Writing a set of functions
$begingroup$
Let
$A$ be a countable set of subsets of $mathbb{R}$
$f:A rightarrow mathbb{Z}$ be a bijection
$O= {x_k:mathbb{R}rightarrow mathbb{R} mid kin [1,dots, K]subset mathbb{Z}}$, for some $K$ and some given functions $x_k$.
I want to define a set $R$ which contain functions $g:mathbb{Z}rightarrow mathbb{Z}, g(n)= i$ if $x(n) in f^{-1}(i)$, where $xin O$.
I believe that I can write the set $R$ like this
$$
R = left{g:mathbb{Z} rightarrow mathbb{Z}, g(n)= i text{ if } x(n)in f^{-1}(i) mid xin Oright}.
$$
but I'm also considering to write the set like this
$$
R = left{n in mathbb{Z} mapsto i in mathbb{Z} text{ if } x(n)in f^{-1}(i) mid xin O right},
$$
avoiding the introduction of the symbol $g$. Are these ways of writing understandable? Do you have any other suggestions?
functions elementary-set-theory article-writing
edited Dec 5 '18 at 12:37
Andrés E. Caicedo
65k8158246
asked Dec 5 '18 at 8:36
AngelosAngelos
788
$endgroup$
add a comment |
$begingroup$
Let
$A$ be a countable set of subsets of $mathbb{R}$
$f:A rightarrow mathbb{Z}$ be a bijection
$O= {x_k:mathbb{R}rightarrow mathbb{R} mid kin [1,dots, K]subset mathbb{Z}}$, for some $K$ and some given functions $x_k$.
I want to define a set $R$ which contain functions $g:mathbb{Z}rightarrow mathbb{Z}, g(n)= i$ if $x(n) in f^{-1}(i)$, where $xin O$.
I believe that I can write the set $R$ like this
$$
R = left{g:mathbb{Z} rightarrow mathbb{Z}, g(n)= i text{ if } x(n)in f^{-1}(i) mid xin Oright}.
$$
but I'm also considering to write the set like this
$$
R = left{n in mathbb{Z} mapsto i in mathbb{Z} text{ if } x(n)in f^{-1}(i) mid xin O right},
$$
avoiding the introduction of the symbol $g$. Are these ways of writing understandable? Do you have any other suggestions?
functions elementary-set-theory article-writing
edited Dec 5 '18 at 12:37
Andrés E. Caicedo
65k8158246
asked Dec 5 '18 at 8:36
AngelosAngelos
788
$endgroup$
add a comment |
$begingroup$
Let
$A$ be a countable set of subsets of $mathbb{R}$
$f:A rightarrow mathbb{Z}$ be a bijection
$O= {x_k:mathbb{R}rightarrow mathbb{R} mid kin [1,dots, K]subset mathbb{Z}}$, for some $K$ and some given functions $x_k$.
I want to define a set $R$ which contain functions $g:mathbb{Z}rightarrow mathbb{Z}, g(n)= i$ if $x(n) in f^{-1}(i)$, where $xin O$.
I believe that I can write the set $R$ like this
$$
R = left{g:mathbb{Z} rightarrow mathbb{Z}, g(n)= i text{ if } x(n)in f^{-1}(i) mid xin Oright}.
$$
but I'm also considering to write the set like this
$$
R = left{n in mathbb{Z} mapsto i in mathbb{Z} text{ if } x(n)in f^{-1}(i) mid xin O right},
$$
avoiding the introduction of the symbol $g$. Are these ways of writing understandable? Do you have any other suggestions?
functions elementary-set-theory article-writing
edited Dec 5 '18 at 12:37
Andrés E. Caicedo
65k8158246
asked Dec 5 '18 at 8:36
AngelosAngelos
788
$endgroup$
Let
$A$ be a countable set of subsets of $mathbb{R}$
$f:A rightarrow mathbb{Z}$ be a bijection
$O= {x_k:mathbb{R}rightarrow mathbb{R} mid kin [1,dots, K]subset mathbb{Z}}$, for some $K$ and some given functions $x_k$.
I want to define a set $R$ which contain functions $g:mathbb{Z}rightarrow mathbb{Z}, g(n)= i$ if $x(n) in f^{-1}(i)$, where $xin O$.
