Mixing
How is an empty Set an initial object
up vote
3
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(I'm using Haskell syntax)
So I have an initial object, which I call $text{Void}$. The prerequisite for an inital object is $forall X in text{Hask.} exists ! f : text{Void} to X$. But how is that true?
Can't I just say:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
(...)
The same applies to Set, with a set $I in text{Set}$ such that $forall X in text{Set}. exists! f : I rightarrow X$, where $I = emptyset$::
$$
f: emptyset rightarrow mathbb{N} \
f(x) = 1 \
f': emptyset rightarrow mathbb{N} \
f'(x) = 2 \
...
$$
Of course I can't call any of them, because the prerequisites of having an element of type Void will never hold true, but still, I can create as many functions as I want.
And if pattern matching is illegal, then how can I even create a single function?
f'' :: Void -> a
-- how is this ok?
category-theory
asked Jun 13 '17 at 15:27
hgiesel
569312
|
show 2 more comments
up vote
3
down vote
favorite
(I'm using Haskell syntax)
So I have an initial object, which I call $text{Void}$. The prerequisite for an inital object is $forall X in text{Hask.} exists ! f : text{Void} to X$. But how is that true?
Can't I just say:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
(...)
The same applies to Set, with a set $I in text{Set}$ such that $forall X in text{Set}. exists! f : I rightarrow X$, where $I = emptyset$::
$$
f: emptyset rightarrow mathbb{N} \
f(x) = 1 \
f': emptyset rightarrow mathbb{N} \
f'(x) = 2 \
...
$$
Of course I can't call any of them, because the prerequisites of having an element of type Void will never hold true, but still, I can create as many functions as I want.
And if pattern matching is illegal, then how can I even create a single function?
f'' :: Void -> a
-- how is this ok?
category-theory
asked Jun 13 '17 at 15:27
hgiesel
569312
Who claimed that Hask has an initial object? Anyways, f and f' are 'equal' in the sense that they give the same value on each term of type Void if Void is 'empty', trivially.
– Ben
Jun 13 '17 at 15:40
@Ben Bartosz Milewski does, but he also kind of disregards the existence of Bottom in Hask. But even without Bottom I don't see how my question would change.
– hgiesel
Jun 13 '17 at 15:43
2
You give a caveat that you're "using Haskell syntax." Are you just using that syntax to ask a question about the category $operatorname{Set}$, or are you actually asking about the category of Haskell types and terms? The answers will be very different in those cases.
– Mike Haskel
Jun 13 '17 at 15:44
@MikeHaskel Actually both answers would be of interest to me. I will clarify my question.
– hgiesel
Jun 13 '17 at 15:46
1
@Pinocchio You don't need to post to every comment... Wait a second, I will create a new answer, where I explain it myself
– hgiesel
10 hours ago
|
show 2 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
(I'm using Haskell syntax)
So I have an initial object, which I call $text{Void}$. The prerequisite for an inital object is $forall X in text{Hask.} exists ! f : text{Void} to X$. But how is that true?
Can't I just say:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
(...)
The same applies to Set, with a set $I in text{Set}$ such that $forall X in text{Set}. exists! f : I rightarrow X$, where $I = emptyset$::
$$
f: emptyset rightarrow mathbb{N} \
f(x) = 1 \
f': emptyset rightarrow mathbb{N} \
f'(x) = 2 \
...
$$
Of course I can't call any of them, because the prerequisites of having an element of type Void will never hold true, but still, I can create as many functions as I want.
And if pattern matching is illegal, then how can I even create a single function?
f'' :: Void -> a
-- how is this ok?
category-theory
asked Jun 13 '17 at 15:27
hgiesel
569312
(I'm using Haskell syntax)
So I have an initial object, which I call $text{Void}$. The prerequisite for an inital object is $forall X in text{Hask.} exists ! f : text{Void} to X$. But how is that true?
Can't I just say:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
(...)
The same applies to Set, with a set $I in text{Set}$ such that $forall X in text{Set}. exists! f : I rightarrow X$, where $I = emptyset$::
$$
f: emptyset rightarrow mathbb{N} \
f(x) = 1 \
f': emptyset rightarrow mathbb{N} \
f'(x) = 2 \
...
$$
Of course I can't call any of them, because the prerequisites of having an element of type Void will never hold true, but still, I can create as many functions as I want.
