Question about proof from a guy with a highschool education

[reenmachine]

That is correct. It might be a good practice to draw those relations in a diagram.

Would this be okay?

Red for R and blue for R'.

Basically , the blue arrows (R') are supposed to complete the red ones (R).

And the inverse would be the red arrows that switched directions?

Are the blue arrows that switched directions the inverse of R' ?

 

[micromass]

They define R-1 by the bolded part.Do they justify <b,a> being R-1 by stating that <a,b> ∈ R? If that's the case , if ,

A = {1,2} and B = {2,3} , isn't <2,2> both ∈ R and ∈ R'?

Not necessarily. If (2,2) is in R, then it is also in $R^{-1}$. But if (2,2) is not in R, then it's also not in $R^{-1}$.

For example, take $A=\{1,2\}$ and $B=\{2,3\}$ and $R=\{(1,2),(2,2)\}$, then $R^{-1} = \{(2,1),(2,2)\}$.

Graphically, if you have drawn R, then you obtain $R^{-1}$ by just exchanging A and B.

R' = ''≥ than'' on N (i.e., on N × N)

?



R -1 = ''> than'' on N (i.e., on N × N)

?

Yes.

 

[micromass]

Would this be okay?

Red for R and blue for R'.

Basically , the blue arrows (R') are supposed to complete the red ones (R).

And the inverse would be the red arrows but going in the other direction?

Are the blue arrows (going in the other direction) the inverse of the R' ?

Yes.

 

[reenmachine]

Suppose A = {1,2} and B = {1,2} , and R in A × B = {<1,2>,<1,1>,<2,1>,<2,2>} , what would R' be?

Would R-1 = R ?

 

[micromass]

Suppose A = {1,2} and B = {1,2} , and R in A × B = {<1,2>,<1,1>,<2,1>,<2,2>} , what would R' be?

It would be the empty set. Yes, the empty set is a valid relation. Any subset of $A\times B$ is a valid relation, and the empty set is one of those subsets.

Would R-1 = R ?

Yes.

 

[reenmachine]

Not necessarily. If (2,2) is in R, then it is also in $R^{-1}$. But if (2,2) is not in R, then it's also not in $R^{-1}$.

For example, take $A=\{1,2\}$ and $B=\{2,3\}$ and $R=\{(1,2),(2,2)\}$, then $R^{-1} = \{(2,1),(2,2)\}$.

Good thing you picked up my typo , meant R-1 instead of R' but no harm done you responded to my question anyway.

thanks!

 

[reenmachine]

The textbook conclude the chapter on relations with this last paragraph:

We have focused so far on binary relations, i.e., sets of ordered pairs. In a similar way we could define ternary , quaternary or just n-place relations consisting respectively of ordered triples, quadruples or n-tuples. A unary relation R on a set A is just a subset of the set A

I think I randomly asked about this concept earlier in the thread.Basically they're just saying that:

A binary relation = ''1 thing that bears relation to 1 other thing'' ,

but that it's possible for as many things as you can imagine to bears relations with each others , which would give us something like <1,2,3,4,...,∞> or <∞,...,3,2,1> ( in any possible orders) instead of ordered pairs?

 

[Fredrik]

R-1

If you don't use LaTeX, you should at least use vBulletin's sub and sup tags: R^-1, x₀, (Quote to see the code).

but that it's possible for as many things as you can imagine to bears relations with each others , which would give us something like <1,2,3,4,...,∞> or <∞,...,3,2,1> ( in any possible orders) instead of ordered pairs?

They're just saying that the concept of "binary relation" can be generalized by saying that for each positive integer n, an n-ary relation is a subset of a Cartesian product of n sets. I haven't thought about whether it's also meaningful to generalize the definition further by considering Cartesian products of infinitely many sets.

 

[reenmachine]

If you don't use LaTeX, you should at least use vBulletin's sub and sup tags: R^-1, x₀, (Quote to see the code).

I'm sorry for not using LaTeX yet , I will start to use it and try to get comfortable with it soon enough.

I now have completed the relations chapter in the textbook , the next one is about functions.

Hopefully I can finish the textbook by tomorrow night and receive/start my books by monday.

 

[reenmachine]

Functions

2.3. Functions

Examples of functions:

f(x) =x^2 + 1
f(x) = the mother of x

Intuitively a function may be thought of as a “process” or as a correspondence.A function is generally represented in set-theoretic terms as a special kind of relation.

In the example above , the f in ''f(x) = x^2 + 1'' means that for element x , f squares it and add 1 to it?

