Mixing
what is “Minimal Uncountable well-ordered set”?
$begingroup$
Can anyone make me understand what is "Minimal Uncountable well-ordered set"?
I know what is Uncountablity and Well ordered set.
Thank You in Advance.
general-topology
asked Apr 20 '18 at 11:54
cmicmi
1,136312
$endgroup$
|
show 1 more comment
$begingroup$
Can anyone make me understand what is "Minimal Uncountable well-ordered set"?
I know what is Uncountablity and Well ordered set.
Thank You in Advance.
general-topology
asked Apr 20 '18 at 11:54
cmicmi
1,136312
$endgroup$
2
$begingroup$
An ordered set order-isomorphic to the first uncountable ordinal?
$endgroup$
– Lord Shark the Unknown
Apr 20 '18 at 11:57
$begingroup$
I think context will be needed
$endgroup$
– Jonathan Hebert
Apr 20 '18 at 11:57
$begingroup$
Can you please explain in simple terms?@LordSharktheUnknown
$endgroup$
– cmi
Apr 20 '18 at 12:09
4
$begingroup$
There are four concepts involved in your question. Do you know what a set is? Do you know what it means for it to be well-ordered? Do you know what it means for a set to be uncountable? Do you know what minimal (in this context) means? If you clarify where your difficulty is, there is a greater chance you'll get a useful answer.
$endgroup$
– Ittay Weiss
Apr 20 '18 at 12:17
1
$begingroup$
I know the meaning of every terms But I found it difficult to understand when they are together.@IttayWeiss
$endgroup$
– cmi
Apr 20 '18 at 13:33
|
show 1 more comment
$begingroup$
Can anyone make me understand what is "Minimal Uncountable well-ordered set"?
I know what is Uncountablity and Well ordered set.
Thank You in Advance.
general-topology
asked Apr 20 '18 at 11:54
cmicmi
1,136312
$endgroup$
Can anyone make me understand what is "Minimal Uncountable well-ordered set"?
I know what is Uncountablity and Well ordered set.
Thank You in Advance.
general-topology
general-topology
asked Apr 20 '18 at 11:54
cmicmi
1,136312
asked Apr 20 '18 at 11:54
cmicmi
1,136312
asked Apr 20 '18 at 11:54
cmicmi
1,136312
asked Apr 20 '18 at 11:54
cmicmi
1,136312
asked Apr 20 '18 at 11:54
cmicmi
1,136312
1,136312
2
$begingroup$
An ordered set order-isomorphic to the first uncountable ordinal?
$endgroup$
– Lord Shark the Unknown
Apr 20 '18 at 11:57
$begingroup$
I think context will be needed
$endgroup$
– Jonathan Hebert
Apr 20 '18 at 11:57
$begingroup$
Can you please explain in simple terms?@LordSharktheUnknown
$endgroup$
– cmi
Apr 20 '18 at 12:09
4
$begingroup$
There are four concepts involved in your question. Do you know what a set is? Do you know what it means for it to be well-ordered? Do you know what it means for a set to be uncountable? Do you know what minimal (in this context) means? If you clarify where your difficulty is, there is a greater chance you'll get a useful answer.
$endgroup$
– Ittay Weiss
Apr 20 '18 at 12:17
1
$begingroup$
I know the meaning of every terms But I found it difficult to understand when they are together.@IttayWeiss
$endgroup$
– cmi
Apr 20 '18 at 13:33
|
show 1 more comment
2
$begingroup$
An ordered set order-isomorphic to the first uncountable ordinal?
$endgroup$
– Lord Shark the Unknown
Apr 20 '18 at 11:57
$begingroup$
I think context will be needed
$endgroup$
– Jonathan Hebert
Apr 20 '18 at 11:57
$begingroup$
Can you please explain in simple terms?@LordSharktheUnknown
$endgroup$
– cmi
Apr 20 '18 at 12:09
4
$begingroup$
There are four concepts involved in your question. Do you know what a set is? Do you know what it means for it to be well-ordered? Do you know what it means for a set to be uncountable? Do you know what minimal (in this context) means? If you clarify where your difficulty is, there is a greater chance you'll get a useful answer.
$endgroup$
– Ittay Weiss
Apr 20 '18 at 12:17
1
$begingroup$
I know the meaning of every terms But I found it difficult to understand when they are together.@IttayWeiss
$endgroup$
– cmi
Apr 20 '18 at 13:33
2
2
$begingroup$
An ordered set order-isomorphic to the first uncountable ordinal?