And if pattern matching is illegal, then how can I even create a single function?
f'' :: Void -> a
-- how is this ok?
category-theory
category-theory
asked Jun 13 '17 at 15:27
hgiesel
569312
asked Jun 13 '17 at 15:27
hgiesel
569312
edited Jun 13 '17 at 15:53
asked Jun 13 '17 at 15:27
hgiesel
569312
asked Jun 13 '17 at 15:27
hgiesel
569312
asked Jun 13 '17 at 15:27
hgiesel
569312
569312
Who claimed that Hask has an initial object? Anyways, f and f' are 'equal' in the sense that they give the same value on each term of type Void if Void is 'empty', trivially.
– Ben
Jun 13 '17 at 15:40
@Ben Bartosz Milewski does, but he also kind of disregards the existence of Bottom in Hask. But even without Bottom I don't see how my question would change.
– hgiesel
Jun 13 '17 at 15:43
2
You give a caveat that you're "using Haskell syntax." Are you just using that syntax to ask a question about the category $operatorname{Set}$, or are you actually asking about the category of Haskell types and terms? The answers will be very different in those cases.
– Mike Haskel
Jun 13 '17 at 15:44
@MikeHaskel Actually both answers would be of interest to me. I will clarify my question.
– hgiesel
Jun 13 '17 at 15:46
1
@Pinocchio You don't need to post to every comment... Wait a second, I will create a new answer, where I explain it myself
– hgiesel
10 hours ago
|
show 2 more comments
Who claimed that Hask has an initial object? Anyways, f and f' are 'equal' in the sense that they give the same value on each term of type Void if Void is 'empty', trivially.
– Ben
Jun 13 '17 at 15:40
@Ben Bartosz Milewski does, but he also kind of disregards the existence of Bottom in Hask. But even without Bottom I don't see how my question would change.
– hgiesel
Jun 13 '17 at 15:43
2
You give a caveat that you're "using Haskell syntax." Are you just using that syntax to ask a question about the category $operatorname{Set}$, or are you actually asking about the category of Haskell types and terms? The answers will be very different in those cases.
– Mike Haskel
Jun 13 '17 at 15:44
@MikeHaskel Actually both answers would be of interest to me. I will clarify my question.
– hgiesel
Jun 13 '17 at 15:46
1
@Pinocchio You don't need to post to every comment... Wait a second, I will create a new answer, where I explain it myself
– hgiesel
10 hours ago
Who claimed that Hask has an initial object? Anyways, f and f' are 'equal' in the sense that they give the same value on each term of type Void if Void is 'empty', trivially.
– Ben
Jun 13 '17 at 15:40
Who claimed that Hask has an initial object? Anyways, f and f' are 'equal' in the sense that they give the same value on each term of type Void if Void is 'empty', trivially.
– Ben
Jun 13 '17 at 15:40
@Ben Bartosz Milewski does, but he also kind of disregards the existence of Bottom in Hask. But even without Bottom I don't see how my question would change.
– hgiesel
Jun 13 '17 at 15:43
@Ben Bartosz Milewski does, but he also kind of disregards the existence of Bottom in Hask. But even without Bottom I don't see how my question would change.
– hgiesel
Jun 13 '17 at 15:43
2
2
You give a caveat that you're "using Haskell syntax." Are you just using that syntax to ask a question about the category $operatorname{Set}$, or are you actually asking about the category of Haskell types and terms? The answers will be very different in those cases.
– Mike Haskel
Jun 13 '17 at 15:44
You give a caveat that you're "using Haskell syntax." Are you just using that syntax to ask a question about the category $operatorname{Set}$, or are you actually asking about the category of Haskell types and terms? The answers will be very different in those cases.
– Mike Haskel
Jun 13 '17 at 15:44
@MikeHaskel Actually both answers would be of interest to me. I will clarify my question.
– hgiesel
Jun 13 '17 at 15:46
@MikeHaskel Actually both answers would be of interest to me. I will clarify my question.
