A couple of questions on set theory

[HallsofIvy]

A formal definition of surjective (bearing in mind the earlier definition of range):

A function $f:A\rightarrow B$ is surjective if:

$$B=f(A) = \{b\in B~\vert~\exists a \in A:~(a,b)\in f\}$$

Two slight notation questions-

Is $f(A)$ clear enough to just have:

$$B=f(A)$$?

For "f surjective" it is sufficient to say
$$B\subseteq f(A)$$

Also, equality means every element of $B$ is in $B=f(A)$ and vice versa. But by definition every element of $f(A)$ is in $B$. So couldn't we define surjective by:

$$\nexists b \in B:~b \notin f(A)$$?

I assume the \nexists operator means "there does not exist", I just modified the \exists operator until something worked.

 

[Dods]

Ok, I see - you can say $B\subseteq f(A)$ because that's equivalent to saying that there doesn't exist an element in $B$ that is not in $f(A)$.

Thanks HallsofIvy.


Definition of "is a straight line in $\mathbb{R^2}$":

s is a straight line in $\mathbb{R^2}$ if -

$$(1)~ s\subseteq \mathbb{R^2}:~ \forall a\in \mathbb{R}:~\exists ! b\in \mathbb{R}:~(a,b)\in s \\ (2)~ \forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y,m,b \in \mathbb{R} :~(p = (x,y) \wedge y = mx+b)]$$

Is that right?

 

[micromass]

Definition of injective:

a function $f$ is injective if -

$$\forall (x,y), (a,b) \in f:~ x \neq a \rightarrow y \neq b$$

How's that?

That's ok.

Ok, I see - you can say $B\subseteq f(A)$ because that's equivalent to saying that there doesn't exist an element in $B$ that is not in $f(A)$.

Thanks HallsofIvy.


Definition of "is a straight line in $\mathbb{R^2}$":

s is a straight line in $\mathbb{R^2}$ if -

$$(1)~ s\subseteq \mathbb{R^2}:~ \forall a\in \mathbb{R}:~\exists ! b\in \mathbb{R}:~(a,b)\in s \\ (2)~ \forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y,m,b \in \mathbb{R} :~(p = (x,y) \wedge y = mx+b)]$$

Is that right?

I'm not seeing the significance of (1). Why is it needed? It's not correct by the way, since a line like x = 0 doesn't satisfy (1) or (2). So you need to fix this so that this line counts as well.

 

[Dods]

For some reason I was thinking of defining a "linear function". That was a silly mistake on my part. (Would the definition work then?)

Other than splitting up "is a straight line" into two cases, one for lines like y=mx+b and one for lines like x=c, no definition jumps out at me..any advice?

 

[micromass]

For some reason I was thinking of defining a "linear function". That was a silly mistake on my part. (Would the definition work then?)

Yes, it works then.

Other than splitting up "is a straight line" into two cases, one for lines like y=mx+b and one for lines like x=c, no definition jumps out at me..any advice?

What about describing a line as $ax + by = c$ ?

 

[Dods]

At first glance I thought:
$$s\subseteq \mathbb{R^2}:~ \forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y,a,b,c \in \mathbb{R} :~(p = (x,y) \wedge ax+by = c)]$$

But then, let's say we had a set $$S=\{p \in \mathbb{R^2}~\vert~\exists x,y,a,b,c \in \mathbb{R}:~(p=(x,y) \wedge ax+by=c)\}$$

This obviously satisfies these criteria for being a "straight line". However, we could then have the points which are all in S:

(1,1) - because for a=5 and b=2, there is a suitable c (7 ->5*1 + 2*1 =7).
(2,1) - because for a=3 and b=3, there is a c (9 -> 3*2+3*1=9)
(1,6) - because for a=7 and b=0.5 there is a c(10-> 7*1+0.5*6=10)

and we definitely don't want that...

