A couple of questions on set theory

[micromass]

Why 0? Is there an underlying theoretical reason to define it like this?

$0$ is just a notation. There is an underlying reason to define it like this, but it requires some set theory that you're not familiar with.

Recall that $B^A$ denotes the set of all functions from $A$ to $B$. This is precisely what $\mathbb{R}^n$ means. It is the set of all functions from $n$ to $\mathbb{R}$.

But, $n$ is not a set, so how does that make sense? Well, the truth is that natural numbers are sets. They are defined as follows:

$0$ is by definition the empty set. So $0=\emptyset$.
If $0,...,n$ are defined, then we define $n+1 = \{0,...,n\}$.

So for example, $1$ is defined as $1 = \{0\} = \{\emptyset\}$ and $2 = \{0,1\} = \{\emptyset, \{\emptyset\}\}$.

So $\mathbb{R}^n$ is the set of functions from $\{0,...,n-1\}$ to $\mathbb{R}$.

Let's look in some special cases:
If $n=1$, then we have $\mathbb{R}^0 = \{f:\{0\}\rightarrow \mathbb{R}~\vert~\text{f is a function}\}$. But such a function $f$ is uniquely determined by giving $f(0)$. Indeed, if I say (for example) that $f(0) = 2$, then my function $f$ is uniquely defined by
f = \{(0,2)\}
(recall that functions are subsets of the cartesian product).
So we often identify the function $\{(0,x)\}$ with the real number $x$. So this leads us to saying that $\mathbb{R}^1 = \mathbb{R}$. This equality is of course false (the left hand side contains functions, the right hand side contains real numbers). But it should not be read as an equality, but as an identification.

If $n=2$, then we have $\mathbb{R}^2$ is the set of functions from $\{0,1\}$ to $\mathbb{R}$. A typical such function is
f = \{(0,x), (1,y)\}
Again there is a canonical identification of this function $f$ and the ordered pair $(x,y)$. So we often say that $\mathbb{R}$ is the set of ordered pairs (although it actually are functions).

Now, if $n=0$, then we have $\mathbb{R}^0$ is the set of functions $f:\emptyset\rightarrow \mathbb{R}$. What in Earth is a function from the emptyset to $\mathbb{R}$? Does such a beast exist? Yes, it does. Recall that a function is just a subset of the cartesian product (with some properties that don't matter here). So such a function $f$ would be a subset of $f\subseteq \emptyset \times \mathbb{R}$. But one can prove that $\emptyset \times \mathbb{R} = \emptyset$. So a function $f:\emptyset \rightarrow \mathbb{R}$ is just a subset from the empty set. There is only one subset from the empty set, namely the empty set itself. So $\emptyset$ is a function from $\emptyset$ to $\mathbb{R}$ (as crazy as it sounds). So $\mathbb{R}^0$ is then the set of all such functions, so

\mathbb{R}^0 = \{\emptyset\} = \{0\}

 

[Dods]

Those are some very interesting definitions! Thank you for explaining this. :) It raises a few questions though.

If you can define R^n in two differnt ways, the way defined below, and as a cartesian product (much like we did for R^2 at the start of this thread), and as you noted below these sets are not equal (one is with functions and one is with ordered pairs, or real numbers, or whatever), is the definition you're using just obvious from context? Is one definition more correct than the other?

You say there is a "canonical identification" between the two definitions. I think "canonical" here means something like accepted by tradition, I'm not sure (English isn't my first language) and I know what identification means, only I'm unsure what technical or mathematical meaning it has here. Could you explain?

Also, when you say:

Let's look in some special cases:
If $n=1$, then we have $\mathbb{R}^0 = \{f:\{0\}\rightarrow \mathbb{R}~\vert~\text{f is a function}\}$.

Do you mean $\mathbb{R}^1$? Or have I missed something?

And similarly,

If $n=2$, then we have $\mathbb{R}^2$ is the set of functions from $\{0,1\}$ to $\mathbb{R}$. A typical such function is
f = \{(0,x), (1,y)\}
Again there is a canonical identification of this function $f$ and the ordered pair $(x,y)$. So we often say that $\mathbb{R}$ is the set of ordered pairs (although it actually are functions).

Wouldn't the second $\mathbb{R}$ be $\mathbb{R}^2$?