What do they mean by ''f(x) = the mother of x''?

Definition:A relation F from A to B is a function from A to B if and only if it meets both of the following conditions:

1. Each element in the domain of F is paired with just one element in the range, i.e., from <a,b> ∈ F and <a,c> ∈ F follows that b=c.

2. The domain of F is equal to A , domF= A.

Suppose A = {1,2} and B = {3,4,5} , what would a function following the bolded condition look like?

Let's take for example {<1,3>,<2,3>} , how am I suppose to write or express it? <1,3> & <2,3> ∈ F? How am I formulating something to include both the F symbol (for function) and the ordered pairs?

Now let's take {<1,3>,<2,4>} , again from the universe A × B , the relation < wouldn't be a function because 3 and 4 aren't the same element in the range?

About the 2nd condition , why do domF have to be = to A for the relation from A × B to be a function?

Take the sets A and B used in the examples above and just switch their side for the sake of my argument , meaning that the domF would have to be equal to B for the relation from B × A to be a function.

If {<3,1>,<4,1>} , you have the same element in the range , yet the domF is 3 and 4 while B = {3,4,5}.But what's the difference between {<3,1>,<4,1>} and {<3,1>,<4,1>,<5,1>} as far as being a relation vs a function? Basically my question is why the 2nd condition exist?

 

[Fredrik]

In the example above , the f in ''f(x) = x^2 + 1'' means that for element x , f squares it and add 1 to it?

Yes.

What do they mean by ''f(x) = the mother of x''?

For example if x=Joffrey, then f(x)=Cersei, and if x=Arya, then f(x)=Catelyn. We have f(Joffrey)=Cersei, and f(Arya)=f(Sansa)=f(Bran)=Catelyn. Obviously they're just trying to help you develop an intuition about functions at this point. This isn't even real mathematics. The reason why this illustrates the idea of a function pretty well is that it's easy to understand that every person has exactly one biological mother.

Suppose A = {1,2} and B = {3,4,5} , what would a function following the bolded condition look like?

A function f from A into B can be defined e.g. by specifying f(1)=3, f(2)=5. This f can also be defined by $f=\left\{\langle 1,3\rangle,\langle 2,5\rangle\right\}$. However, if you e.g. define $g=\left\{\langle 1,3\rangle,\langle 1,5\rangle,\langle 2,5\rangle\right\}$, then g fails to satisfy condition 1, and is therefore not a function. It fails because the set g contains two ordered pairs that have the same first component but different second components. (This is like one person having two biological mothers). Note that the statement $\langle x,y\rangle\in g$ is also written as $g(x)=y$. So if g contains <1,3> and <1,5>, then things get weird, because it would imply that g(1)=3 and g(1)=5.

Let's take for example {<1,3>,<2,3>} , how am I suppose to write or express it? <1,3> & <2,3> ∈ F? How am I formulating something to include both the F symbol (for function) and the ordered pairs?

If you want to say that $F=\left\{\langle 1,3\rangle,\langle 2,3\rangle\right\}$, you can do it by saying that $F(1)=3$ and $F(2)=3$.

Now let's take {<1,3>,<2,4>} , again from the universe A × B , the relation < wouldn't be a function because 3 and 4 aren't the same element in the range?

Do you mean that in this example, you denote {<1,3>,<2,4>} by <? In that case, < is definitely a function. The fact that 3 and 4 aren't the same only means that < isn't a constant function.

I said that symbols like < are often used for binary relations, but they're almost never used for functions, so this notation looks odd here. It gets especially weird if we rewrite $\langle 1,3\rangle\in <$ as $<(1)=3$. This is a good time to denote the relation by a letter.

About the 2nd condition , why do domF have to be = to A for the relation from A × B to be a function?
...
Basically my question is why the 2nd condition exist?

Because we want to define something that associates exactly one element of B with each element of A, and if dom F is a proper subset of A, then there's an element of A that doesn't get associated with an element of B (a person that doesn't have a biological mother).

 

[micromass]

I want to add to Fredrik's excellent response that not every author demands condition (2) in functions. For example, in high school, we often worked with functions whose domain is not the entire set. For example, something like

f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow \frac{1}{x}

was a valid function.

However, the majority of mathematicians do demand conditionn (2). In that case, the above is not a function (since f(0) does not exist). The map

f:\mathbb{R}\setminus \{0\}\rightarrow \mathbb{R}:x\rightarrow \frac{1}{x}

is a function however.

