An infinite set has always a countably infinite subset?How to prove countably infinite?Infinite set always has a countably infinite subsetHow to complete this proof? Union of a countably infinite set and a finite set is countably infiniteinfinite countably cartesian productIs the Cartesian product of two countably infinite sets also countably infinite?How many infinite subsets does $mathbb N$ have?What is the condition on a infinite subset to have a infinite relative complement?Define a countably infinite subset of $X$$sigma$-fields can not be countably infiniteLet $A$ be an infinite set and B a countable set, show that $vert A cup B vert = vert A vert$
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An infinite set has always a countably infinite subset?
How to prove countably infinite?Infinite set always has a countably infinite subsetHow to complete this proof? Union of a countably infinite set and a finite set is countably infiniteinfinite countably cartesian productIs the Cartesian product of two countably infinite sets also countably infinite?How many infinite subsets does $mathbb N$ have?What is the condition on a infinite subset to have a infinite relative complement?Define a countably infinite subset of $X$$sigma$-fields can not be countably infiniteLet $A$ be an infinite set and B a countable set, show that $vert A cup B vert = vert A vert$
$begingroup$
Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:
Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:
$I$ is countably infinite: That is, to say, it has the same cardinality as $mathbbN$.
$$ vert I vert = vert mathbbN vert $$
Then it must be that $vert I vert leq vert mathbbN vert$ and $vert mathbbN vert leq vert I vert$ i.e. there is a bijection, say $f$, such that $f: mathbbN mapsto I$.This implies $I subseteq mathbbN$ and $mathbbN subseteq I$.
Therefore, $mathbbN$ is a subset of $I$.
$I$ is not countably infinite:
Now, since $I neq emptyset$; we choose and take out some element $i_1$ from $I$.
As $I setminus i_1$ is also not empty($vert I vert = 1$ otherwise), we choose and take out some $i_2$ from $I setminus i_1$.
Similarly, continuing we end up with a countably infinite set: $$ I_c = i_1, i_2, i_3, dots$$ from $I$ such that $vert I_c vert = vert mathbbN vert$. Then it must be that $vert I_c vert leq vert mathbbN vert $ and $vert mathbbN vert leq vert I_c vert $.
Therefore, a countably infinite set is a subset of $I_c$.
All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.
proof-verification elementary-set-theory proof-writing alternative-proof
asked 12 hours ago
HKTHKT
486317
$endgroup$
add a comment |
$begingroup$
Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:
Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:
$I$ is countably infinite: That is, to say, it has the same cardinality as $mathbbN$.
$$ vert I vert = vert mathbbN vert $$
Then it must be that $vert I vert leq vert mathbbN vert$ and $vert mathbbN vert leq vert I vert$ i.e. there is a bijection, say $f$, such that $f: mathbbN mapsto I$.This implies $I subseteq mathbbN$ and $mathbbN subseteq I$.
Therefore, $mathbbN$ is a subset of $I$.
$I$ is not countably infinite:
Now, since $I neq emptyset$; we choose and take out some element $i_1$ from $I$.
As $I setminus i_1$ is also not empty($vert I vert = 1$ otherwise), we choose and take out some $i_2$ from $I setminus i_1$.
Similarly, continuing we end up with a countably infinite set: $$ I_c = i_1, i_2, i_3, dots$$ from $I$ such that $vert I_c vert = vert mathbbN vert$. Then it must be that $vert I_c vert leq vert mathbbN vert $ and $vert mathbbN vert leq vert I_c vert $.
Therefore, a countably infinite set is a subset of $I_c$.
All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.
proof-verification elementary-set-theory proof-writing alternative-proof
asked 12 hours ago
HKTHKT
486317
$endgroup$
$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
12 hours ago
3
$begingroup$
It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
$endgroup$
– 5xum
12 hours ago
1
$begingroup$
Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
$endgroup$
– Ilmari Karonen
11 hours ago
$begingroup$
At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbbN$ really a subset of $mathbbR$? I mean, is $3=*,*,*$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbbN$.
$endgroup$
– Chrystomath
3 hours ago
add a comment |
$begingroup$
Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:
Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:
$I$ is countably infinite: That is, to say, it has the same cardinality as $mathbbN$.
$$ vert I vert = vert mathbbN vert $$
Then it must be that $vert I vert leq vert mathbbN vert$ and $vert mathbbN vert leq vert I vert$ i.e. there is a bijection, say $f$, such that $f: mathbbN mapsto I$.This implies $I subseteq mathbbN$ and $mathbbN subseteq I$.
Therefore, $mathbbN$ is a subset of $I$.
$I$ is not countably infinite:
Now, since $I neq emptyset$; we choose and take out some element $i_1$ from $I$.