– hgiesel
Jun 13 '17 at 15:46
1
1
@Pinocchio You don't need to post to every comment... Wait a second, I will create a new answer, where I explain it myself
– hgiesel
10 hours ago
@Pinocchio You don't need to post to every comment... Wait a second, I will create a new answer, where I explain it myself
– hgiesel
10 hours ago
|
show 2 more comments
3 Answers
3
active
oldest
votes
up vote
2
down vote
accepted
I'm not very familiar with Haskell, but let me give an answer from category theory; I believe that this should transfer over to a large extent.
The function "$f(x)=2$" isn't really a function from $emptyset$ - rather, its restriction to $emptyset$ is. And the restrictions of $xmapsto 2$ and $xmapsto 1$ to $emptyset$ are the same.
This becomes clear when we think of functions set-theoretically: a function from $A$ to $B$ is a susbet $f$ of $Atimes B$ such that for each $ain A$ there is exactly one $bin B$ with $(a, b)in f$.
Now of course, when working with Haskell the phrase "thinking of functions set-theoretically" is probably a bit cringe-inducing; but it nonetheless has its place here. In the category Set, a morphism from $A$ to $B$ is a triple $(f, A, B)$ where $fsubseteq Atimes B$ is a set-theoretic function from $A$ to $B$.
The key here is that, in both category theory (or rather, the specific category Set - there are other "categories of sets") and set theory, we identify it with its graph, not its intensional definition; so "map everything to $2$" and "map everything to $1$," while intensionally different, yield the same set-theoretic function.
answered Jun 13 '17 at 16:33
Noah Schweber
117k10144275
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
@Pinocchio You're conflating the definition of a function (e.g. "send everything to $2$") with the function itself (namely, the triple (set of ordered pairs, domain, codomain)). The only function from ${}$ to ${1,2,3}$ is $({},{}, {1,2,3})$, since there is nothing in the domain to "pair up" in the first place. This is in fact exactly what I was saying in my answer: read the last sentence in particular.
– Noah Schweber
2 hours ago
It might be better to consider a non-empty situation first. Consider the definitions of functions "divide by four" and "subtract three." Do you see why these define the same function from ${4}$ to ${1}$ (and that there is in fact only one function from ${4}$ to ${1}$)? The same thing is going on here: even if you write different definitions, they're not necessarily naming different functions.
– Noah Schweber
2 hours ago
@Pinocchio Similarly to quantification over the emptyset, the "no counterexample" interpretation can be useful. We have two functions $f,g$ from ${}$ to ${1,2,3}$, and I claim $f=g$. Since $f$ and $g$ clearly have the same domain and codomain, I'm correct exactly as long as there is no counterexample to my claim - that is, $$f=gquadiffquadforall xin{}(f(x)=g(x)).$$ But this statement is vacuously true, and so indeed $f=g$ - that is, any two functions from $emptyset$ to ${1,2,3}$ are equal.
– Noah Schweber
1 hour ago
add a comment |
up vote
3
down vote
For $mathsf{Set}$, the morphism $0 to X$ is simply the empty function (i.e. view a function as a subset of the cartesian product, choose the empty subset).
I do not know if this is ideal, but you can express this in Haskell as:
data Void
i :: Void -> a
i x = case x of {}
You need to run GHC with -XEmptyCase.
answered Jun 13 '17 at 16:14
user454822
311
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
add a comment |
up vote
0
down vote
I think what confused me when I wrote the question is the Haskell syntax.
You got to remember that the _ character is just a syntactic sugar. It means "for all remaining cases, assign that value". If I had realized that, I probably wouldn't have posed that question. Because "technically" it should look like this:
f :: Void -> Int
Because there is not even a first case. Saying "all remaining cases", although there are zero cases is like saying, "Every person in this room is a doctor.", where the room is empty. There is still no doctor in the room.
Also another thing that is important to remember is that in a mathematical sense, functions are equal, if they produce the same output for every possible output. This is definitely true of the functions I've written above:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
They both give the same output for all imaginable input (which is none).