Thank you very much! :)

 

[micromass]

If you can define R^n in two differnt ways, the way defined below, and as a cartesian product (much like we did for R^2 at the start of this thread), and as you noted below these sets are not equal (one is with functions and one is with ordered pairs, or real numbers, or whatever), is the definition you're using just obvious from context? Is one definition more correct than the other?

The key is that we don't really care about the exact definition. What we care are the properties that the set has. Whether we use $(x,y)$, or a function $f:\{0,1\}\rightarrow \mathbb{R}$ with $f(0) = x$ and $f(1) = y$ doesn't matter in practice. So we can just choose on definition (I prefer the function definition), and keep using that. The actual definition of the set won't be important.

The same with the definition $(x,y) =\{\{x\},\{x,y\}\}$. This is not the only way to define orderer pairs. There are other definitions which are as good. But we never really care about the definition, we just care about the properties of the ordered pair. How exactly it is defined is irrelevant, as long as it can be defined.

You say there is a "canonical identification" between the two definitions. I think "canonical" here means something like accepted by tradition, I'm not sure (English isn't my first language) and I know what identification means, only I'm unsure what technical or mathematical meaning it has here. Could you explain?

It has not technical meaning. Canonical means natural or straightforward (in this context). For example, the two sets $A\times B$ and $B\times A$ are not equal. But there is a bijection between them. What is the first thing you think of, well, it is $f(a,b) = (b,a)$. It is the straightforward thing to do. This is called canonical.

Also, when you say:



Do you mean $\mathbb{R}^1$? Or have I missed something?

No, but it should be $n=0$ in front.

And similarly,



Wouldn't the second $\mathbb{R}$ be $\mathbb{R}^2$?

No, I think this is right.

 

[Dods]

If $n=1$, then we have $\mathbb{R}^0 = \{f:\{0\}\rightarrow \mathbb{R}~\vert~\text{f is a function}\}$.

Now, if $n=0$, then we have $\mathbb{R}^0$ is the set of functions $f:\emptyset\rightarrow \mathbb{R}$. What in Earth is a function from the emptyset to $\mathbb{R}$? Does such a beast exist? Yes, it does. Recall that a function is just a subset of the cartesian product (with some properties that don't matter here). So such a function $f$ would be a subset of $f\subseteq \emptyset \times \mathbb{R}$. But one can prove that $\emptyset \times \mathbb{R} = \emptyset$. So a function $f:\emptyset \rightarrow \mathbb{R}$ is just a subset from the empty set. There is only one subset from the empty set, namely the empty set itself. So $\emptyset$ is a function from $\emptyset$ to $\mathbb{R}$ (as crazy as it sounds). So $\mathbb{R}^0$ is then the set of all such functions, so

\mathbb{R}^0 = \{\emptyset\} = \{0\}

You said that n = 1 should be n =0.
We would then have:

First quote:
If n = 0, $\mathbb{R}^0 = \{f:\{0\}\rightarrow \mathbb{R}~\vert~\text{f is a function}\}$

Second quote:
If n = 0, $\mathbb{R}^0 = \{f:\emptyset \rightarrow \mathbb{R}~\vert~\text{f is a function}\}$

It seems I don't understand this properly then.

 

[micromass]

You said that n = 1 should be n =0.
We would then have:

First quote:
If n = 0, $\mathbb{R}^0 = \{f:\{0\}\rightarrow \mathbb{R}~\vert~\text{f is a function}\}$

Second quote:
If n = 0, $\mathbb{R}^0 = \{f:\emptyset \rightarrow \mathbb{R}~\vert~\text{f is a function}\}$

It seems I don't understand this properly then.

Oops, sorry for the confusion. You're right. It should be

\mathbb{R}^1 = \{f:\{0\}\rightarrow \mathbb{R}~\vert~\text{f is a function}\}

and

\mathbb{R}^0 = \{f:\emptyset \rightarrow \mathbb{R}~\vert~\text{f is a function}\}

 

[Dods]

Ok, thanks micromass :) Your explanations are great! they're really helping me.
The function related definition is really pretty in my opinion.
I see what you mean - you mean that for something like an ordered pair we have a good idea of what it's properties should be, and the set-builder definition is a way of formalizing it.

So if I understand correctly, we intuitively want R^2 to be the 2-dimensional plane/set of all real valued ordered pairs, so it doesn't matter too much how we formalize this idea, and one way of representing the set of all real valued ordered pairs is

\mathbb{R}^2 = \{f:\{0,1\} \rightarrow \mathbb{R}~\vert~\text{f is a function}\}

and we'd have

f = \{(0,x),(1,y)\}

 

[micromass]

Right, I think you got it.

