A couple of questions on set theory

[Dods]

Those look interesting to prove - I already have an idea of how to prove the infinite set property - can I assume the set of all natural numbers is infinite?

Just to make sure I understand the definitions-

Surjective means that the range is equal to the codomain.

Injective means that no element in the range is paired with multiple elements in the codomain, that is, you don't have (a, b), (c,b) with c \neq a

The function f:\mathbb{R} \rightarrow \mathbb{R} that takes x to x^2 (by the way, what is the shorthand for this? f(x)=x^2?) is not surjective (because there are elements in the codomain \mathbb{R} that are not in the range, for example -3) and is not injective because we have for example (2, 4), (-2,4).

The function f:\mathbb{R} \rightarrow \mathbb{R_+} that takes x to x^2 is surjective because the range equals the codomain, and is not injective because we have (2, 4), (-2,4).*

The function f:\mathbb{R} \rightarrow \mathbb{R} that takes x to \sqrt{x} is injective because there are not any two distinct real numbers whose square root is equal, and is not surjective because there are elements in the codomain (the negative real numbers) that are not in the range.

The identity function is surjective and injective and therefore bijective.

* \mathbb{R_+} would be the set on nonnegative real numbers.

 

[micromass]

Those look interesting to prove - I already have an idea of how to prove the infinite set property - can I assume the set of all natural numbers is infinite?

It's very difficult to define an infinite set. People have already called me evil for asking you this. But I'm very interested what you come up with. Let's say that the natural numbers are indeed infinite, how would you use this to define infinite?

Just to make sure I understand the definitions-

Surjective means that the range is equal to the codomain.

Right

Injective means that no element in the range is paired with multiple elements in the codomain, that is, you don't have (a, b), (c,b) with c \neq a

The word in bold should be domain, but I think you meant that.

The function f:\mathbb{R} \rightarrow \mathbb{R} that takes x to x^2 (by the way, what is the shorthand for this? f(x)=x^2?) is not surjective (because there are elements in the codomain \mathbb{R} that are not in the range, for example -3) and is not injective because we have for example (2, 4), (-2,4).

Right. The right shorthand is $f:\mathbb{R}\rightarrow \mathbb{R}: x\rightarrow x^2$. The shorthand $f(x) = x^2$ is also used but I don't like it. I dislike it because it gives no information on what the domain and the codomain is exactly. And you notice now already that the domain and the codomain are pretty crucial things when you determine surjective or injective.

The function f:\mathbb{R} \rightarrow \mathbb{R_+} that takes x to x^2 is surjective because the range equals the codomain, and is not injective because we have (2, 4), (-2,4).*

Right.

The function f:\mathbb{R} \rightarrow \mathbb{R} that takes x to \sqrt{x} is injective because there are not any two distinct real numbers whose square root is equal, and is not surjective because there are elements in the codomain (the negative real numbers) that are not in the range.

Right. Another such example would be $f:\mathbb{R}_+ \rightarrow \mathbb{R}:x\rightarrow x^2$. This is injective but not surjective.

The identity function is surjective and injective and therefore bijective.

Right. Another example is $f:\mathbb{R}_+ \rightarrow \mathbb{R}_+:x\rightarrow x^2$.

 

[Dods]

Yes, I did mean domain and not codomain.

I want to tackle the "infinite set" definition first, but I'll need some properties I haven't defined yet. Can I use a property like "is surjective" (for example) on the understanding I'll define it rigorously later?

 

[micromass]

Yes, I did mean domain and not codomain.

I want to tackle the "infinite set" definition first, but I'll need some properties I haven't defined yet. Can I use a property like "is surjective" (for example) on the understanding I'll define it rigorously later?

Sure, you can do that.

 

[Dods]

Ok then. :) I'll give a definition in words first not with logical symbols (no point taking ages to get the Latex right and having the wrong definition :) ).

A set A is an infinite set if there exists a surjective f:A \rightarrow \mathbb{N} or a injective f:\mathbb{N} \rightarrow A.

Or if there exists a bijection between the two sets, although this seems like a different kind of infinity - I'd think the real numbers could be infinite according to the first definition, but I don't see a bijection between them and the natural numbers (I worked out stuff about this).

If I'm right (I'm probably not :P) I'll go into my whole train of thought..

...

 

[micromass]

Ok then. :) I'll give a definition in words first not with logical symbols (no point taking ages to get the Latex right and having the wrong definition :) ).

A set A is an infinite set if there exists a surjective f:A \rightarrow \mathbb{N} or a injective f:\mathbb{N} \rightarrow A.

Or if there exists a bijection between the two sets, although this seems like a different kind of infinity - I'd think the real numbers could be infinite according to the first definition, but I don't see a bijection between them and the natural numbers (I worked out stuff about this).

If I'm right (I'm probably not :P) I'll go into my whole train of thought..

...

That's indeed a correct definition of infinite, and the most simple definition. This definition was historically given by Dedekind, hence we call it "Dedekind infinite". It is equivalent to other notions of infinite though.

You are right that the real numbers are infinite using this definition (there is an injection $\mathbb{N}\rightarrow \mathbb{R}$), but there is no bijection between the naturals and the reals. This was proven by Cantor. Sets which have a bijection between them and the naturals are called "countably infinite".

 

[Dods]

You have no idea how pleased I am that I got it right :) :)

 

[Dods]

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)?

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.

 

[micromass]

Yes to all questions.

 

[Dods]

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?

 

[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?