Another way to think about it is visually in the category of $Set$. I've drawn a quick picture for that purpose:
Yet another fun fact is that you can use $n^k$ to calculate the number of morphisms of type $K rightarrow N, |K| = k, |N| = n$. And as we now, $n^0$ is 1 for all $n$.
answered 10 hours ago
hgiesel
569312
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I'm not very familiar with Haskell, but let me give an answer from category theory; I believe that this should transfer over to a large extent.
The function "$f(x)=2$" isn't really a function from $emptyset$ - rather, its restriction to $emptyset$ is. And the restrictions of $xmapsto 2$ and $xmapsto 1$ to $emptyset$ are the same.
This becomes clear when we think of functions set-theoretically: a function from $A$ to $B$ is a susbet $f$ of $Atimes B$ such that for each $ain A$ there is exactly one $bin B$ with $(a, b)in f$.
Now of course, when working with Haskell the phrase "thinking of functions set-theoretically" is probably a bit cringe-inducing; but it nonetheless has its place here. In the category Set, a morphism from $A$ to $B$ is a triple $(f, A, B)$ where $fsubseteq Atimes B$ is a set-theoretic function from $A$ to $B$.
The key here is that, in both category theory (or rather, the specific category Set - there are other "categories of sets") and set theory, we identify it with its graph, not its intensional definition; so "map everything to $2$" and "map everything to $1$," while intensionally different, yield the same set-theoretic function.
answered Jun 13 '17 at 16:33
Noah Schweber
117k10144275
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
@Pinocchio You're conflating the definition of a function (e.g. "send everything to $2$") with the function itself (namely, the triple (set of ordered pairs, domain, codomain)). The only function from ${}$ to ${1,2,3}$ is $({},{}, {1,2,3})$, since there is nothing in the domain to "pair up" in the first place. This is in fact exactly what I was saying in my answer: read the last sentence in particular.
– Noah Schweber
2 hours ago
It might be better to consider a non-empty situation first. Consider the definitions of functions "divide by four" and "subtract three." Do you see why these define the same function from ${4}$ to ${1}$ (and that there is in fact only one function from ${4}$ to ${1}$)? The same thing is going on here: even if you write different definitions, they're not necessarily naming different functions.
– Noah Schweber
2 hours ago
@Pinocchio Similarly to quantification over the emptyset, the "no counterexample" interpretation can be useful. We have two functions $f,g$ from ${}$ to ${1,2,3}$, and I claim $f=g$. Since $f$ and $g$ clearly have the same domain and codomain, I'm correct exactly as long as there is no counterexample to my claim - that is, $$f=gquadiffquadforall xin{}(f(x)=g(x)).$$ But this statement is vacuously true, and so indeed $f=g$ - that is, any two functions from $emptyset$ to ${1,2,3}$ are equal.
– Noah Schweber
1 hour ago
add a comment |
up vote
2
down vote
accepted
I'm not very familiar with Haskell, but let me give an answer from category theory; I believe that this should transfer over to a large extent.
The function "$f(x)=2$" isn't really a function from $emptyset$ - rather, its restriction to $emptyset$ is. And the restrictions of $xmapsto 2$ and $xmapsto 1$ to $emptyset$ are the same.
This becomes clear when we think of functions set-theoretically: a function from $A$ to $B$ is a susbet $f$ of $Atimes B$ such that for each $ain A$ there is exactly one $bin B$ with $(a, b)in f$.
Now of course, when working with Haskell the phrase "thinking of functions set-theoretically" is probably a bit cringe-inducing; but it nonetheless has its place here. In the category Set, a morphism from $A$ to $B$ is a triple $(f, A, B)$ where $fsubseteq Atimes B$ is a set-theoretic function from $A$ to $B$.
The key here is that, in both category theory (or rather, the specific category Set - there are other "categories of sets") and set theory, we identify it with its graph, not its intensional definition; so "map everything to $2$" and "map everything to $1$," while intensionally different, yield the same set-theoretic function.
answered Jun 13 '17 at 16:33
Noah Schweber
117k10144275
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
@Pinocchio You're conflating the definition of a function (e.g. "send everything to $2$") with the function itself (namely, the triple (set of ordered pairs, domain, codomain)). The only function from ${}$ to ${1,2,3}$ is $({},{}, {1,2,3})$, since there is nothing in the domain to "pair up" in the first place. This is in fact exactly what I was saying in my answer: read the last sentence in particular.
