Is there any well-ordered uncountable set of real numbers under the original ordering?
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I know that the usual ordering of $mathbb R$ is not a well-ordering but is there an uncountable $Ssubset mathbb R$ such that S is well-ordered by $<_mathbb R$?
Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set.
set-theory order-theory
asked Apr 22 '16 at 17:29
akkarin
668311
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show 6 more comments
up vote
2
down vote
favorite
I know that the usual ordering of $mathbb R$ is not a well-ordering but is there an uncountable $Ssubset mathbb R$ such that S is well-ordered by $<_mathbb R$?
Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set.
set-theory order-theory
asked Apr 22 '16 at 17:29
akkarin
668311
2
FYI, a frequently asked question on the Indiana University (Bloomington) Ph.D. real variables qualifying exams back in the late 1960s to the mid 1970s was to prove that any well-ordered subset of the reals is countable. (Sometimes the question was whether an uncountable well-ordered subset of the reals exists, and you were to prove your answer.)
– Dave L. Renfro
Apr 22 '16 at 17:44
2
@Matt Dyer: akkarin is asking whether there exists such a set on the real line, not whether there exists such a set somewhere.
– Dave L. Renfro
Apr 22 '16 at 17:45
1
Hint: You can get a pairwise disjoint collection of open intervals by picking one open interval lying between each of the points and the next point of the well ordered set.
– Dave L. Renfro
Apr 22 '16 at 17:51
1
@Matt Dyer: True, but the ordering you get may not have the numbers being in the same order as they appear on the number line, which is what akkarin wants.
– Dave L. Renfro
Apr 22 '16 at 17:52
2
Why the vote to close? It's possible this is a duplicate (I vaguely remember the same question being asked earlier), but it's certainly not off-topic.
– Noah Schweber
Apr 22 '16 at 18:52
|
show 6 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I know that the usual ordering of $mathbb R$ is not a well-ordering but is there an uncountable $Ssubset mathbb R$ such that S is well-ordered by $<_mathbb R$?
Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set.
set-theory order-theory
asked Apr 22 '16 at 17:29
akkarin
668311
I know that the usual ordering of $mathbb R$ is not a well-ordering but is there an uncountable $Ssubset mathbb R$ such that S is well-ordered by $<_mathbb R$?
Intuitively I'd say there is no such set but intuitively I'd also say there is no well-ordered uncountable set at all, which is obviously wrong. I still struggle to grasp the idea of an uncountable, well-ordered set.
set-theory order-theory
set-theory order-theory
asked Apr 22 '16 at 17:29
akkarin
668311
asked Apr 22 '16 at 17:29
akkarin
668311
asked Apr 22 '16 at 17:29
akkarin
668311
asked Apr 22 '16 at 17:29
akkarin
668311
asked Apr 22 '16 at 17:29
akkarin
668311
668311
2
FYI, a frequently asked question on the Indiana University (Bloomington) Ph.D. real variables qualifying exams back in the late 1960s to the mid 1970s was to prove that any well-ordered subset of the reals is countable. (Sometimes the question was whether an uncountable well-ordered subset of the reals exists, and you were to prove your answer.)
– Dave L. Renfro
Apr 22 '16 at 17:44
2
@Matt Dyer: akkarin is asking whether there exists such a set on the real line, not whether there exists such a set somewhere.
– Dave L. Renfro
Apr 22 '16 at 17:45
1
Hint: You can get a pairwise disjoint collection of open intervals by picking one open interval lying between each of the points and the next point of the well ordered set.
– Dave L. Renfro
Apr 22 '16 at 17:51
1
@Matt Dyer: True, but the ordering you get may not have the numbers being in the same order as they appear on the number line, which is what akkarin wants.
– Dave L. Renfro
Apr 22 '16 at 17:52
2
Why the vote to close? It's possible this is a duplicate (I vaguely remember the same question being asked earlier), but it's certainly not off-topic.
– Noah Schweber
Apr 22 '16 at 18:52
|
show 6 more comments
2
FYI, a frequently asked question on the Indiana University (Bloomington) Ph.D. real variables qualifying exams back in the late 1960s to the mid 1970s was to prove that any well-ordered subset of the reals is countable. (Sometimes the question was whether an uncountable well-ordered subset of the reals exists, and you were to prove your answer.)