We can of course also make sense of $\mathbb{R}^\mathbb{N} = \{f:\mathbb{N}\rightarrow \mathbb{R}~\vert~f~\text{is a function}\}.$

We usually denote $f(n) = x_n$ and we can write $f$ as $(x_n)_n$. So the sequences you know from calculus are actually functions from $\mathbb{N}$ to $\mathbb{R}$.
This works for any set, and not just for $\mathbb{R}$ and $\mathbb{N}$!

 

[Dods]

The definition of the natural numbers got me thinking, how would addition be defined in set theory?
Do you think I should try coming up with something or would it be way over my head (in which case, I'd be very interested if you could please explain it)?

:)

 

[micromass]

The definition of the natural numbers got me thinking, how would addition be defined in set theory?
Do you think I should try coming up with something or would it be way over my head (in which case, I'd be very interested if you could please explain it)?

:)

Now we're getting deep into the realm of set theory. Recall that we made the following definitions: $0 = \emptyset$ and if $n$ is defined, then $n+1 = n\cup \{n\}$. We want to define

\mathbb{N} = \{0,1,2,3,4,...\}.

Notice that $\mathbb{N}$ should satisfy the following:
1) $\emptyset \in \mathbb{N}$
2) If $x\in \mathbb{N}$, then $x\cup \{x\}\in \mathbb{N}$.
Furthermore, we intuitively want this to determine $\mathbb{N}$ entirely.

For this, we make a definition. We call a set A to be "inductive" if it satisfies (1) and (2) (with A replaced for $\mathbb{N}$. It is an axiom of set theory that at least one inductive set exists (this is called the axiom of infinity, if we would not accept this axiom, then infinite sets would not exist in mathematics).

Now we can define
\mathbb{N} = \bigcap \{B\subseteq A~\vert~B~\text{is inductive}\}
The intersection is well-defined because the set is nonempty (recall that $\bigcap\emptyset$ is not defined in mathematics).

Now we can easily prove that $\mathbb{N}$ is the smallest inductive set. So if $B$ is any inductive set (not necessarily part of A), then $\mathbb{N}\subseteq B$.

You may not recognize it, but the above statement is actually the principle of mathematical induction. Indeed, assume that P is a property such that P is true for $0$ and such that if P is true for $n$, then it is true for $n+1$. Then we can form the set
B=\{x\in \mathbb{N}~\vert~P~\text{is true for}~x\}
Because of the conditions we put on P, we see that B is an inductive set. Thus we have that $\mathbb{N}\subseteq B$ (and actually equality holds). So we see that any $n\in \mathbb{N}$ is in $B$, so P holds for any $n\in \mathbb{N}$.

Now we can go on and put the usual operations of + and . on the natural numbers. And we can also put an ordering relation $\leq$ on the natural numbers. All of this is done in great detail in set theory. Afterwards, we can construct the integers, the rational numbers, the real numbers and the complex numbers.

 

[Dods]

The things we discussed about R^1 and R^2 got me thinking about defining R^n...surely it wouldn't be a matter of just replacing "2" with n? You'd need to define an an ordered n-tuple and put some qualifer(s) on n, wouldn't you?

Also, in general I don't know how to discuss sets (or unions, or whatever) with n elements. Somewhere I saw a format something like "all X_i :~i \in \mathbb{N}, 1 \leq i \leq n" and I can't remember exactly! It's driving me nuts. Anyway, do you have any idea what this should look like?
Thanks.

 

[micromass]

The things we discussed about R^1 and R^2 got me thinking about defining R^n...surely it wouldn't be a matter of just replacing "2" with n? You'd need to define an an ordered n-tuple and put some qualifer(s) on n, wouldn't you?

Why doesn't this work?

\mathbb{R}^n = \{f:\{0,...,n-1\}\rightarrow \mathbb{R}~\vert~f~\text{is a function}\}.

Also, in general I don't know how to discuss sets (or unions, or whatever) with n elements. Somewhere I saw a format something like "all X_i :~i \in \mathbb{N}, 1 \leq i \leq n" and I can't remember exactly! It's driving me nuts. Anyway, do you have any idea what this should look like?