How about:

$$s\subseteq \mathbb{R^2}:~ \forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y \in \mathbb{R} \wedge \exists! a,b,c :~(p = (x,y) \wedge ax+by = c)]$$

 

[micromass]

That won't solve it. Notice that $a x + by = c$ if and only if $2a + 2b y = 2c$. So the $a,b,c$ will never be unique.

I think the solution is moving the $a,b,c$ to the front. So:

$$\exists a,b,c\in \mathbb{R}:~\forall p\in \mathbb{R}^2:~...$$

 

[Dods]

I'm probably just tired, but I don't see how that helps...

Sorry for being slow on the uptake here.
Dods.

 

[micromass]

You want your a,b,c to be independent of the point p in s.

Under your definition

$$\forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y \in \mathbb{R} \wedge \exists a,b,c :~(p = (x,y) \wedge ax+by = c)]$$

we can prove that $s=\mathbb{R}^2$. Indeed, given any $p\in \mathbb{R}^2$, then we can write $p = (x,y)$. Furthermore, we can take $a = y$, $b = -x$ and $c=0$. Then it is certainly satisfied that $ax+ by = c$.

The problem with the above is that my $a,b,c$ were depended of my $x$ and $y$. That is: if I change $x$ and $y$, then my $a,b,c$ change too. For example, if $p=(1,2)$, then it satisfies $2x - y = 0$ and $p=(0,3)$ satisfies $3x = 0$.

What we want is to find $a,b,c$ that can not use $x,y$. This is why you move them in front:

$$\exists a,b,c\in \mathbb{R}:~\forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y \in \mathbb{R}:~(p = (x,y) \wedge ax+by = c)]$$

Now I have to find $a,b,c$ such that ALL $(x,y)\in s$ satisfy $ax+ by = c$. This will give the right answer.
Previously, I had to find $a,b,c$ for each individual $(x,y)$.

 

[Dods]

micromass, congratulations on becoming a Mentor!

You want your a,b,c to be independent of the point p in s.

Under your definition

$$\forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y \in \mathbb{R} \wedge \exists a,b,c :~(p = (x,y) \wedge ax+by = c)]$$

we can prove that $s=\mathbb{R}^2$. Indeed, given any $p\in \mathbb{R}^2$, then we can write $p = (x,y)$. Furthermore, we can take $a = y$, $b = -x$ and $c=0$. Then it is certainly satisfied that $ax+ by = c$.

The problem with the above is that my $a,b,c$ were depended of my $x$ and $y$. That is: if I change $x$ and $y$, then my $a,b,c$ change too. For example, if $p=(1,2)$, then it satisfies $2x - y = 0$ and $p=(0,3)$ satisfies $3x = 0$.

What we want is to find $a,b,c$ that can not use $x,y$. This is why you move them in front:

$$\exists a,b,c\in \mathbb{R}:~\forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y \in \mathbb{R}:~(p = (x,y) \wedge ax+by = c)]$$

Now I have to find $a,b,c$ such that ALL $(x,y)\in s$ satisfy $ax+ by = c$. This will give the right answer.
Previously, I had to find $a,b,c$ for each individual $(x,y)$.

I understand your point (I think) that all our $(x,y)$ need to be constrained by $a,b,c$ and that if we do that then our definition only accepts as points in our line those $(x,y)$ for which our $a,b,c$ make sense.

It isn't obvious to me that the statement $$\exists a,b,c\in \mathbb{R}:~\forall p \in \mathbb{R^2}:~p\in s~\leftrightarrow [\exists x,y \in \mathbb{R}:~(p = (x,y) \wedge ax+by = c)]$$ would imply this.

Hmm. I may be reading it wrong, paying to much attention to the ":"s.

Ok - Is the difference in saying, "There exists a,b,c for all p" rather than "for all p there exists a,b,c"? because then the first statement requires all p to be according to a,b,c?

* By the way, how does the Thanks button system work? I just noticed it.

Congratz again. :)

 

[micromass]

micromass, congratulations on becoming a Mentor!

Thanks!