– Dave L. Renfro
Apr 22 '16 at 17:44
2
@Matt Dyer: akkarin is asking whether there exists such a set on the real line, not whether there exists such a set somewhere.
– Dave L. Renfro
Apr 22 '16 at 17:45
1
Hint: You can get a pairwise disjoint collection of open intervals by picking one open interval lying between each of the points and the next point of the well ordered set.
– Dave L. Renfro
Apr 22 '16 at 17:51
1
@Matt Dyer: True, but the ordering you get may not have the numbers being in the same order as they appear on the number line, which is what akkarin wants.
– Dave L. Renfro
Apr 22 '16 at 17:52
2
Why the vote to close? It's possible this is a duplicate (I vaguely remember the same question being asked earlier), but it's certainly not off-topic.
– Noah Schweber
Apr 22 '16 at 18:52
2
2
FYI, a frequently asked question on the Indiana University (Bloomington) Ph.D. real variables qualifying exams back in the late 1960s to the mid 1970s was to prove that any well-ordered subset of the reals is countable. (Sometimes the question was whether an uncountable well-ordered subset of the reals exists, and you were to prove your answer.)
– Dave L. Renfro
Apr 22 '16 at 17:44
FYI, a frequently asked question on the Indiana University (Bloomington) Ph.D. real variables qualifying exams back in the late 1960s to the mid 1970s was to prove that any well-ordered subset of the reals is countable. (Sometimes the question was whether an uncountable well-ordered subset of the reals exists, and you were to prove your answer.)
– Dave L. Renfro
Apr 22 '16 at 17:44
2
2
@Matt Dyer: akkarin is asking whether there exists such a set on the real line, not whether there exists such a set somewhere.
– Dave L. Renfro
Apr 22 '16 at 17:45
@Matt Dyer: akkarin is asking whether there exists such a set on the real line, not whether there exists such a set somewhere.
– Dave L. Renfro
Apr 22 '16 at 17:45
1
1
Hint: You can get a pairwise disjoint collection of open intervals by picking one open interval lying between each of the points and the next point of the well ordered set.
– Dave L. Renfro
Apr 22 '16 at 17:51
Hint: You can get a pairwise disjoint collection of open intervals by picking one open interval lying between each of the points and the next point of the well ordered set.
– Dave L. Renfro
Apr 22 '16 at 17:51
1
1
@Matt Dyer: True, but the ordering you get may not have the numbers being in the same order as they appear on the number line, which is what akkarin wants.
– Dave L. Renfro
Apr 22 '16 at 17:52
@Matt Dyer: True, but the ordering you get may not have the numbers being in the same order as they appear on the number line, which is what akkarin wants.
– Dave L. Renfro
Apr 22 '16 at 17:52
2
2
Why the vote to close? It's possible this is a duplicate (I vaguely remember the same question being asked earlier), but it's certainly not off-topic.
– Noah Schweber
Apr 22 '16 at 18:52
Why the vote to close? It's possible this is a duplicate (I vaguely remember the same question being asked earlier), but it's certainly not off-topic.
– Noah Schweber
Apr 22 '16 at 18:52
|
show 6 more comments
1 Answer
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There can't be. If $Ssubseteq Bbb R$ is well-ordered by the usual ordering, for every element $s_{alpha}in S$ that has an immediate successor $s_{alpha+1}in S$ (every element of $S$ except the greatest element if there is one), the set of rationals $Q_{alpha}$ between the element and its successor is nonempty: $(s_{alpha}, s_{alpha+1}) cap Bbb Q ne emptyset$, and the $Q_{alpha}$ are disjoint. If $S$ were uncountable, then $bigcup_{alpha < length(S)} Q_{alpha}$ would also be uncountable — impossible, as it's a subset of $Bbb Q$.