Sorry, I have no idea what you're trying to say here. Can you describe what you are trying to write down?

 

[Dods]

Why doesn't this work?

\mathbb{R}^n = \{f:\{0,...,n-1\}\rightarrow \mathbb{R}~\vert~f~\text{is a function}\}.

Right, that makes sense, I totally missed that. :)

If we wanted that in \{x \in B~\vert~P(x)\} format we could write \{f \in \mathcal{P}(\{0,...,n-1\}\times \mathbb{R})~\vert~\text{f is a function}\} right?

Sorry, I have no idea what you're trying to say here. Can you describe what you are trying to write down?

Yeah, sorry - I've seen this somewhere and I can't remember where, Basically instead of listing the elements X_1, X_2, X_3, X_4 \in T they did something like X_i \in T \wedge i \in \mathbb{N}, 1 \leq i \leq 4

 

[micromass]

Yeah, sorry - I've seen this somewhere and I can't remember where, Basically instead of listing the elements X_1, X_2, X_3, X_4 \in T they did something like X_i \in T \wedge i \in \mathbb{N}, 1 \leq i \leq 4

Something like $X_i,~1\leq i \leq 4$?

 

[Dods]

Yes. Thanks. :) (You don't need to specify that i is in the natural numbers?)

I quite want to try constructing an area function (that is, a function that would satisfy the axioms specified in wiki's page of Area http://en.wikipedia.org/wiki/Area#Formal_definition).

Do you think that is something I could try doing on PF? (possibly in another thread if needed).

 

[micromass]

Yes. Thanks. :) (You don't need to specify that i is in the natural numbers?)

Formally, you do. But it is usually clear from context.

I quite want to try constructing an area function (that is, a function that would satisfy the axioms specified in wiki's page of Area http://en.wikipedia.org/wiki/Area#Formal_definition).

Do you think that is something I could try doing on PF? (possibly in another thread if needed).

Sure, you can do that. Although the area function in that link isn't very useful. The useful concept is the 2-dimensional Lebesgue measure. that is an area function that is extremely useful in mathematics. But go ahead and construct it.

 

[Dods]

Ok, great :) I'll get on it.

I have a question though. There are shapes whose area has to be found through integration. Will this area function also have to account for them? That would basically mean I'd need to construct an integral.

 

[micromass]

Ok, great :) I'll get on it.

I have a question though. There are shapes whose area has to be found through integration. Will this area function also have to account for them? That would basically mean I'd need to construct an integral.

The Lebesgue measure can be constructed without any notion of integral. So the definition of area doesn't need integration. However, if you want to explicitely find the area, then integrals might be handy. I mean: with measures you can show that something has an area, but you might not be able to find it.

The logical procedure is: first you construct measures, then you construct the associated integral.

 

[Dods]

Right, so if I just want to show that an object has area I wouldn't need integrals, although for figuring out what that area is they might be needed.

I though of starting with defining a rectangles area, and then expanding that definition to other shapes. Do you think that's a good idea or would I need to start at a level where I define some concepts and then use them to show that a rectangle has an area, and then expand the definition?

I very much appreciate you advice on this :)

 

[micromass]

Right, so if I just want to show that an object has area I wouldn't need integrals, although for figuring out what that area is they might be needed.

I though of starting with defining a rectangles area, and then expanding that definition to other shapes. Do you think that's a good idea or would I need to start at a level where I define some concepts and then use them to show that a rectangle has an area, and then expand the definition?

I very much appreciate you advice on this :)

Sure starting with rectangles is a good idea. In fact, try to start with rectangles whose sides are parallel to the x-axis and y-axis. So rectangles of the form $(a_1,a_2]\times (b_1,b_2]$ (I exclude the sides because it is technically easier).

 

[Dods]

Thanks micromass!

I don't think I'm reading that cartesian product right - wouldn't you need open intervals, not half-open ones to exclude the sides?

Also, at some point we defined a rectangle-like object before in this thread.

Anyway, I'll start coming up with it, I'll post the start of it soon. :)

 

[micromass]

Thanks micromass!

I don't think I'm reading that cartesian product right - wouldn't you need open intervals, not half-open ones to exclude the sides?

I only exclude two of the sides.

I do it this way because things end up to be easier. Why? Because you can easily "paste" together rectangles this way. For example, the union of $(0,1]\times (0,1]$ and $(1,2]\times (0,1]$ is again a rectangle of the same sort and we have no overlapping. If we take closed or open rectangles, then we have the sides to worry about.

