Does this formalism adequately describe functions of one variable?
Clash Royale CLAN TAG#URR8PPP
$begingroup$
Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.
Can this statement be effectively formalized by
$forall a (ain X implies f(a) in Y)$
What logical aspects of functionality, if any, would not be captured by this statement.
functions elementary-set-theory logic
$endgroup$
|
show 4 more comments
$begingroup$
Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.
Can this statement be effectively formalized by
$forall a (ain X implies f(a) in Y)$
What logical aspects of functionality, if any, would not be captured by this statement.
functions elementary-set-theory logic
$endgroup$
$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
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– bounceback
Feb 20 at 1:24
2
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36
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Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38
$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
Feb 20 at 1:45
|
show 4 more comments
$begingroup$
Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.
Can this statement be effectively formalized by
$forall a (ain X implies f(a) in Y)$
What logical aspects of functionality, if any, would not be captured by this statement.
functions elementary-set-theory logic
$endgroup$
Let $f$ be a function mapping every element of a set $X$ to a unique element denoted by $f(x)$ in a set $Y$.
Can this statement be effectively formalized by
$forall a (ain X implies f(a) in Y)$
What logical aspects of functionality, if any, would not be captured by this statement.
functions elementary-set-theory logic
functions elementary-set-theory logic
edited Feb 20 at 1:56
Dan Christensen
asked Feb 20 at 1:15
Dan ChristensenDan Christensen
8,64321835
8,64321835
$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
Feb 20 at 1:24
2
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38
$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
Feb 20 at 1:45
|
show 4 more comments
$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
Feb 20 at 1:24
2
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38
$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
Feb 20 at 1:45
$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
Feb 20 at 1:24
$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
Feb 20 at 1:24
2
2
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
Feb 20 at 1:36
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38
$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
Feb 20 at 1:45
$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
Feb 20 at 1:45
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.
Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.
$f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.
If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)
1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.
$endgroup$
add a comment |
$begingroup$
Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
beginalign*
(forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
endalign*
$endgroup$
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
Feb 20 at 1:54
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
Feb 20 at 1:57
add a comment |
$begingroup$
If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.
$endgroup$
$begingroup$
What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
$endgroup$
– Dan Christensen
Feb 21 at 15:43
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.
Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.
$f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.
If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)
1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.
$endgroup$
add a comment |
$begingroup$
For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.
Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.
$f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.
If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)
1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.
$endgroup$
add a comment |
$begingroup$
For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.
Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.
$f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.
If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)
1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.
$endgroup$
For a more or less standard FOL, as is typically used to formalize, e.g., ZFC set theory, writing $f(a)$ already implies that $f$ is a function. Your statement would merely add that the image of $X$ is contained in $Y$. If that's all you want, then this is fine.
Fixing $ain X$, if $f$ is a set-theoretic function, as opposed to a logical function symbol, then $f(a)in Y$ would mean something like $exists yin Y.(a,y)in f$. This would imply $f$ is a total relation on $X$, but not that it is functional which would just add that such a $y$ has to be unique. This is the point of Randall's comments. Without the uniqueness constraint, $f$ could readily be a relation that relates $a$ to multiple elements of $Y$. Your argument that you can "trivially" show $f(a)=f(a)$ doesn't work when $f$ is not a function symbol. There is no term $f(a)$ in this context.1 Instead $f(x)=y$ is interpreted as $(x,y)in f$, so in your case you'd get $f(a)=f(a)$ means $(a,f(a))in f$ means $exists b.(a,b)in fland(a,b)in f$ which is, of course, equivalent to just $exists b.(a,b)in f$ which in no way states that that $b$ is unique.
$f$ may still not be a function since we haven't said anything about how it behaves outside of $X$. You can add an additional constraint: $forall a,b.(a,b)in fimplies ain X$ which would say that $f$ is only "defined" on $X$. In APC89's answer, this is accomplished by requiring $fsubset Xtimes Y$.
If you just want to specify that a binary relation $f$ is a function $X$ to $Y$, then APC89's answer provides the more or less standard definition for this, except for the unnecessary insistence that $X$ and $Y$ be non-empty. (Of course, if $Y$ is empty and $X$ is non-empty, there won't actually be any functions from $X$ to $Y$. On the other hand, $X$ being empty causes no problems at all.)
1 If your logic has a definite description quantifier, you could say it is the term $iota b.(a,b)in f$, but this also would require $f$ to separately be shown to be a functional relation for this to give a useful result.
edited Feb 20 at 2:59
answered Feb 20 at 2:46
Derek ElkinsDerek Elkins
17.3k11437
17.3k11437
add a comment |
add a comment |
$begingroup$
Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
beginalign*
(forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
endalign*
$endgroup$
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
Feb 20 at 1:54
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
Feb 20 at 1:57
add a comment |
$begingroup$
Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
beginalign*
(forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
endalign*
$endgroup$
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
Feb 20 at 1:54
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
Feb 20 at 1:57
add a comment |
$begingroup$
Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
beginalign*
(forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
endalign*
$endgroup$
Given two non-empty sets $A$ and $B$, we say that the binary relation $fsubset Atimes B$ is a function from $A$ to $B$ if and only if
beginalign*
(forall ain A)(exists!bin B)quadtextsuch thatquad(a,b)in f
endalign*
answered Feb 20 at 1:22
APC89APC89
2,341720
2,341720
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
Feb 20 at 1:54
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
Feb 20 at 1:57
add a comment |
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
Feb 20 at 1:54
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
Feb 20 at 1:57
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
Feb 20 at 1:54
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
Feb 20 at 1:54
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
Feb 20 at 1:57
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
Feb 20 at 1:57
add a comment |
$begingroup$
If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.
$endgroup$
$begingroup$
What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
$endgroup$
– Dan Christensen
Feb 21 at 15:43
add a comment |
$begingroup$
If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.
$endgroup$
$begingroup$
What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
$endgroup$
– Dan Christensen
Feb 21 at 15:43
add a comment |
$begingroup$
If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.
$endgroup$
If the symbols $f$, $X$, and $Y$ are established to be interpreted so that $f$ is a function and $X,Y$ are sets, then your formalism captures $Xsubseteq mathrmdom(f)$ and $mathrmran(fbig|_X)subseteq Y$. It does not establish that $f$ is a function and $X,Y$ are sets---that must be established before writing the formalism.
answered Feb 20 at 9:07
Alberto TakaseAlberto Takase
2,380719
2,380719
$begingroup$
What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
$endgroup$
– Dan Christensen
Feb 21 at 15:43
add a comment |
$begingroup$
What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
$endgroup$
– Dan Christensen
Feb 21 at 15:43
$begingroup$
What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
$endgroup$
– Dan Christensen
Feb 21 at 15:43
$begingroup$
What aspect of functionality is not captured by my formalism? Uniqueness is built into the notation. From f(x)=a and f(x)=b, we can infer by substitution that a=b.
$endgroup$
– Dan Christensen
Feb 21 at 15:43
add a comment |
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$begingroup$
Do you just mean that $f$ is well-defined (not one-to-many), or that $f$ is a constant function, that is, all elements of $X$ map to the same unique element of $Y$?
$endgroup$
– bounceback
Feb 20 at 1:24
2
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
Feb 20 at 1:24
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
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– Dan Christensen
Feb 20 at 1:36
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
Feb 20 at 1:38
$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
$endgroup$
– Dan Christensen
Feb 20 at 1:45