Mixing
Defining a sequence [on hold]
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Just a quick question.
Can a sequence, say $s_n$ be defined as a union of two or more sequences, say $s_1,s_2,s_3$?
real-analysis sequences-and-series
asked yesterday
Allorja
298
put on hold as unclear what you're asking by Martin R, vrugtehagel, Mark, Vasya, Jyrki Lahtonen 22 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
0
down vote
favorite
Just a quick question.
Can a sequence, say $s_n$ be defined as a union of two or more sequences, say $s_1,s_2,s_3$?
real-analysis sequences-and-series
asked yesterday
Allorja
298
put on hold as unclear what you're asking by Martin R, vrugtehagel, Mark, Vasya, Jyrki Lahtonen 22 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
You can define whatever you like.
– Martin R
yesterday
but the concept of sequence defined without ordering
– ehsan eskandari
yesterday
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Just a quick question.
Can a sequence, say $s_n$ be defined as a union of two or more sequences, say $s_1,s_2,s_3$?
real-analysis sequences-and-series
asked yesterday
Allorja
298
Just a quick question.
Can a sequence, say $s_n$ be defined as a union of two or more sequences, say $s_1,s_2,s_3$?
real-analysis sequences-and-series
real-analysis sequences-and-series
asked yesterday
Allorja
298
asked yesterday
Allorja
298
asked yesterday
Allorja
298
asked yesterday
Allorja
298
asked yesterday
Allorja
298
298
put on hold as unclear what you're asking by Martin R, vrugtehagel, Mark, Vasya, Jyrki Lahtonen 22 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
put on hold as unclear what you're asking by Martin R, vrugtehagel, Mark, Vasya, Jyrki Lahtonen 22 hours ago
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
You can define whatever you like.
– Martin R
yesterday
but the concept of sequence defined without ordering
– ehsan eskandari
yesterday
add a comment |
You can define whatever you like.
– Martin R
yesterday
but the concept of sequence defined without ordering
– ehsan eskandari
yesterday
You can define whatever you like.
– Martin R
yesterday
You can define whatever you like.
– Martin R
yesterday
but the concept of sequence defined without ordering
– ehsan eskandari
yesterday
but the concept of sequence defined without ordering
– ehsan eskandari
yesterday
add a comment |
2 Answers
2
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0
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A sequence is a function
$$s_n: Asubseteqmathbb{N} to mathbb{R}$$
then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.
Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.
answered yesterday
gimusi
85.5k74294
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It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence
answered yesterday
ehsan eskandari
885
This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
– Qeeko
yesterday
yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
– ehsan eskandari
yesterday
I cannot make sense of your reply.
– Qeeko
yesterday
I get an example right now
– ehsan eskandari
yesterday
We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
– ehsan eskandari
yesterday
|
show 1 more comment
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
A sequence is a function
$$s_n: Asubseteqmathbb{N} to mathbb{R}$$
then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.
Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.
answered yesterday
gimusi
85.5k74294
add a comment |
up vote
0
down vote
A sequence is a function
$$s_n: Asubseteqmathbb{N} to mathbb{R}$$
then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.
Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.
answered yesterday
gimusi
85.5k74294
add a comment |
up vote
0
down vote
up vote
0
down vote
A sequence is a function
$$s_n: Asubseteqmathbb{N} to mathbb{R}$$
then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.
Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.
answered yesterday
gimusi
85.5k74294
A sequence is a function
$$s_n: Asubseteqmathbb{N} to mathbb{R}$$
then we need to specify what we mean with union of $s_1$, $s_2$, $s_3$.
Therefore, if we are referring to the definition used for the union of sets, we can't apply that to sequences since those are not sets but functions.
answered yesterday
gimusi
85.5k74294
answered yesterday
gimusi
85.5k74294
answered yesterday
gimusi
85.5k74294
answered yesterday
gimusi
85.5k74294
85.5k74294
add a comment |
add a comment |
up vote
0
down vote
It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence
answered yesterday
ehsan eskandari
885
This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
– Qeeko
yesterday
yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
– ehsan eskandari
yesterday
I cannot make sense of your reply.
– Qeeko
yesterday
I get an example right now
– ehsan eskandari
yesterday
We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
– ehsan eskandari
yesterday
|
show 1 more comment
up vote
0
down vote
It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence
answered yesterday
ehsan eskandari
885
This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
– Qeeko
yesterday
yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
– ehsan eskandari
yesterday
I cannot make sense of your reply.
– Qeeko
yesterday
I get an example right now
– ehsan eskandari
yesterday
We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
– ehsan eskandari
yesterday
|
show 1 more comment
up vote
0
down vote
up vote
0
down vote
It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence
answered yesterday
ehsan eskandari
885
It should be noted that a sequence is a range of a function like $f:N rightarrow X$(where $X$ is any space)and if we put $f(n)=s_n$ then {$s_n$} is a sequence.Hence each sequence is a countable set...now if{$a_n$},{$b_n$},{$c_n$} are sequences then the union of them is acountable set and hence it is asequence
answered yesterday
ehsan eskandari
885
answered yesterday
ehsan eskandari
885
answered yesterday
ehsan eskandari
885
answered yesterday
ehsan eskandari
885
885
This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
– Qeeko
yesterday
yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
– ehsan eskandari
yesterday
I cannot make sense of your reply.
– Qeeko
yesterday
I get an example right now
– ehsan eskandari
yesterday
We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
– ehsan eskandari
yesterday
|
show 1 more comment
This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
– Qeeko
yesterday
yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
– ehsan eskandari
yesterday
I cannot make sense of your reply.
– Qeeko
yesterday
I get an example right now
– ehsan eskandari
yesterday
We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
– ehsan eskandari
yesterday
This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
– Qeeko
yesterday
This is not true. A sequence is not the range of a mapping of the natural numbers into some set; it is the mapping itself. It is therefore also incorrect to say that a sequence is a countable set. Even though the union of the images of two sequences is countably infinite and may therefore be put in one-to-one correspondence with the natural numbers, there is no canonical bijection from the natural numbers to this union of sets.
– Qeeko
yesterday
yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
– ehsan eskandari
yesterday
yeah......but we exactly we show a sequence with{$s_n$} where $S:N rightarrow X$ and $S(n)=s_n$
– ehsan eskandari
yesterday
I cannot make sense of your reply.
– Qeeko
yesterday
I cannot make sense of your reply.
– Qeeko
yesterday
I get an example right now
– ehsan eskandari
yesterday
I get an example right now
– ehsan eskandari
yesterday
We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
– ehsan eskandari
yesterday
We know that {$2n$},{$4n+10$},...are sequences.in fact the function $f:N rightarrow R$ s.t $f(n)=2n$.......so each countable set is a sequence
– ehsan eskandari
yesterday
|
show 1 more comment
You can define whatever you like.
– Martin R
yesterday
but the concept of sequence defined without ordering
– ehsan eskandari
yesterday