Also, at some point we defined a rectangle-like object before in this thread.

OK, but that's not the same rectangle as we'll use here.

 

[Dods]

I've been doing another round of exams (last day of high school on Sunday! :]). I'll come back and update soon. I'm trying to come up with definitions that are readily generalizable to n-dimensional area/volume/equivalent.

Dods

 

[Dods]

Heya,

I finally have time to come back to this :) !

I have a question:

If I have a set that is a union of several sets, say Y = A \cup B \cup C is there a way using set-builder notation to refer to the set (let's call it S_Y) of all the sets that are part of the union?

That is S_Y = \{A, B, C\} in this case. I'm not sure how to approach this, especially given that the same set can be described by different unions: we can have Y = A \cup B \cup C = E \cup F.

Thanks!

 

[micromass]

If you're only given the set $Y$, then there is no way to find the sets whose union give $Y$.

I mean, you could find all sets whose union is $Y$, but you're not going to find the specific sets.

 

[Dods]

By the way, in general if I have a list of elements where each successive element is connected to the one before it by the same operation, like:


x_1 + x_2 + x_3 + x_4 + x_5 + x_6

or


x_1 \cdot x_2 \cdot x_3 \cdot x_4 \cdot x_5

or even

x_1 \times x_2 \times x_3 \times x_4

and you want to talk about the same list with n elements, it considered rigorous to express it as something like:

x_1 + ... + x_n

or

x_1 \times ... \times x_n

? Or is this shorthand for a more rigorous statement?

The issue just came up with some sets I was considering.

Thanks :)

 

[micromass]

Good question. Let's look at the definition of addition. Addition is defined as a binary operation. What this means is that it is a function $+:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$. So to $(a,b)$, we associate $+(a,b)$. Since this looks weird, we just write it as $a+b$.

From this definition, we see that something like $a+b+c$ does not make sense. We can't define it as $+(a,b,c)$, since $+$ only operates on $2$ terms. However, the following is defined:

(a+b)+c

In more primitive notation, we have $+(+(a,b),c)$. So we let $+$ operate on $+(a,b)$ and $c$. But we can also define

a+(b+c)

which of course means $+(a,+(b,c))$. And it turns out that both are equal. So, we have

+(+(a,b),c) = +(a,+(b,c))

It is for this reason, that we can invent the notation $a+b+c$ to mean the value of the above. This is an abuse of notation. Purely rigorously, we can't write it like $a+b+c$. But of course, the abuse is very convenient.

There is an operation, called the Lie bracket (denoted by $[a,b]$). What it is, is of no importance. But it is also a binary operation. However, it is not associative in general. Thus we have

[a,[b,c]] \neq [a,[b,c]]

This is a major example of a nonassociative operation. Because it is nonassociative, we do not write it as $[a,b,c]$. We always write it as a binary operation.

The cartesian product of sets is a bit more subtle. There, it does not holds that

A\times (B\times C) = (A\times B)\times C

So the operation fails to be associative. However, we do write it as $A\times B\times C$. The reason is that there is a natural bijection between the two sets. That is, we can identity the element $(a,(b,c))\in A\times (B\times C)$ with the element $((a,b),c)\in (A\times B)\times C$. This is a canonical or natural identificiation. So when we write $A\times (B\times C) = (A\times B)\times C$, we always keep this identification in mind.

 

[Dods]

Thank you for an excellent explanation!

Good question. Let's look at the definition of addition. Addition is defined as a binary operation. What this means is that it is a function $+:\mathbb{R}\times\mathbb{R}\rightarrow\mathbb{R}$.

Could you say that for any_1 binary operation *, we can define it as $*:S \times S\rightarrow S$?

That is, a function that takes two elements of the same set to another element of that set?

_1 I mean as a basic definition - I assume that you'd need to amend this for an operation like division over the real numbers (but not for division over the "real numbers not including zero") ?

In more primitive notation, we have $+(+(a,b),c)$. So we let $+$ operate on $+(a,b)$ and $c$. But we can also define

a+(b+c)

which of course means $+(a,+(b,c))$. And it turns out that both are equal. So, we have

+(+(a,b),c) = +(a,+(b,c))

In Spivak's Calculus book he shows this for a + b + c + d using properties of real numbers (I did it as an exercise and then checked the book). Is this how it turns out both are equal?

