discontinuity of certain sums of goniometric series, and he needed ordinals to formulate his results, which led him to develop set theory for the first time.
answered May 3 '17 at 21:54
Henno BrandsmaHenno Brandsma
114k349125
$endgroup$
$begingroup$
Thanks for your answer! Good to know that it motivated Cantor to invent ordinals.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:10
add a comment |
$begingroup$
The book by Zemadeni (IIRC) Banach spaces of continuous functions, has a full proof of the standard classification of countable subspaces of the reals (or in fact countable metric spaces). The compact ones are countable ordinal numbers in essence, the others versions of this with $mathbb{Q}$ as a dense in itself kernel.
The idea is to consider the scattering sequence for a countable space $X$. $X^{(1)} = X'$ (the set of limit points of $X$), $X^{(alpha+1)} = (X^{(alpha)})'$, for successor ordinals $alpha +1$, and $X^{(beta)} =bigcap {X^{(alpha)} alpha < beta}$ for limit ordinals $beta$. This process stops on some countable ordinal $gamma$ such that $X^{(gamma)} = X^{(gamma+1)}$. Then $X$ is uniquely determined by this $gamma$ and if we end with the empty set, how many points there were on the stage before. (This is finite for a compact $X$ but can be countable discrete as well), or whether it ended in a countable dense in itself space (which is $mathbb{Q}$ essentially.)
So in that sense limit points are involved and important but we have to iterate the process. In fact, for this very reason ordinals were invented by Cantor. He came from analysis trying to analyse the possible countable sets of discontinuity of certain sums of goniometric series, and he needed ordinals to formulate his results, which led him to develop set theory for the first time.
answered May 3 '17 at 21:54
Henno BrandsmaHenno Brandsma
114k349125
$endgroup$
The book by Zemadeni (IIRC) Banach spaces of continuous functions, has a full proof of the standard classification of countable subspaces of the reals (or in fact countable metric spaces). The compact ones are countable ordinal numbers in essence, the others versions of this with $mathbb{Q}$ as a dense in itself kernel.
The idea is to consider the scattering sequence for a countable space $X$. $X^{(1)} = X'$ (the set of limit points of $X$), $X^{(alpha+1)} = (X^{(alpha)})'$, for successor ordinals $alpha +1$, and $X^{(beta)} =bigcap {X^{(alpha)} alpha < beta}$ for limit ordinals $beta$. This process stops on some countable ordinal $gamma$ such that $X^{(gamma)} = X^{(gamma+1)}$. Then $X$ is uniquely determined by this $gamma$ and if we end with the empty set, how many points there were on the stage before. (This is finite for a compact $X$ but can be countable discrete as well), or whether it ended in a countable dense in itself space (which is $mathbb{Q}$ essentially.)
So in that sense limit points are involved and important but we have to iterate the process. In fact, for this very reason ordinals were invented by Cantor. He came from analysis trying to analyse the possible countable sets of discontinuity of certain sums of goniometric series, and he needed ordinals to formulate his results, which led him to develop set theory for the first time.
answered May 3 '17 at 21:54
Henno BrandsmaHenno Brandsma
114k349125
edited Jan 29 at 22:11
answered May 3 '17 at 21:54
Henno BrandsmaHenno Brandsma
114k349125
answered May 3 '17 at 21:54
Henno BrandsmaHenno Brandsma
114k349125
answered May 3 '17 at 21:54
Henno BrandsmaHenno Brandsma
114k349125
114k349125
$begingroup$
Thanks for your answer! Good to know that it motivated Cantor to invent ordinals.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:10
add a comment |
$begingroup$
Thanks for your answer! Good to know that it motivated Cantor to invent ordinals.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:10
$begingroup$
Thanks for your answer! Good to know that it motivated Cantor to invent ordinals.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:10
$begingroup$
Thanks for your answer! Good to know that it motivated Cantor to invent ordinals.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:10
add a comment |
$begingroup$
It does not suffice to record which points are limit points. For example, the subspace given by
$${ -n, n in mathbb{N} } cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
and the subspace given by points of the form
$$left{ -frac{1}{n}, n in mathbb{N} right} cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
both have the same limit point structure (one limit point "in the middle") but are not homeomorphic.
answered May 3 '17 at 20:28
Qiaochu YuanQiaochu Yuan
281k32595940
$endgroup$
$begingroup$
Thanks so much for your answer. So, limit points are irrelevant. But how can we classify?
$endgroup$
– Math.StackExchange
May 3 '17 at 20:31
$begingroup$
Well, they're not irrelevant, they're just not enough. At each limit point you can in addition keep track of whether it is being approached from the left, from the right, or both. I don't know if this is enough or not.
$endgroup$
– Qiaochu Yuan
May 3 '17 at 20:54
$begingroup$
Sorry for my sloppy use of the word. Thanks for input regarding the direction of the limit, which is really important. If this question is truly non-trivial and will not receive a decisive answer, I will migrate it to MO.
$endgroup$
– Math.StackExchange
May 3 '17 at 21:04
add a comment |
$begingroup$
It does not suffice to record which points are limit points. For example, the subspace given by
$${ -n, n in mathbb{N} } cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
and the subspace given by points of the form
$$left{ -frac{1}{n}, n in mathbb{N} right} cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
both have the same limit point structure (one limit point "in the middle") but are not homeomorphic.
answered May 3 '17 at 20:28
Qiaochu YuanQiaochu Yuan
281k32595940
$endgroup$
$begingroup$
Thanks so much for your answer. So, limit points are irrelevant. But how can we classify?
