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Cardinality of set of sequences from $mathbb{Q}$ that converge to $0$
$begingroup$
I'm looking for cardinality of a set of sequences from $mathbb{Q}$ which are convergent to $0$.
I think the answer is continuum, but I don't know how to prove it.
sequences-and-series cardinals rational-numbers
edited Jan 20 at 15:28
Jakobian
2,650721
asked Jan 20 at 15:25
René AccardoRené Accardo
214
$endgroup$
add a comment |
$begingroup$
I'm looking for cardinality of a set of sequences from $mathbb{Q}$ which are convergent to $0$.
I think the answer is continuum, but I don't know how to prove it.
sequences-and-series cardinals rational-numbers
edited Jan 20 at 15:28
Jakobian
2,650721
asked Jan 20 at 15:25
René AccardoRené Accardo
214
$endgroup$
add a comment |
$begingroup$
I'm looking for cardinality of a set of sequences from $mathbb{Q}$ which are convergent to $0$.
I think the answer is continuum, but I don't know how to prove it.
sequences-and-series cardinals rational-numbers
edited Jan 20 at 15:28
Jakobian
2,650721
asked Jan 20 at 15:25
René AccardoRené Accardo
214
$endgroup$
I'm looking for cardinality of a set of sequences from $mathbb{Q}$ which are convergent to $0$.
I think the answer is continuum, but I don't know how to prove it.
sequences-and-series cardinals rational-numbers
sequences-and-series cardinals rational-numbers
edited Jan 20 at 15:28
Jakobian
2,650721
asked Jan 20 at 15:25
René AccardoRené Accardo
214
edited Jan 20 at 15:28
Jakobian
2,650721
asked Jan 20 at 15:25
René AccardoRené Accardo
214
edited Jan 20 at 15:28
Jakobian
2,650721
edited Jan 20 at 15:28
Jakobian
2,650721
edited Jan 20 at 15:28
Jakobian
2,650721
2,650721
asked Jan 20 at 15:25
René AccardoRené Accardo
214
asked Jan 20 at 15:25
René AccardoRené Accardo
214
asked Jan 20 at 15:25
René AccardoRené Accardo
214
214
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Consider the set $L$ of sequences $(x_n)_{ninmathbb{N}}$, with $x_n in {0, frac{1}{n}}$. Then $L$ has lesser cardinality than the set of all rational sequences that converge to $0$ which we will denote by $M$, which has lesser cardinality than $mathbb{Q}^{mathbb{N}}$.
Then $|L| = |2^{mathbb{N}}| = |mathbb{R}| leq |M| leq |mathbb{Q}^{mathbb{N}}| leq |(2^{mathbb{N}})^{mathbb{N}}| = |2^{mathbb{N}^2}| = |2^{mathbb{N}}| = |mathbb{R}|$, so $|M| = |mathbb{R}|$.
answered Jan 20 at 15:39
JakobianJakobian
2,650721
$endgroup$
add a comment |
$begingroup$
There are certainly at most $c$ of them, since there are only $aleph_0^{aleph_0}=c$ sequences in $Bbb Q$ to begin with. But there are also at least $c$, because each infinite set $S$ of positive integers gives a unique example of such a sequence, viz. $2^{-s_n}$ with $s_n$ the $n$th smallest element of $S$. Therefore, the cardinality is indeed $c$.
answered Jan 20 at 15:34
J.G.J.G.
28.7k22845
$endgroup$
add a comment |
$begingroup$
It's at most the continuum since there are only continuum many sequences of reals, let alone rationals, let alone which converge to $0$. It's at least the continuum since given just one such sequence with no repeated entries, there are continuum many infinite subsequences of it, all of which still converge to $0$.
answered Jan 20 at 15:36
NedNed
2,048910
$endgroup$
$begingroup$
Indeed, the examples I gave are infinite subsequences of the sequence $2^{-n}$.
$endgroup$
– J.G.