I like to see has condition (2) as a technical condition that makes things easier. The essential condition is condition (1). (2) is less essential because we can always modify things such that it is satisfied. Indeed, like I've done above: I just restricted $\mathbb{R}$ to $\mathbb{R}\setminus \{0\}$.

 

[lugita15]

The terminology I learned is a relation that satisfies 1 alone is called a partial function, and a relation satisfying both is called a total function.

 

[reenmachine]

For example if x=Joffrey, then f(x)=Cersei, and if x=Arya, then f(x)=Catelyn. We have f(Joffrey)=Cersei, and f(Arya)=f(Sansa)=f(Bran)=Catelyn. Obviously they're just trying to help you develop an intuition about functions at this point. This isn't even real mathematics. The reason why this illustrates the idea of a function pretty well is that it's easy to understand that every person has exactly one biological mother.

So Arya , Bran and Sansa are brothers and sisters?

A function f from A into B can be defined e.g. by specifying f(1)=3, f(2)=5. This f can also be defined by $f=\left\{\langle 1,3\rangle,\langle 2,5\rangle\right\}$. However, if you e.g. define $g=\left\{\langle 1,3\rangle,\langle 1,5\rangle,\langle 2,5\rangle\right\}$, then g fails to satisfy condition 1, and is therefore not a function. It fails because the set g contains two ordered pairs that have the same first component but different second components. (This is like one person having two biological mothers). Note that the statement $\langle x,y\rangle\in g$ is also written as $g(x)=y$. So if g contains <1,3> and <1,5>, then things get weird, because it would imply that g(1)=3 and g(1)=5.

So g is just a letter you randomly used to describe a set?

The rest is very clear , thanks!

If you want to say that $F=\left\{\langle 1,3\rangle,\langle 2,3\rangle\right\}$, you can do it by saying that $F(1)=3$ and $F(2)=3$.

What if F transforms every single natural numbers into 3.Would something like:

N = set of all natural numbers
x ∈ N
F(x) = 3

...be a good way to express it?

Do you mean that in this example, you denote {<1,3>,<2,4>} by <? In that case, < is definitely a function. The fact that 3 and 4 aren't the same only means that < isn't a constant function.

So a constant function is a function that gives the same result no matter which element of the domain you throw at it while a function like the one above has different results depending which elements of the domain you throw into it?

Because we want to define something that associates exactly one element of B with each element of A, and if dom F is a proper subset of A, then there's an element of A that doesn't get associated with an element of B (a person that doesn't have a biological mother).

Wait , if this ''thing'' associates exactly one element of B with each element of A , then in the following example F = {<1,3>,<2,4>} , here 4 can't be associated with 1 if the function is +2 , yet it is said to be a function even if an element of B wasn't associated with each elements of A.

That was a very helpful post!

Thanks a lot again!

 

[reenmachine]

I want to add to Fredrik's excellent response that not every author demands condition (2) in functions. For example, in high school, we often worked with functions whose domain is not the entire set. For example, something like

f:\mathbb{R}\rightarrow \mathbb{R}:x\rightarrow \frac{1}{x}

First of all I need a remainder on what the arrow means.Does it mean function?

edit: if it does mean a function , I'm not sure I understand the way they expressed it.

Is it: Function = real number into a real number = x into 1/x , meaning that real number is a function to another real number is describing what x should be and x into 1/x is describing which real numbers can apply?

thanks

 

[Fredrik]

The notation $f:X\to Y$ means that f is a function from X into Y. The second arrow tells you what member of the codomain is associated with an arbitrary x in the domain. Many people use the "mapsto" arrow ($\mapsto$) instead of the "to" arrow ($\to$) for that.

So instead of saying "f is the function from ℝ-{0} into ℝ such that f(x)=1/x for all x in R-{0}", you can just say $f:\mathbb R-\{0\}\to\mathbb R:x\mapsto 1/x$.

(Edit: I had typed the domain as $\mathbb R$. I have corrected it now, to $\mathbb R-\{0\}$).

The point of the posts above is to explain that the formula f(x)=1/x only defines a function from ℝ into ℝ if condition 2 is omitted from the definition of "function". If you keep condition 2, then the formula defines a function from ℝ-{0} into ℝ, and a partial function from ℝ into ℝ.

 

[reenmachine]

The notation $f:X\to Y$ means that f is a function from X into Y. The second arrow tells you what member of the codomain is associated with an arbitrary x in the domain. Many people use the "mapsto" arrow ($\mapsto$) instead of the "to" arrow ($\to$) for that.