$endgroup$
– Math.StackExchange
May 3 '17 at 20:31
$begingroup$
Well, they're not irrelevant, they're just not enough. At each limit point you can in addition keep track of whether it is being approached from the left, from the right, or both. I don't know if this is enough or not.
$endgroup$
– Qiaochu Yuan
May 3 '17 at 20:54
$begingroup$
Sorry for my sloppy use of the word. Thanks for input regarding the direction of the limit, which is really important. If this question is truly non-trivial and will not receive a decisive answer, I will migrate it to MO.
$endgroup$
– Math.StackExchange
May 3 '17 at 21:04
add a comment |
$begingroup$
It does not suffice to record which points are limit points. For example, the subspace given by
$${ -n, n in mathbb{N} } cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
and the subspace given by points of the form
$$left{ -frac{1}{n}, n in mathbb{N} right} cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
both have the same limit point structure (one limit point "in the middle") but are not homeomorphic.
answered May 3 '17 at 20:28
Qiaochu YuanQiaochu Yuan
281k32595940
$endgroup$
It does not suffice to record which points are limit points. For example, the subspace given by
$${ -n, n in mathbb{N} } cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
and the subspace given by points of the form
$$left{ -frac{1}{n}, n in mathbb{N} right} cup { 0 } cup left{ frac{1}{n}, n in mathbb{N} right}$$
both have the same limit point structure (one limit point "in the middle") but are not homeomorphic.
answered May 3 '17 at 20:28
Qiaochu YuanQiaochu Yuan
281k32595940
answered May 3 '17 at 20:28
Qiaochu YuanQiaochu Yuan
281k32595940
answered May 3 '17 at 20:28
Qiaochu YuanQiaochu Yuan
281k32595940
answered May 3 '17 at 20:28
Qiaochu YuanQiaochu Yuan
281k32595940
281k32595940
$begingroup$
Thanks so much for your answer. So, limit points are irrelevant. But how can we classify?
$endgroup$
– Math.StackExchange
May 3 '17 at 20:31
$begingroup$
Well, they're not irrelevant, they're just not enough. At each limit point you can in addition keep track of whether it is being approached from the left, from the right, or both. I don't know if this is enough or not.
$endgroup$
– Qiaochu Yuan
May 3 '17 at 20:54
$begingroup$
Sorry for my sloppy use of the word. Thanks for input regarding the direction of the limit, which is really important. If this question is truly non-trivial and will not receive a decisive answer, I will migrate it to MO.
$endgroup$
– Math.StackExchange
May 3 '17 at 21:04
add a comment |
$begingroup$
Thanks so much for your answer. So, limit points are irrelevant. But how can we classify?
$endgroup$
– Math.StackExchange
May 3 '17 at 20:31
$begingroup$
Well, they're not irrelevant, they're just not enough. At each limit point you can in addition keep track of whether it is being approached from the left, from the right, or both. I don't know if this is enough or not.
$endgroup$
– Qiaochu Yuan
May 3 '17 at 20:54
$begingroup$
Sorry for my sloppy use of the word. Thanks for input regarding the direction of the limit, which is really important. If this question is truly non-trivial and will not receive a decisive answer, I will migrate it to MO.
$endgroup$
– Math.StackExchange
May 3 '17 at 21:04
$begingroup$
Thanks so much for your answer. So, limit points are irrelevant. But how can we classify?
$endgroup$
– Math.StackExchange
May 3 '17 at 20:31
$begingroup$
Thanks so much for your answer. So, limit points are irrelevant. But how can we classify?
$endgroup$
– Math.StackExchange
May 3 '17 at 20:31
$begingroup$
Well, they're not irrelevant, they're just not enough. At each limit point you can in addition keep track of whether it is being approached from the left, from the right, or both. I don't know if this is enough or not.
$endgroup$
– Qiaochu Yuan
May 3 '17 at 20:54
$begingroup$
Well, they're not irrelevant, they're just not enough. At each limit point you can in addition keep track of whether it is being approached from the left, from the right, or both. I don't know if this is enough or not.
$endgroup$
– Qiaochu Yuan
May 3 '17 at 20:54
$begingroup$
Sorry for my sloppy use of the word. Thanks for input regarding the direction of the limit, which is really important. If this question is truly non-trivial and will not receive a decisive answer, I will migrate it to MO.
$endgroup$
– Math.StackExchange
May 3 '17 at 21:04
$begingroup$
Sorry for my sloppy use of the word. Thanks for input regarding the direction of the limit, which is really important. If this question is truly non-trivial and will not receive a decisive answer, I will migrate it to MO.
$endgroup$
– Math.StackExchange
May 3 '17 at 21:04
add a comment |
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$begingroup$
See math.stackexchange.com/a/416672 for a very useful answer for the compact case.
$endgroup$
– Henno Brandsma
May 3 '17 at 22:01
$begingroup$
Very informative! Thanks so much.
$endgroup$
– Math.StackExchange
May 3 '17 at 22:12