Does this formalism adequately describe functions of one variable?
$begingroup$
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
asked 6 hours ago
Dan ChristensenDan Christensen
8,63821835
$endgroup$
|
show 1 more comment
$begingroup$
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
asked 6 hours ago
Dan ChristensenDan Christensen
8,63821835
$endgroup$
$begingroup$
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 $Y$?
$endgroup$
– bounceback
6 hours ago
1
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
6 hours ago
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
6 hours ago
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
6 hours ago
$begingroup$
@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
5 hours ago
|
show 1 more comment
$begingroup$
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
asked 6 hours ago
Dan ChristensenDan Christensen
8,63821835
$endgroup$
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
asked 6 hours ago
Dan ChristensenDan Christensen
8,63821835
asked 6 hours ago
Dan ChristensenDan Christensen
8,63821835
edited 5 hours ago
Dan Christensen
asked 6 hours ago
Dan ChristensenDan Christensen
8,63821835
asked 6 hours ago
Dan ChristensenDan Christensen
8,63821835
asked 6 hours ago
Dan ChristensenDan Christensen
8,63821835
8,63821835
$begingroup$
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
6 hours ago
1
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
6 hours ago
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
6 hours ago
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
6 hours ago
$begingroup$
@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
5 hours ago
|
show 1 more comment
$begingroup$
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
6 hours ago
1
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
6 hours ago
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
6 hours ago
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
6 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
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
6 hours ago
$begingroup$
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
6 hours ago
1
1
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
6 hours ago
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
6 hours ago
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
6 hours ago
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
6 hours ago
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
6 hours ago
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
6 hours ago
$begingroup$
@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
5 hours ago
$begingroup$
@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
5 hours ago
|
show 1 more comment
2 Answers
2
active
oldest
votes
$begingroup$
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
begin{align*}
(forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
end{align*}
answered 6 hours ago
APC89APC89
2,443420
$endgroup$
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
5 hours ago
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.
answered 4 hours ago
Derek ElkinsDerek Elkins
17k11437
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
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active
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oldest
votes
$begingroup$
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
begin{align*}
(forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
end{align*}
answered 6 hours ago
APC89APC89
2,443420
$endgroup$
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
5 hours ago
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
begin{align*}
(forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
end{align*}
answered 6 hours ago
APC89APC89
2,443420
$endgroup$
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
5 hours ago
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
begin{align*}
(forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
end{align*}
answered 6 hours ago
APC89APC89
2,443420
$endgroup$
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
begin{align*}
(forall ain A)(exists!bin B)quadtext{such that}quad(a,b)in f
end{align*}
answered 6 hours ago
APC89APC89
2,443420
answered 6 hours ago
APC89APC89
2,443420
answered 6 hours ago
APC89APC89
2,443420
answered 6 hours ago
APC89APC89
2,443420
2,443420
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
5 hours ago
add a comment |
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I am familiar with the that particular formalism.Could you comment on the one I presented here?
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
5 hours ago
$begingroup$
I guess the heading wasn't very clear. I have changed it.
$endgroup$
– Dan Christensen
5 hours ago
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.
answered 4 hours ago
Derek ElkinsDerek Elkins
17k11437
$endgroup$
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.
answered 4 hours ago
Derek ElkinsDerek Elkins
17k11437
$endgroup$
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.
answered 4 hours ago
Derek ElkinsDerek Elkins
17k11437
$endgroup$
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.
answered 4 hours ago
Derek ElkinsDerek Elkins
17k11437
edited 4 hours ago
answered 4 hours ago
Derek ElkinsDerek Elkins
17k11437
answered 4 hours ago
Derek ElkinsDerek Elkins
17k11437
answered 4 hours ago
Derek ElkinsDerek Elkins
17k11437
17k11437
add a comment |
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
6 hours ago
1
$begingroup$
Seems like you're missing the uniqueness of $f(a)$ given $a$.
$endgroup$
– Randall
6 hours ago
$begingroup$
@Randall It is trivial to show that $forall a (ain X implies f(a)=f(a))$
$endgroup$
– Dan Christensen
6 hours ago
$begingroup$
Why is that? Why can't I have two different $f(a)$'s, each one making your implication true?
$endgroup$
– Randall
6 hours ago
$begingroup$
@Randall If $f(a)=b$ and $f(a)=c$, by substitution have $b=c$. Thus $f(a)$ can have only a single value.
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– Dan Christensen
5 hours ago