– Noah Schweber
2 hours ago
It might be better to consider a non-empty situation first. Consider the definitions of functions "divide by four" and "subtract three." Do you see why these define the same function from ${4}$ to ${1}$ (and that there is in fact only one function from ${4}$ to ${1}$)? The same thing is going on here: even if you write different definitions, they're not necessarily naming different functions.
– Noah Schweber
2 hours ago
@Pinocchio Similarly to quantification over the emptyset, the "no counterexample" interpretation can be useful. We have two functions $f,g$ from ${}$ to ${1,2,3}$, and I claim $f=g$. Since $f$ and $g$ clearly have the same domain and codomain, I'm correct exactly as long as there is no counterexample to my claim - that is, $$f=gquadiffquadforall xin{}(f(x)=g(x)).$$ But this statement is vacuously true, and so indeed $f=g$ - that is, any two functions from $emptyset$ to ${1,2,3}$ are equal.
– Noah Schweber
1 hour ago
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I'm not very familiar with Haskell, but let me give an answer from category theory; I believe that this should transfer over to a large extent.
The function "$f(x)=2$" isn't really a function from $emptyset$ - rather, its restriction to $emptyset$ is. And the restrictions of $xmapsto 2$ and $xmapsto 1$ to $emptyset$ are the same.
This becomes clear when we think of functions set-theoretically: a function from $A$ to $B$ is a susbet $f$ of $Atimes B$ such that for each $ain A$ there is exactly one $bin B$ with $(a, b)in f$.
Now of course, when working with Haskell the phrase "thinking of functions set-theoretically" is probably a bit cringe-inducing; but it nonetheless has its place here. In the category Set, a morphism from $A$ to $B$ is a triple $(f, A, B)$ where $fsubseteq Atimes B$ is a set-theoretic function from $A$ to $B$.
The key here is that, in both category theory (or rather, the specific category Set - there are other "categories of sets") and set theory, we identify it with its graph, not its intensional definition; so "map everything to $2$" and "map everything to $1$," while intensionally different, yield the same set-theoretic function.
answered Jun 13 '17 at 16:33
Noah Schweber
117k10144275
I'm not very familiar with Haskell, but let me give an answer from category theory; I believe that this should transfer over to a large extent.
The function "$f(x)=2$" isn't really a function from $emptyset$ - rather, its restriction to $emptyset$ is. And the restrictions of $xmapsto 2$ and $xmapsto 1$ to $emptyset$ are the same.
This becomes clear when we think of functions set-theoretically: a function from $A$ to $B$ is a susbet $f$ of $Atimes B$ such that for each $ain A$ there is exactly one $bin B$ with $(a, b)in f$.
Now of course, when working with Haskell the phrase "thinking of functions set-theoretically" is probably a bit cringe-inducing; but it nonetheless has its place here. In the category Set, a morphism from $A$ to $B$ is a triple $(f, A, B)$ where $fsubseteq Atimes B$ is a set-theoretic function from $A$ to $B$.
The key here is that, in both category theory (or rather, the specific category Set - there are other "categories of sets") and set theory, we identify it with its graph, not its intensional definition; so "map everything to $2$" and "map everything to $1$," while intensionally different, yield the same set-theoretic function.
answered Jun 13 '17 at 16:33
Noah Schweber
117k10144275
answered Jun 13 '17 at 16:33
Noah Schweber
117k10144275
answered Jun 13 '17 at 16:33
Noah Schweber
117k10144275
answered Jun 13 '17 at 16:33
Noah Schweber
117k10144275
117k10144275
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
@Pinocchio You're conflating the definition of a function (e.g. "send everything to $2$") with the function itself (namely, the triple (set of ordered pairs, domain, codomain)). The only function from ${}$ to ${1,2,3}$ is $({},{}, {1,2,3})$, since there is nothing in the domain to "pair up" in the first place. This is in fact exactly what I was saying in my answer: read the last sentence in particular.