answered Apr 22 '16 at 18:39
BrianO
14.3k1822
add a comment |
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1 Answer
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There can't be. If $Ssubseteq Bbb R$ is well-ordered by the usual ordering, for every element $s_{alpha}in S$ that has an immediate successor $s_{alpha+1}in S$ (every element of $S$ except the greatest element if there is one), the set of rationals $Q_{alpha}$ between the element and its successor is nonempty: $(s_{alpha}, s_{alpha+1}) cap Bbb Q ne emptyset$, and the $Q_{alpha}$ are disjoint. If $S$ were uncountable, then $bigcup_{alpha < length(S)} Q_{alpha}$ would also be uncountable — impossible, as it's a subset of $Bbb Q$.
answered Apr 22 '16 at 18:39
BrianO
14.3k1822
add a comment |
up vote
6
down vote
accepted
There can't be. If $Ssubseteq Bbb R$ is well-ordered by the usual ordering, for every element $s_{alpha}in S$ that has an immediate successor $s_{alpha+1}in S$ (every element of $S$ except the greatest element if there is one), the set of rationals $Q_{alpha}$ between the element and its successor is nonempty: $(s_{alpha}, s_{alpha+1}) cap Bbb Q ne emptyset$, and the $Q_{alpha}$ are disjoint. If $S$ were uncountable, then $bigcup_{alpha < length(S)} Q_{alpha}$ would also be uncountable — impossible, as it's a subset of $Bbb Q$.
answered Apr 22 '16 at 18:39
BrianO
14.3k1822
add a comment |
up vote
6
down vote
accepted
up vote
6
down vote
accepted
There can't be. If $Ssubseteq Bbb R$ is well-ordered by the usual ordering, for every element $s_{alpha}in S$ that has an immediate successor $s_{alpha+1}in S$ (every element of $S$ except the greatest element if there is one), the set of rationals $Q_{alpha}$ between the element and its successor is nonempty: $(s_{alpha}, s_{alpha+1}) cap Bbb Q ne emptyset$, and the $Q_{alpha}$ are disjoint. If $S$ were uncountable, then $bigcup_{alpha < length(S)} Q_{alpha}$ would also be uncountable — impossible, as it's a subset of $Bbb Q$.
answered Apr 22 '16 at 18:39
BrianO
14.3k1822
There can't be. If $Ssubseteq Bbb R$ is well-ordered by the usual ordering, for every element $s_{alpha}in S$ that has an immediate successor $s_{alpha+1}in S$ (every element of $S$ except the greatest element if there is one), the set of rationals $Q_{alpha}$ between the element and its successor is nonempty: $(s_{alpha}, s_{alpha+1}) cap Bbb Q ne emptyset$, and the $Q_{alpha}$ are disjoint. If $S$ were uncountable, then $bigcup_{alpha < length(S)} Q_{alpha}$ would also be uncountable — impossible, as it's a subset of $Bbb Q$.
answered Apr 22 '16 at 18:39
BrianO
14.3k1822
answered Apr 22 '16 at 18:39
BrianO
14.3k1822
answered Apr 22 '16 at 18:39
BrianO
14.3k1822
answered Apr 22 '16 at 18:39
BrianO
14.3k1822
14.3k1822
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FYI, a frequently asked question on the Indiana University (Bloomington) Ph.D. real variables qualifying exams back in the late 1960s to the mid 1970s was to prove that any well-ordered subset of the reals is countable. (Sometimes the question was whether an uncountable well-ordered subset of the reals exists, and you were to prove your answer.)
– Dave L. Renfro
Apr 22 '16 at 17:44
2
@Matt Dyer: akkarin is asking whether there exists such a set on the real line, not whether there exists such a set somewhere.
– Dave L. Renfro
Apr 22 '16 at 17:45
1
Hint: You can get a pairwise disjoint collection of open intervals by picking one open interval lying between each of the points and the next point of the well ordered set.
– Dave L. Renfro
Apr 22 '16 at 17:51
1
@Matt Dyer: True, but the ordering you get may not have the numbers being in the same order as they appear on the number line, which is what akkarin wants.
– Dave L. Renfro
Apr 22 '16 at 17:52
2
Why the vote to close? It's possible this is a duplicate (I vaguely remember the same question being asked earlier), but it's certainly not off-topic.
– Noah Schweber
Apr 22 '16 at 18:52