As $I setminus i_1$ is also not empty($vert I vert = 1$ otherwise), we choose and take out some $i_2$ from $I setminus i_1$.
Similarly, continuing we end up with a countably infinite set: $$ I_c = i_1, i_2, i_3, dots$$ from $I$ such that $vert I_c vert = vert mathbbN vert$. Then it must be that $vert I_c vert leq vert mathbbN vert $ and $vert mathbbN vert leq vert I_c vert $.
Therefore, a countably infinite set is a subset of $I_c$.
All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.
proof-verification elementary-set-theory proof-writing alternative-proof
asked 12 hours ago
HKTHKT
486317
$endgroup$
Upon studying cardinality, the statement "an infinite set has always a countably infinite subset" seems intuitive to me. Is it really true though? If so, could someone provide a complete, formal proof or even an alternative one? Here is my reasoning:
Let $I$ be an infinite set. Now there are two types of infinite sets so we will divide our proof in two cases:
$I$ is countably infinite: That is, to say, it has the same cardinality as $mathbbN$.
$$ vert I vert = vert mathbbN vert $$
Then it must be that $vert I vert leq vert mathbbN vert$ and $vert mathbbN vert leq vert I vert$ i.e. there is a bijection, say $f$, such that $f: mathbbN mapsto I$.This implies $I subseteq mathbbN$ and $mathbbN subseteq I$.
Therefore, $mathbbN$ is a subset of $I$.
$I$ is not countably infinite:
Now, since $I neq emptyset$; we choose and take out some element $i_1$ from $I$.
As $I setminus i_1$ is also not empty($vert I vert = 1$ otherwise), we choose and take out some $i_2$ from $I setminus i_1$.
Similarly, continuing we end up with a countably infinite set: $$ I_c = i_1, i_2, i_3, dots$$ from $I$ such that $vert I_c vert = vert mathbbN vert$. Then it must be that $vert I_c vert leq vert mathbbN vert $ and $vert mathbbN vert leq vert I_c vert $.
Therefore, a countably infinite set is a subset of $I_c$.
All in all, a countably infinite set is a subset of $I$ where $I$ is an infinite set.
proof-verification elementary-set-theory proof-writing alternative-proof
proof-verification elementary-set-theory proof-writing alternative-proof
asked 12 hours ago
HKTHKT
486317
asked 12 hours ago
HKTHKT
486317
edited 12 hours ago
HKT
asked 12 hours ago
HKTHKT
486317
asked 12 hours ago
HKTHKT
486317
asked 12 hours ago
HKTHKT
486317
486317
$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
12 hours ago
3
$begingroup$
It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
$endgroup$
– 5xum
12 hours ago
1
$begingroup$
Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
$endgroup$
– Ilmari Karonen
11 hours ago
$begingroup$
At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbbN$ really a subset of $mathbbR$? I mean, is $3=*,*,*$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbbN$.
$endgroup$
– Chrystomath
3 hours ago
add a comment |
$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
12 hours ago
3
$begingroup$
It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
$endgroup$
– 5xum
12 hours ago
1
$begingroup$
Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
$endgroup$
– Ilmari Karonen
11 hours ago
$begingroup$
At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbbN$ really a subset of $mathbbR$? I mean, is $3=*,*,*$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbbN$.
$endgroup$
– Chrystomath
3 hours ago
$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
12 hours ago
$begingroup$
Maybe, an infinite set has always a countably infinite subset, would be a better title.
$endgroup$
– dmtri
12 hours ago
3
3
$begingroup$
It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
$endgroup$
– 5xum
12 hours ago
$begingroup$
It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
$endgroup$
– 5xum
12 hours ago
1
1
$begingroup$
Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
$endgroup$
– Ilmari Karonen
11 hours ago
$begingroup$
Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
$endgroup$
– Ilmari Karonen
11 hours ago
$begingroup$
At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbbN$ really a subset of $mathbbR$? I mean, is $3=*,*,*$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbbN$.
$endgroup$
– Chrystomath
3 hours ago
$begingroup$
At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbbN$ really a subset of $mathbbR$? I mean, is $3=*,*,*$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbbN$.
$endgroup$
– Chrystomath
3 hours ago
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.
The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.
Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.
answered 12 hours ago
Dan UznanskiDan Uznanski
7,14821528
$endgroup$
add a comment |
$begingroup$
Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?
No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.
Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.
answered 12 hours ago
Eevee TrainerEevee Trainer
8,67031540
$endgroup$
add a comment |
$begingroup$
In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
Counterexample is set of all irrational numbers.
The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.
answered 12 hours ago
Slepecky MamutSlepecky Mamut
695313
$endgroup$
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.
The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.
Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.
answered 12 hours ago
Dan UznanskiDan Uznanski
7,14821528
$endgroup$
add a comment |
$begingroup$
Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.