– Noah Schweber
2 hours ago
It might be better to consider a non-empty situation first. Consider the definitions of functions "divide by four" and "subtract three." Do you see why these define the same function from ${4}$ to ${1}$ (and that there is in fact only one function from ${4}$ to ${1}$)? The same thing is going on here: even if you write different definitions, they're not necessarily naming different functions.
– Noah Schweber
2 hours ago
@Pinocchio Similarly to quantification over the emptyset, the "no counterexample" interpretation can be useful. We have two functions $f,g$ from ${}$ to ${1,2,3}$, and I claim $f=g$. Since $f$ and $g$ clearly have the same domain and codomain, I'm correct exactly as long as there is no counterexample to my claim - that is, $$f=gquadiffquadforall xin{}(f(x)=g(x)).$$ But this statement is vacuously true, and so indeed $f=g$ - that is, any two functions from $emptyset$ to ${1,2,3}$ are equal.
– Noah Schweber
1 hour ago
add a comment |
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
@Pinocchio You're conflating the definition of a function (e.g. "send everything to $2$") with the function itself (namely, the triple (set of ordered pairs, domain, codomain)). The only function from ${}$ to ${1,2,3}$ is $({},{}, {1,2,3})$, since there is nothing in the domain to "pair up" in the first place. This is in fact exactly what I was saying in my answer: read the last sentence in particular.
– Noah Schweber
2 hours ago
It might be better to consider a non-empty situation first. Consider the definitions of functions "divide by four" and "subtract three." Do you see why these define the same function from ${4}$ to ${1}$ (and that there is in fact only one function from ${4}$ to ${1}$)? The same thing is going on here: even if you write different definitions, they're not necessarily naming different functions.
– Noah Schweber
2 hours ago
@Pinocchio Similarly to quantification over the emptyset, the "no counterexample" interpretation can be useful. We have two functions $f,g$ from ${}$ to ${1,2,3}$, and I claim $f=g$. Since $f$ and $g$ clearly have the same domain and codomain, I'm correct exactly as long as there is no counterexample to my claim - that is, $$f=gquadiffquadforall xin{}(f(x)=g(x)).$$ But this statement is vacuously true, and so indeed $f=g$ - that is, any two functions from $emptyset$ to ${1,2,3}$ are equal.
– Noah Schweber
1 hour ago
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
@Pinocchio You're conflating the definition of a function (e.g. "send everything to $2$") with the function itself (namely, the triple (set of ordered pairs, domain, codomain)). The only function from ${}$ to ${1,2,3}$ is $({},{}, {1,2,3})$, since there is nothing in the domain to "pair up" in the first place. This is in fact exactly what I was saying in my answer: read the last sentence in particular.
– Noah Schweber
2 hours ago
@Pinocchio You're conflating the definition of a function (e.g. "send everything to $2$") with the function itself (namely, the triple (set of ordered pairs, domain, codomain)). The only function from ${}$ to ${1,2,3}$ is $({},{}, {1,2,3})$, since there is nothing in the domain to "pair up" in the first place. This is in fact exactly what I was saying in my answer: read the last sentence in particular.
– Noah Schweber
2 hours ago
It might be better to consider a non-empty situation first. Consider the definitions of functions "divide by four" and "subtract three." Do you see why these define the same function from ${4}$ to ${1}$ (and that there is in fact only one function from ${4}$ to ${1}$)? The same thing is going on here: even if you write different definitions, they're not necessarily naming different functions.
– Noah Schweber
2 hours ago
It might be better to consider a non-empty situation first. Consider the definitions of functions "divide by four" and "subtract three." Do you see why these define the same function from ${4}$ to ${1}$ (and that there is in fact only one function from ${4}$ to ${1}$)? The same thing is going on here: even if you write different definitions, they're not necessarily naming different functions.
– Noah Schweber
2 hours ago
@Pinocchio Similarly to quantification over the emptyset, the "no counterexample" interpretation can be useful. We have two functions $f,g$ from ${}$ to ${1,2,3}$, and I claim $f=g$. Since $f$ and $g$ clearly have the same domain and codomain, I'm correct exactly as long as there is no counterexample to my claim - that is, $$f=gquadiffquadforall xin{}(f(x)=g(x)).$$ But this statement is vacuously true, and so indeed $f=g$ - that is, any two functions from $emptyset$ to ${1,2,3}$ are equal.