So instead of saying "f is the function from ℝ-{0} into ℝ such that f(x)=1/x for all x in R-{0}", you can just say $f:\mathbb R\to\mathbb R:x\mapsto 1/x$.

The point of the posts above is to explain that the formula f(x)=1/x only defines a function from ℝ into ℝ if condition 2 is omitted from the definition of "function". If you keep condition 2, then the formula defines a function from ℝ-{0} into ℝ, and a partial function from ℝ into ℝ.

very clear thank you!

edit: just a small question , why ℝ-{0} instead of ℝ?

 

[Fredrik]

So Arya , Bran and Sansa are brothers and sisters?

OMG, I thought everyone watched Game of Thrones. Yes, they are all children of Eddard "Ned" Stark and Catelyn Stark.

So g is just a letter you randomly used to describe a set?

Yes.

What if F transforms every single natural numbers into 3.Would something like:

N = set of all natural numbers
x ∈ N
F(x) = 3

...be a good way to express it?

Yes, but it's clearer if you include the words "for all" or "for each" somewhere. (They mean the same thing, but "for each" sounds better in definitions). For each x ∈ N, we define F(x) = 3.

So a constant function is a function that gives the same result no matter which element of the domain you throw at it while a function like the one above has different results depending which elements of the domain you throw into it?

Yes.

Wait , if this ''thing'' associates exactly one element of B with each element of A , then in the following example F = {<1,3>,<2,4>} , here 4 can't be associated with 1 if the function is +2 , yet it is said to be a function even if an element of B wasn't associated with each elements of A.

But A={1,2}, right? Then F associates 3 with 1 and 4 with 2, so both 1 and 2 have been associated with something.

Not sure what you mean by "if the function is +2".

If what you're concerned about is that some members of B are not associated with members of A, then you have discovered that the codomain B and the range $f(A)=\{f(x)\in B|x\in A\}$ don't have to be equal. The latter can be a proper subset of the former. If f(A)=B, the function is said to be surjective. Edit: I had typed the definition of the range completely wrong. It's fixed now.

 

[Fredrik]

very clear thank you!

edit: just a small question , why ℝ-{0} instead of ℝ?

Because 1/0 is undefined. So the formula f(x)=1/x can't define f(0).

I have to leave the computer for a few hours now. Edit:I have corrected a mistake in post #166 and one in post #168. I had typed ℝ in one place where it should be ℝ-{0}, and typed the definition of f(A) completely wrong.

 

[reenmachine]

Because 1/0 is undefined. So the formula f(x)=1/x can't define f(0).

I have to leave the computer for a few hours now. Edit:I have corrected a mistake in post #166 and one in post #168. I had typed ℝ in one place where it should be ℝ-{0}, and typed the definition of f(A) completely wrong.

Note: You guys have been so helpful in 2 weeks that I upgraded my membership.Hopefully I can learn more and more on this website and one day become an helper.

So basically , ℝ-{0} means that the function is from ℝ into ℝ except that you take out the subset {0} from the first set ℝ?

That way , every members of ℝ could be defined if we throw them into f(x)=1/x , which wasn't the case with {0}.

Just a minor question , why is it ℝ-{0} instead of ℝ-0 ?

 

[reenmachine]

OMG, I thought everyone watched Game of Thrones. Yes, they are all children of Eddard "Ned" Stark and Catelyn Stark.

Heard about this TV show , might watch it one day but TV series are extremely time consuming.

If what you're concerned about is that some members of B are not associated with members of A, then you have discovered that the codomain B and the range $f(A)=\{f(x)\in B|x\in A\}$ don't have to be equal. The latter can be a proper subset of the former. If f(A)=B, the function is said to be surjective. Edit: I had typed the definition of the range completely wrong. It's fixed now.

Yeah this is what concerned me.I'm still having difficulties understanding clearly what $f(A)=\{f(x)\in B|x\in A\}$ is suppose to mean exactly.

Does it mean that x is originally a member of A , but that if you throw it into the function , then x is a member of B?

 

[Fredrik]

Yes. The right-hand side of $f(A)=\{f(x)\in B\,|\,x\in A\}$ should be read as "the set of all f(x) in B such that x is in A". This is another way to say it: $f(A)=\{y\in B\,|\,\exists x\in A~ f(x)=y\}$. "The set of all y in B such that there exists an x in A such that f(x)=y".

 

[Fredrik]

So basically , ℝ-{0} means that the function is from ℝ into ℝ except that you take out the subset {0} from the first set ℝ?