I believe that I can write the set $R$ like this
$$
R = left{g:mathbb{Z} rightarrow mathbb{Z}, g(n)= i text{ if } x(n)in f^{-1}(i) mid xin Oright}.
$$
but I'm also considering to write the set like this
$$
R = left{n in mathbb{Z} mapsto i in mathbb{Z} text{ if } x(n)in f^{-1}(i) mid xin O right},
$$
avoiding the introduction of the symbol $g$. Are these ways of writing understandable? Do you have any other suggestions?
functions elementary-set-theory article-writing
functions elementary-set-theory article-writing
edited Dec 5 '18 at 12:37
Andrés E. Caicedo
65k8158246
asked Dec 5 '18 at 8:36
AngelosAngelos
788
edited Dec 5 '18 at 12:37
Andrés E. Caicedo
65k8158246
asked Dec 5 '18 at 8:36
AngelosAngelos
788
edited Dec 5 '18 at 12:37
Andrés E. Caicedo
65k8158246
edited Dec 5 '18 at 12:37
Andrés E. Caicedo
65k8158246
edited Dec 5 '18 at 12:37
Andrés E. Caicedo
65k8158246
65k8158246
asked Dec 5 '18 at 8:36
AngelosAngelos
788
asked Dec 5 '18 at 8:36
AngelosAngelos
788
asked Dec 5 '18 at 8:36
AngelosAngelos
788
788
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Formally, a set is written in the form ${x :|: P(x) }$ with $P(x)$ describing the properties of a set $x$ to be contained in the set in question. Of course, this is relaxed a bit in practice (for example writing ${x in mathbb{R} :|: x^2 geq 2}$ instead of ${x :|: x in mathbb{R} text{ and } x^2 geq 2}$ or even ${x^2 :|: x in mathbb{R}}$ instead of ${y :|: text{ there exists } x in mathbb{R} text{ such that } y = x^2}$ ), but in general, the properties your elements have to satisfy should still all come after the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: ...}$.
Your description of $R$ is not completely clear, there are two possible in
terpretations coming to my mind:
a) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then $x(n) in f^{-1}(i)$.
b) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ we have $x(n) in f^{-1}(g(n))$.
a) We have $g in R$ if for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then there exists $x in O$ such that we have $x(n) in f^{-1}(i)$.
b) We have $g in R$ if for all $n in mathbb{Z}$ there exists $x in O$ such that we have $x(n) in f^{-1}(g(n))$.
(the descriptions 1 a) and b) describe the same set as do 2 a) and 2 b), the a) parts are more in the flavor you used in your question, whereas the b) parts are a bit shorter)
If one of these two adequately describes what you want, then you can just take the part of the sentence beginning after if and write the part behind the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: text{ there exists } x in O ...}$.
answered Dec 5 '18 at 9:23
Matthias KlupschMatthias Klupsch
6,1591227
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
Formally, a set is written in the form ${x :|: P(x) }$ with $P(x)$ describing the properties of a set $x$ to be contained in the set in question. Of course, this is relaxed a bit in practice (for example writing ${x in mathbb{R} :|: x^2 geq 2}$ instead of ${x :|: x in mathbb{R} text{ and } x^2 geq 2}$ or even ${x^2 :|: x in mathbb{R}}$ instead of ${y :|: text{ there exists } x in mathbb{R} text{ such that } y = x^2}$ ), but in general, the properties your elements have to satisfy should still all come after the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: ...}$.
Your description of $R$ is not completely clear, there are two possible in
terpretations coming to my mind:
a) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then $x(n) in f^{-1}(i)$.
b) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ we have $x(n) in f^{-1}(g(n))$.
a) We have $g in R$ if for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then there exists $x in O$ such that we have $x(n) in f^{-1}(i)$.
b) We have $g in R$ if for all $n in mathbb{Z}$ there exists $x in O$ such that we have $x(n) in f^{-1}(g(n))$.
(the descriptions 1 a) and b) describe the same set as do 2 a) and 2 b), the a) parts are more in the flavor you used in your question, whereas the b) parts are a bit shorter)
If one of these two adequately describes what you want, then you can just take the part of the sentence beginning after if and write the part behind the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: text{ there exists } x in O ...}$.
answered Dec 5 '18 at 9:23
Matthias KlupschMatthias Klupsch
6,1591227
$endgroup$
add a comment |
$begingroup$
Formally, a set is written in the form ${x :|: P(x) }$ with $P(x)$ describing the properties of a set $x$ to be contained in the set in question. Of course, this is relaxed a bit in practice (for example writing ${x in mathbb{R} :|: x^2 geq 2}$ instead of ${x :|: x in mathbb{R} text{ and } x^2 geq 2}$ or even ${x^2 :|: x in mathbb{R}}$ instead of ${y :|: text{ there exists } x in mathbb{R} text{ such that } y = x^2}$ ), but in general, the properties your elements have to satisfy should still all come after the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: ...}$.