– Noah Schweber
1 hour ago
@Pinocchio Similarly to quantification over the emptyset, the "no counterexample" interpretation can be useful. We have two functions $f,g$ from ${}$ to ${1,2,3}$, and I claim $f=g$. Since $f$ and $g$ clearly have the same domain and codomain, I'm correct exactly as long as there is no counterexample to my claim - that is, $$f=gquadiffquadforall xin{}(f(x)=g(x)).$$ But this statement is vacuously true, and so indeed $f=g$ - that is, any two functions from $emptyset$ to ${1,2,3}$ are equal.
– Noah Schweber
1 hour ago
add a comment |
up vote
3
down vote
For $mathsf{Set}$, the morphism $0 to X$ is simply the empty function (i.e. view a function as a subset of the cartesian product, choose the empty subset).
I do not know if this is ideal, but you can express this in Haskell as:
data Void
i :: Void -> a
i x = case x of {}
You need to run GHC with -XEmptyCase.
answered Jun 13 '17 at 16:14
user454822
311
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
add a comment |
up vote
3
down vote
For $mathsf{Set}$, the morphism $0 to X$ is simply the empty function (i.e. view a function as a subset of the cartesian product, choose the empty subset).
I do not know if this is ideal, but you can express this in Haskell as:
data Void
i :: Void -> a
i x = case x of {}
You need to run GHC with -XEmptyCase.
answered Jun 13 '17 at 16:14
user454822
311
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
add a comment |
up vote
3
down vote
up vote
3
down vote
For $mathsf{Set}$, the morphism $0 to X$ is simply the empty function (i.e. view a function as a subset of the cartesian product, choose the empty subset).
I do not know if this is ideal, but you can express this in Haskell as:
data Void
i :: Void -> a
i x = case x of {}
You need to run GHC with -XEmptyCase.
answered Jun 13 '17 at 16:14
user454822
311
For $mathsf{Set}$, the morphism $0 to X$ is simply the empty function (i.e. view a function as a subset of the cartesian product, choose the empty subset).
I do not know if this is ideal, but you can express this in Haskell as:
data Void
i :: Void -> a
i x = case x of {}
You need to run GHC with -XEmptyCase.
answered Jun 13 '17 at 16:14
user454822
311
answered Jun 13 '17 at 16:14
user454822
311
answered Jun 13 '17 at 16:14
user454822
311
answered Jun 13 '17 at 16:14
user454822
311
311
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
add a comment |
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
Why the Empty set is initial still doesn't make sense to me. The function f() can be mapped to any non-empty set (object in this case) in many different ways (so the morphisms are NOT unique). Say consider the functions f:{}->{1,2,3}. Let f()=1 or g()=2 or h()=3 which are different morphisms to the same object {1,2,3} from the empty set. Thus, the empty set is not initial. Where did I go wrong?
– Pinocchio
11 hours ago
add a comment |
up vote
0
down vote
I think what confused me when I wrote the question is the Haskell syntax.
You got to remember that the _ character is just a syntactic sugar. It means "for all remaining cases, assign that value". If I had realized that, I probably wouldn't have posed that question. Because "technically" it should look like this:
f :: Void -> Int
Because there is not even a first case. Saying "all remaining cases", although there are zero cases is like saying, "Every person in this room is a doctor.", where the room is empty. There is still no doctor in the room.
Also another thing that is important to remember is that in a mathematical sense, functions are equal, if they produce the same output for every possible output. This is definitely true of the functions I've written above:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
They both give the same output for all imaginable input (which is none).
Another way to think about it is visually in the category of $Set$. I've drawn a quick picture for that purpose:
Yet another fun fact is that you can use $n^k$ to calculate the number of morphisms of type $K rightarrow N, |K| = k, |N| = n$. And as we now, $n^0$ is 1 for all $n$.
answered 10 hours ago
hgiesel
569312
add a comment |
up vote
0
down vote
I think what confused me when I wrote the question is the Haskell syntax.