$endgroup$
– Lord Shark the Unknown
Apr 20 '18 at 11:57
$begingroup$
An ordered set order-isomorphic to the first uncountable ordinal?
$endgroup$
– Lord Shark the Unknown
Apr 20 '18 at 11:57
$begingroup$
I think context will be needed
$endgroup$
– Jonathan Hebert
Apr 20 '18 at 11:57
$begingroup$
I think context will be needed
$endgroup$
– Jonathan Hebert
Apr 20 '18 at 11:57
$begingroup$
Can you please explain in simple terms?@LordSharktheUnknown
$endgroup$
– cmi
Apr 20 '18 at 12:09
$begingroup$
Can you please explain in simple terms?@LordSharktheUnknown
$endgroup$
– cmi
Apr 20 '18 at 12:09
4
4
$begingroup$
There are four concepts involved in your question. Do you know what a set is? Do you know what it means for it to be well-ordered? Do you know what it means for a set to be uncountable? Do you know what minimal (in this context) means? If you clarify where your difficulty is, there is a greater chance you'll get a useful answer.
$endgroup$
– Ittay Weiss
Apr 20 '18 at 12:17
$begingroup$
There are four concepts involved in your question. Do you know what a set is? Do you know what it means for it to be well-ordered? Do you know what it means for a set to be uncountable? Do you know what minimal (in this context) means? If you clarify where your difficulty is, there is a greater chance you'll get a useful answer.
$endgroup$
– Ittay Weiss
Apr 20 '18 at 12:17
1
1
$begingroup$
I know the meaning of every terms But I found it difficult to understand when they are together.@IttayWeiss
$endgroup$
– cmi
Apr 20 '18 at 13:33
$begingroup$
I know the meaning of every terms But I found it difficult to understand when they are together.@IttayWeiss
$endgroup$
– cmi
Apr 20 '18 at 13:33
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
Perhaps what you might be missing is a theorem of set theory which says (roughly speaking) that any set of well ordered sets is well ordered.
To be precise, suppose that ${cal A} ={A_i}_{i in I}$ is a set, each of whose elements $A_i$ is a well ordered set, and let's suppose that no two elements of ${cal A}$ are order isomorphic (i.e. for all $i ne j in I$ there does not exist an order preserving bijection between $A_i$ and $A_j$). Let's define a relation on ${cal A}$, where $A_i < A_j$ means that there exists an order preserving injection $f : A_i to A_j$.
Theorem: This relation is a well-ordering on ${cal A}$. Therefore, ${cal A}$ has a minimal element.
So, to say that a well-ordered set $A$ is a "minimimal uncountable well ordered set" simply means that for any set ${cal A}$ of uncountable well-ordered sets, no two of which are order isomorphic, if $A in {cal A}$ then $A$ is the minimal element of ${cal A}$.
There are a few more theorems to ponder which help to iron out the understanding of this concept:
Theorem: Suppose $A$ is an uncountable well ordered set, and suppose that for each $b in A$ the set ${a in A mid a < b}$ is countable. Then $A$ is a minimal uncountable well ordered set.
Theorem: Any two minimal uncountable well ordered sets are order isomorphic. (So it is fair to speak about the minimal uncountable well ordered set, because it is unique up to order isomorphism)
For a more concrete example of this concept, the set of natural numbers $mathbb{N}$ is the "minimal infinite well ordered set": for any set ${cal A}$ of well ordered infinite sets (no two of which are order isomorphic), if $mathbb{N} in {cal A}$ then $mathbb{N}$ is the minimal element of ${cal A}$. The other theorems mentioned have analogues here:
Theorem: If $A$ is an infinite well ordered set, and if for each $b in A$ the set ${a in A mid a < b}$ is finite, then $A$ is order isomorphic to $mathbb{N}$.
Theorem: Any minimal infinite well ordered set is order isomorphic to $mathbb{N}$.
answered Apr 20 '18 at 14:27
Lee MosherLee Mosher
50.6k33787
$endgroup$
$begingroup$
So $mathbb R$ is not well ordered minimal uncountable set for sure.. As we remove an element , It still remains uncountable...So can you give me a concrete example of $S_{Omega}$?@Lee Mosher
$endgroup$
– cmi
Dec 27 '18 at 16:27
$begingroup$
@cmi : Where is there an "$S_Omega$" anywhere on this page?