Jan 20 at 15:37
add a comment |
$begingroup$
There are (at least) as many such sequences as there are subsets of $Bbb N$. Indeed, given $Ssubseteq Bbb N$, we can let
$$a_n=begin{cases}frac1n&nin S\-frac 1n&nnotin Send{cases} $$
and thus obtain a different zero-sequence for every subset $S$.
On the other hand, the cardinality of the set of all sequences of rationals is the same as that of the set of all sequences of naturals (as $|Bbb Q|=|Bbb N|$). By mapping $(a_n)_{ninBbb N}$ to $(a_1+ldots +a_n)_{nin Bbb N}$, we see that there are just as many strictly increasing sequences of naturals. And such increasing sequences are "the same" as infinite subsets of $Bbb N$, of which there are at most continuum-many.
answered Jan 20 at 15:38
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Consider the set $L$ of sequences $(x_n)_{ninmathbb{N}}$, with $x_n in {0, frac{1}{n}}$. Then $L$ has lesser cardinality than the set of all rational sequences that converge to $0$ which we will denote by $M$, which has lesser cardinality than $mathbb{Q}^{mathbb{N}}$.
Then $|L| = |2^{mathbb{N}}| = |mathbb{R}| leq |M| leq |mathbb{Q}^{mathbb{N}}| leq |(2^{mathbb{N}})^{mathbb{N}}| = |2^{mathbb{N}^2}| = |2^{mathbb{N}}| = |mathbb{R}|$, so $|M| = |mathbb{R}|$.
answered Jan 20 at 15:39
JakobianJakobian
2,650721
$endgroup$
add a comment |
$begingroup$
Consider the set $L$ of sequences $(x_n)_{ninmathbb{N}}$, with $x_n in {0, frac{1}{n}}$. Then $L$ has lesser cardinality than the set of all rational sequences that converge to $0$ which we will denote by $M$, which has lesser cardinality than $mathbb{Q}^{mathbb{N}}$.
Then $|L| = |2^{mathbb{N}}| = |mathbb{R}| leq |M| leq |mathbb{Q}^{mathbb{N}}| leq |(2^{mathbb{N}})^{mathbb{N}}| = |2^{mathbb{N}^2}| = |2^{mathbb{N}}| = |mathbb{R}|$, so $|M| = |mathbb{R}|$.
answered Jan 20 at 15:39
JakobianJakobian
2,650721
$endgroup$
add a comment |
$begingroup$
Consider the set $L$ of sequences $(x_n)_{ninmathbb{N}}$, with $x_n in {0, frac{1}{n}}$. Then $L$ has lesser cardinality than the set of all rational sequences that converge to $0$ which we will denote by $M$, which has lesser cardinality than $mathbb{Q}^{mathbb{N}}$.
Then $|L| = |2^{mathbb{N}}| = |mathbb{R}| leq |M| leq |mathbb{Q}^{mathbb{N}}| leq |(2^{mathbb{N}})^{mathbb{N}}| = |2^{mathbb{N}^2}| = |2^{mathbb{N}}| = |mathbb{R}|$, so $|M| = |mathbb{R}|$.
answered Jan 20 at 15:39
JakobianJakobian
2,650721
$endgroup$
Consider the set $L$ of sequences $(x_n)_{ninmathbb{N}}$, with $x_n in {0, frac{1}{n}}$. Then $L$ has lesser cardinality than the set of all rational sequences that converge to $0$ which we will denote by $M$, which has lesser cardinality than $mathbb{Q}^{mathbb{N}}$.
Then $|L| = |2^{mathbb{N}}| = |mathbb{R}| leq |M| leq |mathbb{Q}^{mathbb{N}}| leq |(2^{mathbb{N}})^{mathbb{N}}| = |2^{mathbb{N}^2}| = |2^{mathbb{N}}| = |mathbb{R}|$, so $|M| = |mathbb{R}|$.
answered Jan 20 at 15:39
JakobianJakobian
2,650721
answered Jan 20 at 15:39
JakobianJakobian
2,650721
answered Jan 20 at 15:39
JakobianJakobian
2,650721
answered Jan 20 at 15:39
JakobianJakobian
2,650721
2,650721
add a comment |
add a comment |
$begingroup$
There are certainly at most $c$ of them, since there are only $aleph_0^{aleph_0}=c$ sequences in $Bbb Q$ to begin with. But there are also at least $c$, because each infinite set $S$ of positive integers gives a unique example of such a sequence, viz. $2^{-s_n}$ with $s_n$ the $n$th smallest element of $S$. Therefore, the cardinality is indeed $c$.
answered Jan 20 at 15:34
J.G.J.G.