The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.
Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.
answered 12 hours ago
Dan UznanskiDan Uznanski
7,14821528
$endgroup$
add a comment |
$begingroup$
Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.
The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.
Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.
answered 12 hours ago
Dan UznanskiDan Uznanski
7,14821528
$endgroup$
Being a subset cares what elements are actually in there: $A subseteq B iff forall x in A, xin B$.
The set of all even numbers, despite being infinite, does not contain $1$, therefore the set of all natural numbers is not a subset of this set.
Similarly, the set of all transcendental numbers, despite being uncountably infinite, does not contain any integers at all, therefore the set of all natural numbers is not a subset of this set.
answered 12 hours ago
Dan UznanskiDan Uznanski
7,14821528
answered 12 hours ago
Dan UznanskiDan Uznanski
7,14821528
answered 12 hours ago
Dan UznanskiDan Uznanski
7,14821528
answered 12 hours ago
Dan UznanskiDan Uznanski
7,14821528
7,14821528
add a comment |
add a comment |
$begingroup$
Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?
No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.
Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.
answered 12 hours ago
Eevee TrainerEevee Trainer
8,67031540
$endgroup$
add a comment |
$begingroup$
Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?
No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.
Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.
answered 12 hours ago
Eevee TrainerEevee Trainer
8,67031540
$endgroup$
add a comment |
$begingroup$
Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?
No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.
Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.
answered 12 hours ago
Eevee TrainerEevee Trainer
8,67031540
$endgroup$
Upon studying cardinality, the statement "the set of natural numbers is a subset of all infinite sets" seems intuitive to me. Is it really true though?
No. Consider the set $Bbb R setminus Bbb N$ (the set of reals that are not natural) or $Bbb R setminus Bbb Q$ (the set of irrationals) for examples. Neither have a single natural number in them, and thus $Bbb N$ is a subset of neither, but both have cardinality of the continuum, which is known to be greater than $aleph_0$ owing to Cantor's diagonal argument.
Perhaps what you mean to refer to is whether every infinite set has a countable subset? That's certainly true.
answered 12 hours ago
Eevee TrainerEevee Trainer
8,67031540
answered 12 hours ago
Eevee TrainerEevee Trainer
8,67031540
answered 12 hours ago
Eevee TrainerEevee Trainer
8,67031540
answered 12 hours ago
Eevee TrainerEevee Trainer
8,67031540
8,67031540
add a comment |
add a comment |
$begingroup$
In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
Counterexample is set of all irrational numbers.
The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.
answered 12 hours ago
Slepecky MamutSlepecky Mamut
695313
$endgroup$
add a comment |
$begingroup$
In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
Counterexample is set of all irrational numbers.
The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.
answered 12 hours ago
Slepecky MamutSlepecky Mamut
695313
$endgroup$
add a comment |
$begingroup$
In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
Counterexample is set of all irrational numbers.
The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.
answered 12 hours ago
Slepecky MamutSlepecky Mamut
695313
$endgroup$
In fact, it is not a subset of every infinite set, but each infinite set contains its isomorphic copy.
Counterexample is set of all irrational numbers.
The statement "the set of natural numbers is a subset of all infinite sets" implied that no two infinite sets are disjoint.
answered 12 hours ago
Slepecky MamutSlepecky Mamut
695313
answered 12 hours ago
Slepecky MamutSlepecky Mamut
695313
answered 12 hours ago
Slepecky MamutSlepecky Mamut
695313
answered 12 hours ago
Slepecky MamutSlepecky Mamut
695313
695313
add a comment |
add a comment |
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Maybe, an infinite set has always a countably infinite subset, would be a better title.
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– dmtri
12 hours ago
3
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It is absolutely not true that $|I_c|leq |mathbb N|$ implies that $I_csubseteq mathbb N$. Where did you even get that idea?
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– 5xum
12 hours ago
1
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Note to readers: this question was significantly edited recently, changing the title from "Is the set of natural numbers a subset of every infinite set?" to the current "An infinite set has always a countably infinite subset?" That's why some of the answers below may seem to be answering a different question. (See also Are questions necessarily “static”? on meta.)
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– Ilmari Karonen
11 hours ago
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At the risk of appearing controversial, and in defence of OP, the answer is not fundamentally wrong. Is $mathbbN$ really a subset of $mathbbR$? I mean, is $3=*,*,*$ an equivalence class of Cauchy sequences (or whatever it is that you define natural and real numbers)? So we also make the same short-cuts that OP is making. If you pick a countable set of points on the real line, are you allowed to call them $0,1,2,3,ldots$? Since we don't have any extra structure involved like distance etc., I think OP is not wrong in calling the countable subset $mathbbN$.
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– Chrystomath
3 hours ago