You got to remember that the _ character is just a syntactic sugar. It means "for all remaining cases, assign that value". If I had realized that, I probably wouldn't have posed that question. Because "technically" it should look like this:
f :: Void -> Int
Because there is not even a first case. Saying "all remaining cases", although there are zero cases is like saying, "Every person in this room is a doctor.", where the room is empty. There is still no doctor in the room.
Also another thing that is important to remember is that in a mathematical sense, functions are equal, if they produce the same output for every possible output. This is definitely true of the functions I've written above:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
They both give the same output for all imaginable input (which is none).
Another way to think about it is visually in the category of $Set$. I've drawn a quick picture for that purpose:
Yet another fun fact is that you can use $n^k$ to calculate the number of morphisms of type $K rightarrow N, |K| = k, |N| = n$. And as we now, $n^0$ is 1 for all $n$.
answered 10 hours ago
hgiesel
569312
add a comment |
up vote
0
down vote
up vote
0
down vote
I think what confused me when I wrote the question is the Haskell syntax.
You got to remember that the _ character is just a syntactic sugar. It means "for all remaining cases, assign that value". If I had realized that, I probably wouldn't have posed that question. Because "technically" it should look like this:
f :: Void -> Int
Because there is not even a first case. Saying "all remaining cases", although there are zero cases is like saying, "Every person in this room is a doctor.", where the room is empty. There is still no doctor in the room.
Also another thing that is important to remember is that in a mathematical sense, functions are equal, if they produce the same output for every possible output. This is definitely true of the functions I've written above:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
They both give the same output for all imaginable input (which is none).
Another way to think about it is visually in the category of $Set$. I've drawn a quick picture for that purpose:
Yet another fun fact is that you can use $n^k$ to calculate the number of morphisms of type $K rightarrow N, |K| = k, |N| = n$. And as we now, $n^0$ is 1 for all $n$.
answered 10 hours ago
hgiesel
569312
I think what confused me when I wrote the question is the Haskell syntax.
You got to remember that the _ character is just a syntactic sugar. It means "for all remaining cases, assign that value". If I had realized that, I probably wouldn't have posed that question. Because "technically" it should look like this:
f :: Void -> Int
Because there is not even a first case. Saying "all remaining cases", although there are zero cases is like saying, "Every person in this room is a doctor.", where the room is empty. There is still no doctor in the room.
Also another thing that is important to remember is that in a mathematical sense, functions are equal, if they produce the same output for every possible output. This is definitely true of the functions I've written above:
f :: Void -> Int
f _ = 1
f' :: Void -> Int
f _ = 2
They both give the same output for all imaginable input (which is none).
Another way to think about it is visually in the category of $Set$. I've drawn a quick picture for that purpose:
Yet another fun fact is that you can use $n^k$ to calculate the number of morphisms of type $K rightarrow N, |K| = k, |N| = n$. And as we now, $n^0$ is 1 for all $n$.
answered 10 hours ago
hgiesel
569312
answered 10 hours ago
hgiesel
569312
answered 10 hours ago
hgiesel
569312
answered 10 hours ago
hgiesel
569312
569312
add a comment |
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Who claimed that Hask has an initial object? Anyways, f and f' are 'equal' in the sense that they give the same value on each term of type Void if Void is 'empty', trivially.
– Ben
Jun 13 '17 at 15:40
@Ben Bartosz Milewski does, but he also kind of disregards the existence of Bottom in Hask. But even without Bottom I don't see how my question would change.
– hgiesel
Jun 13 '17 at 15:43
2
You give a caveat that you're "using Haskell syntax." Are you just using that syntax to ask a question about the category $operatorname{Set}$, or are you actually asking about the category of Haskell types and terms? The answers will be very different in those cases.
– Mike Haskel
Jun 13 '17 at 15:44
@MikeHaskel Actually both answers would be of interest to me. I will clarify my question.
– hgiesel
Jun 13 '17 at 15:46
1
@Pinocchio You don't need to post to every comment... Wait a second, I will create a new answer, where I explain it myself
– hgiesel
10 hours ago