Yes. (Some people write \ instead of -. I prefer the good old minus sign. You can use whatever you want).

Just a minor question , why is it ℝ-{0} instead of ℝ-0 ?

ℝ-0 would be the set of all real numbers except those that are members of the set 0. This is clearly not what we want.
$$X-Y=\{x\in X|x\notin Y\}$$

 

[reenmachine]

Yes. (Some people write \ instead of -. I prefer the good old minus sign. You can use whatever you want).ℝ-0 would be the set of all real numbers except those that are members of the set 0. This is clearly not what we want.
$$X-Y=\{x\in X|x\notin Y\}$$

Forgive this stupid question , but what members does the set 0 have except 0? And what's the difference with {0}?

Maybe I didn't get have enough sleep last night but I'm confused over such a trivial detail.

 

[reenmachine]

Yes. The right-hand side of $f(A)=\{f(x)\in B\,|\,x\in A\}$ should be read as "the set of all f(x) in B such that x is in A". This is another way to say it: $f(A)=\{y\in B\,|\,\exists x\in A~ f(x)=y\}$. "The set of all y in B such that there exists an x in A such that f(x)=y".

what does the reverse E symbol means?

 

[Fredrik]

what does the reverse E symbol means?

"There exists". Similarly, an upside down A ($\forall$) means "for all". This notation is explained in the logic section of the Book of Proof. (I haven't checked, but it seems like a safe assumption).

 

[reenmachine]

"There exists". Similarly, an upside down A ($\forall$) means "for all". This notation is explained in the logic section of the Book of Proof. (I haven't checked, but it seems like a safe assumption).

I'm expecting the book today.Don't know if they're going to deliver it on time though.I thought I was going to get it friday.

Personal learning-method question: Would you guys think it's a good idea if I get started with calculus at the same time I'm learning the naive set theory basics and eventually beyond? Or is that unreasonable to try to learn them simultaneously?

 

[Fredrik]

Forgive this stupid question , but what members does the set 0 have except 0? And what's the difference with {0}?

Maybe I didn't get have enough sleep last night but I'm confused over such a trivial detail.

The answer to the first question is actually very complicated. I can't explain it all here. 0 denotes the additive identity of the Dedekind-complete ordered field that we have chosen to denote by ℝ. So what set 0 is depends on that choice. One of the possible choices starts with a definition of an equivalence relation on the set of Cauchy sequences in the set of rational numbers, and then defines the set ℝ as the set of all equivalence classes that correspond to that equivalence relation. If that's the choice we've made, then 0 denotes the set of all Cauchy sequences in the set of rational numbers that converge to the additive identity of the field of rational numbers. That additive identity is usually also denoted by 0, but I will denote it by 0' here, so that we can distinguish between them. Note that 0 is a set of sequences that all converge to 0'.

$\mathbb R-0$ would be the set of all x such that x is a real number (an equivalence class of Cauchy sequences of rational numbers), and x is not a Cauchy sequence of rational numbers that converges to 0'. Since no equivalence class of Cauchy sequences is a Cauchy sequence, this implies that $\mathbb R-0=\mathbb R$.

However, since $0\in\mathbb R$, we have $\mathbb R-\{0\}\neq\mathbb R$.

Personal learning-method question: Would you guys think it's a good idea if I get started with calculus at the same time I'm learning the naive set theory basics and eventually beyond? Or is that unreasonable to try to learn them simultaneously?

Go ahead. It shouldn't be a problem.

 

[reenmachine]

The answer to the first question is actually very complicated. I can't explain it all here. 0 denotes the additive identity of the Dedekind-complete ordered field that we have chosen to denote by ℝ. So what set 0 is depends on that choice. One of the possible choices starts with a definition of an equivalence relation on the set of Cauchy sequences in the set of rational numbers, and then defines the set ℝ as the set of all equivalence classes that correspond to that equivalence relation. If that's the choice we've made, then 0 denotes the set of all Cauchy sequences in the set of rational numbers that converge to the additive identity of the field of rational numbers. That additive identity is usually also denoted by 0, but I will denote it by 0' here, so that we can distinguish between them.

I think that's over my head.I'll try to re-inforce my basics before going there.

Go ahead. It shouldn't be a problem.

Great , I'll get started this week.

 

[Fredrik]

I think that's over my head.I'll try to re-inforce my basics before going there.

It should be, since it involves some abstract algebra and topology. (Fairly simple concepts from those areas of mathematics, but still).