$endgroup$
– Eric Towers
Dec 28 '18 at 8:41
$begingroup$
If you have something else to ask that is related to this question and answer, it is best that you ask it in a new and fully formulated question, instead of asking it in a comment with new notation, which very few people will stumble across, and fewer still will understand.
$endgroup$
– Lee Mosher
Dec 28 '18 at 16:08
add a comment |
$begingroup$
@cmi ℝ in the usual ordering is not well ordered. And if equipped with proper ordering, ℝ can be isomorphic to the "Minimal Uncountable well-ordered set" (assuming the continuum hypothesis), because under the continuum hypothesis, ℝ has the same cardinality as the "Minimal Uncountable well-ordered set", and therefore a bijection between them exists. We can then just define the ordering on ℝ based on the bijection and the ordering of the "Minimal Uncountable well-ordered set".
So the reason of ℝ (with the usual ordering) being "not well ordered minimal uncountable set" is not the reason you gave (can remove element), but the its ordering.
answered Jan 23 at 18:35
ALifeALife
12
$endgroup$
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
$begingroup$
Perhaps what you might be missing is a theorem of set theory which says (roughly speaking) that any set of well ordered sets is well ordered.
To be precise, suppose that ${cal A} ={A_i}_{i in I}$ is a set, each of whose elements $A_i$ is a well ordered set, and let's suppose that no two elements of ${cal A}$ are order isomorphic (i.e. for all $i ne j in I$ there does not exist an order preserving bijection between $A_i$ and $A_j$). Let's define a relation on ${cal A}$, where $A_i < A_j$ means that there exists an order preserving injection $f : A_i to A_j$.
Theorem: This relation is a well-ordering on ${cal A}$. Therefore, ${cal A}$ has a minimal element.
So, to say that a well-ordered set $A$ is a "minimimal uncountable well ordered set" simply means that for any set ${cal A}$ of uncountable well-ordered sets, no two of which are order isomorphic, if $A in {cal A}$ then $A$ is the minimal element of ${cal A}$.
There are a few more theorems to ponder which help to iron out the understanding of this concept:
Theorem: Suppose $A$ is an uncountable well ordered set, and suppose that for each $b in A$ the set ${a in A mid a < b}$ is countable. Then $A$ is a minimal uncountable well ordered set.
Theorem: Any two minimal uncountable well ordered sets are order isomorphic. (So it is fair to speak about the minimal uncountable well ordered set, because it is unique up to order isomorphism)
For a more concrete example of this concept, the set of natural numbers $mathbb{N}$ is the "minimal infinite well ordered set": for any set ${cal A}$ of well ordered infinite sets (no two of which are order isomorphic), if $mathbb{N} in {cal A}$ then $mathbb{N}$ is the minimal element of ${cal A}$. The other theorems mentioned have analogues here:
Theorem: If $A$ is an infinite well ordered set, and if for each $b in A$ the set ${a in A mid a < b}$ is finite, then $A$ is order isomorphic to $mathbb{N}$.
Theorem: Any minimal infinite well ordered set is order isomorphic to $mathbb{N}$.
answered Apr 20 '18 at 14:27
Lee MosherLee Mosher
50.6k33787
$endgroup$
$begingroup$
So $mathbb R$ is not well ordered minimal uncountable set for sure.. As we remove an element , It still remains uncountable...So can you give me a concrete example of $S_{Omega}$?@Lee Mosher
$endgroup$
– cmi
Dec 27 '18 at 16:27
$begingroup$
@cmi : Where is there an "$S_Omega$" anywhere on this page?
$endgroup$
– Eric Towers
Dec 28 '18 at 8:41
$begingroup$
If you have something else to ask that is related to this question and answer, it is best that you ask it in a new and fully formulated question, instead of asking it in a comment with new notation, which very few people will stumble across, and fewer still will understand.
$endgroup$
– Lee Mosher
Dec 28 '18 at 16:08
add a comment |
$begingroup$
Perhaps what you might be missing is a theorem of set theory which says (roughly speaking) that any set of well ordered sets is well ordered.
To be precise, suppose that ${cal A} ={A_i}_{i in I}$ is a set, each of whose elements $A_i$ is a well ordered set, and let's suppose that no two elements of ${cal A}$ are order isomorphic (i.e. for all $i ne j in I$ there does not exist an order preserving bijection between $A_i$ and $A_j$). Let's define a relation on ${cal A}$, where $A_i < A_j$ means that there exists an order preserving injection $f : A_i to A_j$.
Theorem: This relation is a well-ordering on ${cal A}$. Therefore, ${cal A}$ has a minimal element.
