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).

 

[micromass]

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?

May I ask which book you'll be using for calculus?

 

[reenmachine]

May I ask which book you'll be using for calculus?

Unfortunately Lang's "A first course in calculus'' was too expensive for me at the moment so I bought ''Calculus: An Intuitive and Physical Approach'' by Morris Kline which was the other book you suggested to me in pm.I'll give it a shot and if I don't like it I'll eventually buy Lang's book.It's a gamble but my budget was a little bit tight this week.

My goal is to manage to get through ''Calculus: An Intuitive and Physical Approach'' and then move on to Spivak , which is very expensive but apparently worth it.

 

[WannabeNewton]

You know it's pretty beautiful seeing your progress from page 1 up till now. You are one exceptionally fast learner. Keep it up mate! And remember, topology is the best :D

 

[reenmachine]

You know it's pretty beautiful seeing your progress from page 1 up till now. You are one exceptionally fast learner. Keep it up mate! And remember, topology is the best :D

Thanks , your encouragements means a lot to me!

The biggest reason I'm progressing so smoothly is the quality of the help I'm receiving in the thread.Not just the quality but also the quantity.I can always expect a solid explanation within 12 hours from one of you guys and learning is way easier with some momentum.You guys have been sharp and patient with me as I tend to ask a lot of questions so I'm really grateful for that.Can't thank you guys enough.

Why the topology comment?

 

[micromass]

Thanks , your encouragements means a lot to me!

The biggest reason I'm progressing so smoothly is the quality of the help I'm receiving in the thread.Not just the quality but also the quantity.I can always expect a solid explanation within 12 hours from one of you guys and learning is way easier with some momentum.You guys have been sharp and patient with me as I tend to ask a lot of questions.

Why the topology comment?

Wbn is a bit obsessed by topology. I can't blame him, topology really is the most beautiful part of mathematics. Once you know topology, your definition of beauty changes radically

Now I'm going to get a lot of hate by the people who dislike topology...

 

[WannabeNewton]

Wbn is a bit obsessed by topology. I can't blame him, topology really is the most beautiful part of mathematics. Once you know topology, your definition of beauty changes radically

A bit? Sometimes I dream of null homotopic curves. Also, if I could I would marry Urysohn's lemma :D

 

[pwsnafu]

Thanks , your encouragements means a lot to me!

The biggest reason I'm progressing so smoothly is the quality of the help I'm receiving in the thread.Not just the quality but also the quantity.

I biggest reason is you are asking questions! Too many university students just passively sit in lectures and tutorials, and they never ask anything. There is no such thing as a stupid question.

Why the topology comment?

Topology is very popular with pure mathematicians.

 

[reenmachine]

The only things I know about topology is it deals with shapes , like the difference between a sphere and a torus and of course the poincare conjecture proved by Perelman.

 

[pwsnafu]

The only things I know about topology is it deals with shapes , like the difference between a sphere and a torus and of course the poincare conjecture proved by Perelman.

Topology asks the question "What do continuous functions preserve?" For example f(x) = 2x. It has an inverse g(x) = x/2, which (and this is critical) is also continuous. Now take (say) an open interval. I'll choose (0,1) And now I apply f to it. I get another open interval (0,2). Similarly, I can take (0,2), apply g and obtain (0,1). Hence "Openness" is preserved.

There are other concepts of course.

 

[WannabeNewton]

The only things I know about topology is it deals with shapes , like the difference between a sphere and a torus and of course the poincare conjecture proved by Perelman.

Just to tie things in with the theme of this thread, there is this topology problem involving proving properties of this topological space called the long line which is constructed in a very non-intuitive but elegant way using the Well Ordering Theorem. Anyways, so I still haven't been able to do it and micromass always picks on me for it

 

[Fredrik]

The biggest reason I'm progressing so smoothly is the quality of the help I'm receiving in the thread.Not just the quality but also the quantity.I can always expect a solid explanation within 12 hours from one of you guys and learning is way easier with some momentum.You guys have been sharp and patient with me as I tend to ask a lot of questions so I'm really grateful for that.Can't thank you guys enough.

I think other members here could learn a lot from how you've been handling this. You're getting a lot of quality help because you've been following our recommendations. You have read things we have recommended. You have done exercises that we have recommended. You have answered our questions to you. You have given us feedback on our replies, so that we know what parts you understood and what parts you didn't. Since you're new here, you probably don't realize how rare that is.

A bit? Sometimes I dream of null homotopic curves. Also, if I could I would marry Urysohn's lemma :D

Hahaha...you're funny. I'm not as fanatical about topology as you. It may be because I was naive enough to start reading a book on functional analysis that required topology, before I had studied topology. And then I kept underestimating how much time I would need to spend on topology before I could return to that book. I eventually choose to study other books on functional analysis, even though I did eventually learn some topology.