28.7k22845
$endgroup$
add a comment |
$begingroup$
There are certainly at most $c$ of them, since there are only $aleph_0^{aleph_0}=c$ sequences in $Bbb Q$ to begin with. But there are also at least $c$, because each infinite set $S$ of positive integers gives a unique example of such a sequence, viz. $2^{-s_n}$ with $s_n$ the $n$th smallest element of $S$. Therefore, the cardinality is indeed $c$.
answered Jan 20 at 15:34
J.G.J.G.
28.7k22845
$endgroup$
add a comment |
$begingroup$
There are certainly at most $c$ of them, since there are only $aleph_0^{aleph_0}=c$ sequences in $Bbb Q$ to begin with. But there are also at least $c$, because each infinite set $S$ of positive integers gives a unique example of such a sequence, viz. $2^{-s_n}$ with $s_n$ the $n$th smallest element of $S$. Therefore, the cardinality is indeed $c$.
answered Jan 20 at 15:34
J.G.J.G.
28.7k22845
$endgroup$
There are certainly at most $c$ of them, since there are only $aleph_0^{aleph_0}=c$ sequences in $Bbb Q$ to begin with. But there are also at least $c$, because each infinite set $S$ of positive integers gives a unique example of such a sequence, viz. $2^{-s_n}$ with $s_n$ the $n$th smallest element of $S$. Therefore, the cardinality is indeed $c$.
answered Jan 20 at 15:34
J.G.J.G.
28.7k22845
answered Jan 20 at 15:34
J.G.J.G.
28.7k22845
answered Jan 20 at 15:34
J.G.J.G.
28.7k22845
answered Jan 20 at 15:34
J.G.J.G.
28.7k22845
28.7k22845
add a comment |
add a comment |
$begingroup$
It's at most the continuum since there are only continuum many sequences of reals, let alone rationals, let alone which converge to $0$. It's at least the continuum since given just one such sequence with no repeated entries, there are continuum many infinite subsequences of it, all of which still converge to $0$.
answered Jan 20 at 15:36
NedNed
2,048910
$endgroup$
$begingroup$
Indeed, the examples I gave are infinite subsequences of the sequence $2^{-n}$.
$endgroup$
– J.G.
Jan 20 at 15:37
add a comment |
$begingroup$
It's at most the continuum since there are only continuum many sequences of reals, let alone rationals, let alone which converge to $0$. It's at least the continuum since given just one such sequence with no repeated entries, there are continuum many infinite subsequences of it, all of which still converge to $0$.
answered Jan 20 at 15:36
NedNed
2,048910
$endgroup$
$begingroup$
Indeed, the examples I gave are infinite subsequences of the sequence $2^{-n}$.
$endgroup$
– J.G.
Jan 20 at 15:37
add a comment |
$begingroup$
It's at most the continuum since there are only continuum many sequences of reals, let alone rationals, let alone which converge to $0$. It's at least the continuum since given just one such sequence with no repeated entries, there are continuum many infinite subsequences of it, all of which still converge to $0$.
answered Jan 20 at 15:36
NedNed
2,048910
$endgroup$
It's at most the continuum since there are only continuum many sequences of reals, let alone rationals, let alone which converge to $0$. It's at least the continuum since given just one such sequence with no repeated entries, there are continuum many infinite subsequences of it, all of which still converge to $0$.
answered Jan 20 at 15:36
NedNed
2,048910
answered Jan 20 at 15:36
NedNed
2,048910
answered Jan 20 at 15:36
NedNed
2,048910
answered Jan 20 at 15:36
NedNed
2,048910
2,048910
$begingroup$
Indeed, the examples I gave are infinite subsequences of the sequence $2^{-n}$.