There is an operation, called the Lie bracket (denoted by $[a,b]$). What it is, is of no importance. But it is also a binary operation. However, it is not associative in general. Thus we have

[a,[b,c]] \neq [a,[b,c]]

This is a major example of a nonassociative operation. Because it is nonassociative, we do not write it as $[a,b,c]$. We always write it as a binary operation.

Do you mean [a,[b,c]] \neq [[a,b],c]?

The cartesian product of sets is a bit more subtle. There, it does not holds that

A\times (B\times C) = (A\times B)\times C

So the operation fails to be associative. However, we do write it as $A\times B\times C$. The reason is that there is a natural bijection between the two sets. That is, we can identity the element $(a,(b,c))\in A\times (B\times C)$ with the element $((a,b),c)\in (A\times B)\times C$. This is a canonical or natural identificiation. So when we write $A\times (B\times C) = (A\times B)\times C$, we always keep this identification in mind.

That's very interesting. If for a minute we think about ordered 4-tuples as nested ordered pairs, if the definition "nest"s them from the left

(((x,y)z)w)

or from the right

(x(y(z,w)))

it does change the set's definition but we associate both ways of defining it with the ordered 4-tuple (x,y,z,w)...

Is that relevant or have I started a whole other thing here?

 

[micromass]

Could you say that for any_1 binary operation *, we can define it as $*:S \times S\rightarrow S$?

That is, a function that takes two elements of the same set to another element of that set?

Yes, that is essentially how a binary operation is defined.

_1 I mean as a basic definition - I assume that you'd need to amend this for an operation like division over the real numbers (but not for division over the "real numbers not including zero") ?

Division is usually not seen as a binary operation. It is possible to see it as a binary operation, but this is annoying. In abstract algebra, we usually look at the unary operation

i:\mathbb{R}\setminus \{0\}\rightarrow \mathbb{R}:x\rightarrow 1/x

then division $a/b$ is just $a*i(b)$. This is the usual way to deal with this.

In Spivak's Calculus book he shows this for a + b + c + d using properties of real numbers (I did it as an exercise and then checked the book). Is this how it turns out both are equal?

Yes.

Do you mean [a,[b,c]] \neq [[a,b],c]?

Yes, sorry.

That's very interesting. If for a minute we think about ordered 4-tuples as nested ordered pairs, if the definition "nest"s them from the left

(((x,y)z)w)

or from the right

(x(y(z,w)))

it does change the set's definition but we associate both ways of defining it with the ordered 4-tuple (x,y,z,w)...

Is that relevant or have I started a whole other thing here?

No, what you said is exactly right. We "associate" $(((x,y),z),w)$ and $(x,y,z,w)$. So although they aren't the same thing, we do identify them informally.

 

[Dods]

I've been reviewing some of the material we've covered and I had some miscellaneous questions:

First of all, I've seen A \setminus D type sets and while I've understood them from context, how is the "\" operation defined?

-----

Second, let's say we want the set of all functions from \{0,1\} to \mathbb{R}. We could write \{f: \{0,1\} \rightarrow \mathbb{R} \vert f \ \text{is a function} \} and that'd be clear enough. But if we were being pedantically rigorous and we wanted that in $\{x \in B~\vert~P(x)\}$ format, could we write: \{f \in \mathcal{P}(\{0,1\}\times \mathbb{R})~\vert~\text{f is a function}\}, seeing as our $f$ is always a subset of $\{0,1\} \times \mathbb{R}$?

As always, thanks!

 

[micromass]

I've been reviewing some of the material we've covered and I had some miscellaneous questions:

First of all, I've seen A \setminus D type sets and while I've understood them from context, how is the "\" operation defined?

The definition is A\setminus D = \{x\in A~\vert~x\notin D\}

Second, let's say we want the set of all functions from \{0,1\} to \mathbb{R}. We could write \{f: \{0,1\} \rightarrow \mathbb{R} \vert f \ \text{is a function} \} and that'd be clear enough. But if we were being pedantically rigorous and we wanted that in $\{x \in B~\vert~P(x)\}$ format, could we write: \{f \in \mathcal{P}(\{0,1\}\times \mathbb{R})~\vert~\text{f is a function}\}, seeing as our $f$ is always a subset of $\{0,1\} \times \mathbb{R}$?

Yes, that is correct.