So, to say that a well-ordered set $A$ is a "minimimal uncountable well ordered set" simply means that for any set ${cal A}$ of uncountable well-ordered sets, no two of which are order isomorphic, if $A in {cal A}$ then $A$ is the minimal element of ${cal A}$.
There are a few more theorems to ponder which help to iron out the understanding of this concept:
Theorem: Suppose $A$ is an uncountable well ordered set, and suppose that for each $b in A$ the set ${a in A mid a < b}$ is countable. Then $A$ is a minimal uncountable well ordered set.
Theorem: Any two minimal uncountable well ordered sets are order isomorphic. (So it is fair to speak about the minimal uncountable well ordered set, because it is unique up to order isomorphism)
For a more concrete example of this concept, the set of natural numbers $mathbb{N}$ is the "minimal infinite well ordered set": for any set ${cal A}$ of well ordered infinite sets (no two of which are order isomorphic), if $mathbb{N} in {cal A}$ then $mathbb{N}$ is the minimal element of ${cal A}$. The other theorems mentioned have analogues here:
Theorem: If $A$ is an infinite well ordered set, and if for each $b in A$ the set ${a in A mid a < b}$ is finite, then $A$ is order isomorphic to $mathbb{N}$.
Theorem: Any minimal infinite well ordered set is order isomorphic to $mathbb{N}$.
answered Apr 20 '18 at 14:27
Lee MosherLee Mosher
50.6k33787
$endgroup$
$begingroup$
So $mathbb R$ is not well ordered minimal uncountable set for sure.. As we remove an element , It still remains uncountable...So can you give me a concrete example of $S_{Omega}$?@Lee Mosher
$endgroup$
– cmi
Dec 27 '18 at 16:27
$begingroup$
@cmi : Where is there an "$S_Omega$" anywhere on this page?
$endgroup$
– Eric Towers
Dec 28 '18 at 8:41
$begingroup$
If you have something else to ask that is related to this question and answer, it is best that you ask it in a new and fully formulated question, instead of asking it in a comment with new notation, which very few people will stumble across, and fewer still will understand.
$endgroup$
– Lee Mosher
Dec 28 '18 at 16:08
add a comment |
$begingroup$
Perhaps what you might be missing is a theorem of set theory which says (roughly speaking) that any set of well ordered sets is well ordered.
To be precise, suppose that ${cal A} ={A_i}_{i in I}$ is a set, each of whose elements $A_i$ is a well ordered set, and let's suppose that no two elements of ${cal A}$ are order isomorphic (i.e. for all $i ne j in I$ there does not exist an order preserving bijection between $A_i$ and $A_j$). Let's define a relation on ${cal A}$, where $A_i < A_j$ means that there exists an order preserving injection $f : A_i to A_j$.
Theorem: This relation is a well-ordering on ${cal A}$. Therefore, ${cal A}$ has a minimal element.
So, to say that a well-ordered set $A$ is a "minimimal uncountable well ordered set" simply means that for any set ${cal A}$ of uncountable well-ordered sets, no two of which are order isomorphic, if $A in {cal A}$ then $A$ is the minimal element of ${cal A}$.
There are a few more theorems to ponder which help to iron out the understanding of this concept:
Theorem: Suppose $A$ is an uncountable well ordered set, and suppose that for each $b in A$ the set ${a in A mid a < b}$ is countable. Then $A$ is a minimal uncountable well ordered set.
Theorem: Any two minimal uncountable well ordered sets are order isomorphic. (So it is fair to speak about the minimal uncountable well ordered set, because it is unique up to order isomorphism)
For a more concrete example of this concept, the set of natural numbers $mathbb{N}$ is the "minimal infinite well ordered set": for any set ${cal A}$ of well ordered infinite sets (no two of which are order isomorphic), if $mathbb{N} in {cal A}$ then $mathbb{N}$ is the minimal element of ${cal A}$. The other theorems mentioned have analogues here:
Theorem: If $A$ is an infinite well ordered set, and if for each $b in A$ the set ${a in A mid a < b}$ is finite, then $A$ is order isomorphic to $mathbb{N}$.
Theorem: Any minimal infinite well ordered set is order isomorphic to $mathbb{N}$.
answered Apr 20 '18 at 14:27
Lee MosherLee Mosher
50.6k33787
$endgroup$
Perhaps what you might be missing is a theorem of set theory which says (roughly speaking) that any set of well ordered sets is well ordered.