Your description of $R$ is not completely clear, there are two possible in
terpretations coming to my mind:
a) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then $x(n) in f^{-1}(i)$.
b) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ we have $x(n) in f^{-1}(g(n))$.
a) We have $g in R$ if for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then there exists $x in O$ such that we have $x(n) in f^{-1}(i)$.
b) We have $g in R$ if for all $n in mathbb{Z}$ there exists $x in O$ such that we have $x(n) in f^{-1}(g(n))$.
(the descriptions 1 a) and b) describe the same set as do 2 a) and 2 b), the a) parts are more in the flavor you used in your question, whereas the b) parts are a bit shorter)
If one of these two adequately describes what you want, then you can just take the part of the sentence beginning after if and write the part behind the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: text{ there exists } x in O ...}$.
answered Dec 5 '18 at 9:23
Matthias KlupschMatthias Klupsch
6,1591227
$endgroup$
add a comment |
$begingroup$
Formally, a set is written in the form ${x :|: P(x) }$ with $P(x)$ describing the properties of a set $x$ to be contained in the set in question. Of course, this is relaxed a bit in practice (for example writing ${x in mathbb{R} :|: x^2 geq 2}$ instead of ${x :|: x in mathbb{R} text{ and } x^2 geq 2}$ or even ${x^2 :|: x in mathbb{R}}$ instead of ${y :|: text{ there exists } x in mathbb{R} text{ such that } y = x^2}$ ), but in general, the properties your elements have to satisfy should still all come after the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: ...}$.
Your description of $R$ is not completely clear, there are two possible in
terpretations coming to my mind:
a) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then $x(n) in f^{-1}(i)$.
b) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ we have $x(n) in f^{-1}(g(n))$.
a) We have $g in R$ if for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then there exists $x in O$ such that we have $x(n) in f^{-1}(i)$.
b) We have $g in R$ if for all $n in mathbb{Z}$ there exists $x in O$ such that we have $x(n) in f^{-1}(g(n))$.
(the descriptions 1 a) and b) describe the same set as do 2 a) and 2 b), the a) parts are more in the flavor you used in your question, whereas the b) parts are a bit shorter)
If one of these two adequately describes what you want, then you can just take the part of the sentence beginning after if and write the part behind the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: text{ there exists } x in O ...}$.
answered Dec 5 '18 at 9:23
Matthias KlupschMatthias Klupsch
6,1591227
$endgroup$
Formally, a set is written in the form ${x :|: P(x) }$ with $P(x)$ describing the properties of a set $x$ to be contained in the set in question. Of course, this is relaxed a bit in practice (for example writing ${x in mathbb{R} :|: x^2 geq 2}$ instead of ${x :|: x in mathbb{R} text{ and } x^2 geq 2}$ or even ${x^2 :|: x in mathbb{R}}$ instead of ${y :|: text{ there exists } x in mathbb{R} text{ such that } y = x^2}$ ), but in general, the properties your elements have to satisfy should still all come after the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: ...}$.
Your description of $R$ is not completely clear, there are two possible in
terpretations coming to my mind:
a) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then $x(n) in f^{-1}(i)$.
b) We have $g in R$ if there exists $x in O$ such that for all $n in mathbb{Z}$ we have $x(n) in f^{-1}(g(n))$.
a) We have $g in R$ if for all $n in mathbb{Z}$ and $i in mathbb{Z}$ if we have $g(n) = i$, then there exists $x in O$ such that we have $x(n) in f^{-1}(i)$.
b) We have $g in R$ if for all $n in mathbb{Z}$ there exists $x in O$ such that we have $x(n) in f^{-1}(g(n))$.
(the descriptions 1 a) and b) describe the same set as do 2 a) and 2 b), the a) parts are more in the flavor you used in your question, whereas the b) parts are a bit shorter)
If one of these two adequately describes what you want, then you can just take the part of the sentence beginning after if and write the part behind the bar, i.e. $R = {g : mathbb{Z} to mathbb{Z} :|: text{ there exists } x in O ...}$.
answered Dec 5 '18 at 9:23
Matthias KlupschMatthias Klupsch
6,1591227
answered Dec 5 '18 at 9:23
Matthias KlupschMatthias Klupsch
6,1591227
answered Dec 5 '18 at 9:23
Matthias KlupschMatthias Klupsch
6,1591227
answered Dec 5 '18 at 9:23
Matthias KlupschMatthias Klupsch
6,1591227
6,1591227
add a comment |
add a comment |
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