$endgroup$
– J.G.
Jan 20 at 15:37
add a comment |
$begingroup$
Indeed, the examples I gave are infinite subsequences of the sequence $2^{-n}$.
$endgroup$
– J.G.
Jan 20 at 15:37
$begingroup$
Indeed, the examples I gave are infinite subsequences of the sequence $2^{-n}$.
$endgroup$
– J.G.
Jan 20 at 15:37
$begingroup$
Indeed, the examples I gave are infinite subsequences of the sequence $2^{-n}$.
$endgroup$
– J.G.
Jan 20 at 15:37
add a comment |
$begingroup$
There are (at least) as many such sequences as there are subsets of $Bbb N$. Indeed, given $Ssubseteq Bbb N$, we can let
$$a_n=begin{cases}frac1n&nin S\-frac 1n&nnotin Send{cases} $$
and thus obtain a different zero-sequence for every subset $S$.
On the other hand, the cardinality of the set of all sequences of rationals is the same as that of the set of all sequences of naturals (as $|Bbb Q|=|Bbb N|$). By mapping $(a_n)_{ninBbb N}$ to $(a_1+ldots +a_n)_{nin Bbb N}$, we see that there are just as many strictly increasing sequences of naturals. And such increasing sequences are "the same" as infinite subsets of $Bbb N$, of which there are at most continuum-many.
answered Jan 20 at 15:38
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
add a comment |
$begingroup$
There are (at least) as many such sequences as there are subsets of $Bbb N$. Indeed, given $Ssubseteq Bbb N$, we can let
$$a_n=begin{cases}frac1n&nin S\-frac 1n&nnotin Send{cases} $$
and thus obtain a different zero-sequence for every subset $S$.
On the other hand, the cardinality of the set of all sequences of rationals is the same as that of the set of all sequences of naturals (as $|Bbb Q|=|Bbb N|$). By mapping $(a_n)_{ninBbb N}$ to $(a_1+ldots +a_n)_{nin Bbb N}$, we see that there are just as many strictly increasing sequences of naturals. And such increasing sequences are "the same" as infinite subsets of $Bbb N$, of which there are at most continuum-many.
answered Jan 20 at 15:38
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
add a comment |
$begingroup$
There are (at least) as many such sequences as there are subsets of $Bbb N$. Indeed, given $Ssubseteq Bbb N$, we can let
$$a_n=begin{cases}frac1n&nin S\-frac 1n&nnotin Send{cases} $$
and thus obtain a different zero-sequence for every subset $S$.
On the other hand, the cardinality of the set of all sequences of rationals is the same as that of the set of all sequences of naturals (as $|Bbb Q|=|Bbb N|$). By mapping $(a_n)_{ninBbb N}$ to $(a_1+ldots +a_n)_{nin Bbb N}$, we see that there are just as many strictly increasing sequences of naturals. And such increasing sequences are "the same" as infinite subsets of $Bbb N$, of which there are at most continuum-many.
answered Jan 20 at 15:38
Hagen von EitzenHagen von Eitzen
282k23272505
$endgroup$
There are (at least) as many such sequences as there are subsets of $Bbb N$. Indeed, given $Ssubseteq Bbb N$, we can let
$$a_n=begin{cases}frac1n&nin S\-frac 1n&nnotin Send{cases} $$
and thus obtain a different zero-sequence for every subset $S$.
On the other hand, the cardinality of the set of all sequences of rationals is the same as that of the set of all sequences of naturals (as $|Bbb Q|=|Bbb N|$). By mapping $(a_n)_{ninBbb N}$ to $(a_1+ldots +a_n)_{nin Bbb N}$, we see that there are just as many strictly increasing sequences of naturals. And such increasing sequences are "the same" as infinite subsets of $Bbb N$, of which there are at most continuum-many.
answered Jan 20 at 15:38
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 15:38
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 15:38
Hagen von EitzenHagen von Eitzen
282k23272505
answered Jan 20 at 15:38
Hagen von EitzenHagen von Eitzen
282k23272505
282k23272505
add a comment |
add a comment |
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