To be precise, suppose that ${cal A} ={A_i}_{i in I}$ is a set, each of whose elements $A_i$ is a well ordered set, and let's suppose that no two elements of ${cal A}$ are order isomorphic (i.e. for all $i ne j in I$ there does not exist an order preserving bijection between $A_i$ and $A_j$). Let's define a relation on ${cal A}$, where $A_i < A_j$ means that there exists an order preserving injection $f : A_i to A_j$.
Theorem: This relation is a well-ordering on ${cal A}$. Therefore, ${cal A}$ has a minimal element.
So, to say that a well-ordered set $A$ is a "minimimal uncountable well ordered set" simply means that for any set ${cal A}$ of uncountable well-ordered sets, no two of which are order isomorphic, if $A in {cal A}$ then $A$ is the minimal element of ${cal A}$.
There are a few more theorems to ponder which help to iron out the understanding of this concept:
Theorem: Suppose $A$ is an uncountable well ordered set, and suppose that for each $b in A$ the set ${a in A mid a < b}$ is countable. Then $A$ is a minimal uncountable well ordered set.
Theorem: Any two minimal uncountable well ordered sets are order isomorphic. (So it is fair to speak about the minimal uncountable well ordered set, because it is unique up to order isomorphism)
For a more concrete example of this concept, the set of natural numbers $mathbb{N}$ is the "minimal infinite well ordered set": for any set ${cal A}$ of well ordered infinite sets (no two of which are order isomorphic), if $mathbb{N} in {cal A}$ then $mathbb{N}$ is the minimal element of ${cal A}$. The other theorems mentioned have analogues here:
Theorem: If $A$ is an infinite well ordered set, and if for each $b in A$ the set ${a in A mid a < b}$ is finite, then $A$ is order isomorphic to $mathbb{N}$.
Theorem: Any minimal infinite well ordered set is order isomorphic to $mathbb{N}$.
answered Apr 20 '18 at 14:27
Lee MosherLee Mosher
50.6k33787
edited Apr 20 '18 at 18:51
answered Apr 20 '18 at 14:27
Lee MosherLee Mosher
50.6k33787
answered Apr 20 '18 at 14:27
Lee MosherLee Mosher
50.6k33787
answered Apr 20 '18 at 14:27
Lee MosherLee Mosher
50.6k33787
50.6k33787
$begingroup$
So $mathbb R$ is not well ordered minimal uncountable set for sure.. As we remove an element , It still remains uncountable...So can you give me a concrete example of $S_{Omega}$?@Lee Mosher
$endgroup$
– cmi
Dec 27 '18 at 16:27
$begingroup$
@cmi : Where is there an "$S_Omega$" anywhere on this page?
$endgroup$
– Eric Towers
Dec 28 '18 at 8:41
$begingroup$
If you have something else to ask that is related to this question and answer, it is best that you ask it in a new and fully formulated question, instead of asking it in a comment with new notation, which very few people will stumble across, and fewer still will understand.
$endgroup$
– Lee Mosher
Dec 28 '18 at 16:08
add a comment |
$begingroup$
So $mathbb R$ is not well ordered minimal uncountable set for sure.. As we remove an element , It still remains uncountable...So can you give me a concrete example of $S_{Omega}$?@Lee Mosher
$endgroup$
– cmi
Dec 27 '18 at 16:27
$begingroup$
@cmi : Where is there an "$S_Omega$" anywhere on this page?
$endgroup$
– Eric Towers
Dec 28 '18 at 8:41
$begingroup$
If you have something else to ask that is related to this question and answer, it is best that you ask it in a new and fully formulated question, instead of asking it in a comment with new notation, which very few people will stumble across, and fewer still will understand.
$endgroup$
– Lee Mosher
Dec 28 '18 at 16:08
$begingroup$
So $mathbb R$ is not well ordered minimal uncountable set for sure.. As we remove an element , It still remains uncountable...So can you give me a concrete example of $S_{Omega}$?@Lee Mosher
$endgroup$
– cmi
Dec 27 '18 at 16:27
$begingroup$
So $mathbb R$ is not well ordered minimal uncountable set for sure.. As we remove an element , It still remains uncountable...So can you give me a concrete example of $S_{Omega}$?@Lee Mosher
$endgroup$
– cmi
Dec 27 '18 at 16:27
$begingroup$
@cmi : Where is there an "$S_Omega$" anywhere on this page?