Ok - Is the difference in saying, "There exists a,b,c for all p" rather than "for all p there exists a,b,c"? because then the first statement requires all p to be according to a,b,c?

Yes. Maybe we can make it clear by an easier statement:

$$\forall x\in \mathbb{R}:~\exists y\in \mathbb{R}:~x+y=0$$

this is a true statement. For all $x$, there certainly exists such a $y$. Indeed, we take $y=-x$.

However,

$$\exists y\in \mathbb{R}:~\forall x\in \mathbb{R}:~x+y=0$$

This is false. This says that there exists $y$ such that ALL $x$ must satisfy $x+y=0$. And the $y$ must be good for all $x$. The difference with the previous case, is that previously, we could have a $y$ good for only one $x$. Now it must be good for all $x$.

* By the way, how does the Thanks button system work? I just noticed it.

You go to a post you like, then you click thanks and fill in a reason. It's that simple :)

 

[Dods]

I think I get it now. I was interpreting the symbols as if a statement like

$$\exists y\in \mathbb{R}:~\forall x\in \mathbb{R}:~x+y=0$$

meant, "for all x in R, you can pick a y so that x+y=0"

when it actually means:

"there exists a y such that -> for all x in R, x+y=0 is true"

-------
Example of such a formula:

$$\exists y\in \mathbb{R}:~\forall x\in \mathbb{R}:~x+y=x$$

Any ideas what to try defining or learning about now?

Edit - and I think I got the thanks button figured out. :)

 

[micromass]

Any ideas what to try defining or learning about now?

I think that you know enough logic now. If your goal is set theory, then perhaps it would be a good idea to get a set theory book and work through that. Halmos is a decent book, but it's not axiomatic, which is a shame. Hrbacek and Jech is my favorite book on the topic.

If you're interested in logic, then you can also read a logic book. That's also very interesting. It's up to you, really.

 

[Dods]

Set theory and logic do interest me, but I'm also very interested in using set theory to ground other areas of math - for instance building up the concepts of calculus using set theory, like defining area as a set function or defining integrals. Is that a thing? How would I go about doing/learning about that?

 

[micromass]

Set theory and logic do interest me, but I'm also very interested in using set theory to ground other areas of math - for instance building up the concepts of calculus using set theory, like defining area as a set function or defining integrals. Is that a thing? How would I go about doing/learning about that?

Set theory grounds other areas of math. For example, you define what functions are relations are. You construct the natural numbers and the usual operations. You construct the integers, rational numbers and real numbers. That's all done in set theory.

However, integrals and area is usually not covered in set theory. If you want to define area as a set function, then you should look for measure theory. That's exactly what is done there. Nice books here are Bartle: https://www.amazon.com/dp/0471042226/?tag=pfamazon01-20 and Jones: https://www.amazon.com/dp/0763717088/?tag=pfamazon01-20

 

[Dods]

Thank you :)

 

[Dods]

I got a copy of Lebesgue Integration on Euclidean Space. I've been looking through it (and I had such a huge grin when I skimmed the section on countable sets and saw their definition of "countable", hehe :) ). Right now I'm dealing with the start of the book - unions and other relations between sets (De Morgan's Laws, difference and complements). I'll update once I prove a few things.

:) Dods.

 

[Dods]

$\mathbb{R^2}$ can be interpreted as a 2-dimensional plane. I've seen $\mathbb{R}$ interpreted as the real line. Can we meaningfully interpret $\mathbb{R^0}$? Would it be interpreted as a point? Edit: That raises the question, "point" in/of what?

Thanks

 

[micromass]

$\mathbb{R^2}$ can be interpreted as a 2-dimensional plane. I've seen $\mathbb{R}$ interpreted as the real line. Can we meaningfully interpret $\mathbb{R^0}$? Would it be interpreted as a point? Edit: That raises the question, "point" in/of what?

Thanks

The definition of $\mathbb{R}^0$ is

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

So indeed, $\mathbb{R}^0$ is a point. The unique point in there is called $0$.

 

[Dods]

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

 

[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.