$endgroup$
– Eric Towers
Dec 28 '18 at 8:41
$begingroup$
@cmi : Where is there an "$S_Omega$" anywhere on this page?
$endgroup$
– Eric Towers
Dec 28 '18 at 8:41
$begingroup$
If you have something else to ask that is related to this question and answer, it is best that you ask it in a new and fully formulated question, instead of asking it in a comment with new notation, which very few people will stumble across, and fewer still will understand.
$endgroup$
– Lee Mosher
Dec 28 '18 at 16:08
$begingroup$
If you have something else to ask that is related to this question and answer, it is best that you ask it in a new and fully formulated question, instead of asking it in a comment with new notation, which very few people will stumble across, and fewer still will understand.
$endgroup$
– Lee Mosher
Dec 28 '18 at 16:08
add a comment |
$begingroup$
@cmi ℝ in the usual ordering is not well ordered. And if equipped with proper ordering, ℝ can be isomorphic to the "Minimal Uncountable well-ordered set" (assuming the continuum hypothesis), because under the continuum hypothesis, ℝ has the same cardinality as the "Minimal Uncountable well-ordered set", and therefore a bijection between them exists. We can then just define the ordering on ℝ based on the bijection and the ordering of the "Minimal Uncountable well-ordered set".
So the reason of ℝ (with the usual ordering) being "not well ordered minimal uncountable set" is not the reason you gave (can remove element), but the its ordering.
answered Jan 23 at 18:35
ALifeALife
12
$endgroup$
add a comment |
$begingroup$
@cmi ℝ in the usual ordering is not well ordered. And if equipped with proper ordering, ℝ can be isomorphic to the "Minimal Uncountable well-ordered set" (assuming the continuum hypothesis), because under the continuum hypothesis, ℝ has the same cardinality as the "Minimal Uncountable well-ordered set", and therefore a bijection between them exists. We can then just define the ordering on ℝ based on the bijection and the ordering of the "Minimal Uncountable well-ordered set".
So the reason of ℝ (with the usual ordering) being "not well ordered minimal uncountable set" is not the reason you gave (can remove element), but the its ordering.
answered Jan 23 at 18:35
ALifeALife
12
$endgroup$
add a comment |
$begingroup$
@cmi ℝ in the usual ordering is not well ordered. And if equipped with proper ordering, ℝ can be isomorphic to the "Minimal Uncountable well-ordered set" (assuming the continuum hypothesis), because under the continuum hypothesis, ℝ has the same cardinality as the "Minimal Uncountable well-ordered set", and therefore a bijection between them exists. We can then just define the ordering on ℝ based on the bijection and the ordering of the "Minimal Uncountable well-ordered set".
So the reason of ℝ (with the usual ordering) being "not well ordered minimal uncountable set" is not the reason you gave (can remove element), but the its ordering.
answered Jan 23 at 18:35
ALifeALife
12
$endgroup$
@cmi ℝ in the usual ordering is not well ordered. And if equipped with proper ordering, ℝ can be isomorphic to the "Minimal Uncountable well-ordered set" (assuming the continuum hypothesis), because under the continuum hypothesis, ℝ has the same cardinality as the "Minimal Uncountable well-ordered set", and therefore a bijection between them exists. We can then just define the ordering on ℝ based on the bijection and the ordering of the "Minimal Uncountable well-ordered set".
So the reason of ℝ (with the usual ordering) being "not well ordered minimal uncountable set" is not the reason you gave (can remove element), but the its ordering.
answered Jan 23 at 18:35
ALifeALife
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answered Jan 23 at 18:35
ALifeALife
12
answered Jan 23 at 18:35
ALifeALife
12
answered Jan 23 at 18:35
ALifeALife
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An ordered set order-isomorphic to the first uncountable ordinal?
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– Lord Shark the Unknown
Apr 20 '18 at 11:57
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I think context will be needed
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– Jonathan Hebert
Apr 20 '18 at 11:57
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Can you please explain in simple terms?@LordSharktheUnknown
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– cmi
Apr 20 '18 at 12:09
4
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There are four concepts involved in your question. Do you know what a set is? Do you know what it means for it to be well-ordered? Do you know what it means for a set to be uncountable? Do you know what minimal (in this context) means? If you clarify where your difficulty is, there is a greater chance you'll get a useful answer.
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– Ittay Weiss
Apr 20 '18 at 12:17
1
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I know the meaning of every terms But I found it difficult to understand when they are together.@IttayWeiss
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– cmi
Apr 20 '18 at 13:33