I have to agree about Urysohn's lemma though. When I had studied the proof, I was in awe of how beautiful it was. Well, the main trick anyway. It was a bit tedious to finish the proof after the main part of it.

The only things I know about topology is it deals with shapes , like the difference between a sphere and a torus and of course the poincare conjecture proved by Perelman.

Mathematicians always mention those things (only those things) when they try to explain what topology is. I find that very strange, because if you study topology, you won't be concerned with any of those things for at least a couple of months. I would say that topology is anything that has anything to do with limits. It's a huge subject that includes the things you mentioned (because the definition of continuity involves limits, and because of some things that are too complicated to explain here).

 

[reenmachine]

I think other members here could learn a lot from how you've been handling this. You're getting a lot of quality help because you've been following our recommendations. You have read things we have recommended. You have done exercises that we have recommended. You have answered our questions to you. You have given us feedback on our replies, so that we know what parts you understood and what parts you didn't. Since you're new here, you probably don't realize how rare that is.

I was unaware it was rare , thought it was fairly common.

Hopefully I can maintain my pace and learn more and more math!

 

[reenmachine]

FINALLY I received my books! Yesss!

Calculus: an intuitive and physical approach by Morris Kline is a brick.Intimidating at first , but I'm not one to back down.That is also without knowing if there's some basic high school concepts I'm unaware of that I would need to understand calculus, but that won't stop me even if that's the case , I'll just pick them up by necessity (it's highly likely that the last high school I've been to had a poor math program).

The book of proof looks very nice and a much easier read , but I won't judge a book by it's cover.

I'm very happy to have them , now I should finish the very last section of the textbook I used during the thread and I'll be ready to start these new adventures.

EDIT: Should I try to do some basic exercises on relations and functions? If so , any suggestions?

 

[Fredrik]

Should I try to do some basic exercises on relations and functions? If so , any suggestions?

There are lots of exercises in the book of proof. The more of them you take the time to work through, the better you will become. You will have to choose how many of them you do, but you should at least do a few from each section with exercises. Do as many as it takes to feel comfortable with the concepts.

One of the most useful results involving relations is that if ~ is an equivalence relation on a set X, then each x in X belongs to exactly one equivalence class, so I would recommend that you make sure that you understand the proof of that. (I haven't checked if the book proves it, or leaves it as an exercise. You should either study the book's proof, or do the exercise).

Some of the most useful stuff involving functions is to be able to determine if a function is injective, surjective, bijective, to understand the terms range and preimage, and prove simple theorems about about those things. For example, $f^{-1}(A\cap B)=f^{-1}(A)\cap f^{-1}(B)$. This is really important because books on analysis and topology never include such details in proofs. You are expected to be able to prove those things for yourself.

 

[reenmachine]

Just a little refresher for fun:1) We will attempt to prove that A∪(B∩Z) = (A∪B)∩(A∪Z).

Let x ∈ A∪(B∩Z) be arbitrary.Since x ∈ A∪(B∩Z) , it implies that x ∈ A.Since x ∈ A , it implies that x ∈ (A∪B) , x ∈ (A∪Z) implying that x ∈ (A∪B)∩(A∪Z).

Now let x ∈ (A∪B)∩(A∪Z) be arbitrary.It implies that x ∈ (A∪B) and that x ∈ (A∪Z) , which in turns implies that x ∈ A.Since x ∈ A , it implies that x ∈ A∪(B∩Z) proving that A∪(B∩Z) = (A∪B)∩(A∪Z).2) We will attempt to prove that A–B = A∩B'.

Let x ∈ A and x ∉ B be arbitrary (x ∈ A-B).Since x ∉ B , it implies that x ∈ B' and since x ∈ A , it implies that x ∈ A∩B'.

Now let x ∈ A∩B' be arbitrary.This implies that x ∈ A and that x ∈ B' , which implies that x ∉ B.Since x ∈ A and x ∉ B , it implies that x ∈ A-B proving that A-B = A∩B'.

 

[reenmachine]

(just using this post to play around a little bit)

Let N be the set of all natural numbers.Let x ∈ N be arbitrary.

∀ x ∈ N , we define f(x) = x^3.

If x=2 , then f(2) = 8
If x=3 , then f(3) = 27
If x=4 , then f(4) = 64

-----------------------------------------------------------------------

∀ x and y ∈ N , we define f(x) = (f(y) = y-5) + 3x

If x=2 and y=3 , then f(2) = 4
If x=13 and y=4 , then f(13) = 38

-----------------------------------------------------------------------

Let set B = {1,2}

∀ x ∈ N , we define f(x) = x + B

If x=5 , the f(5) = 8

Here my problem is how can I express that x would add all elements of B? (x+1+2 in this example)

-----------------------------------------------------------------------

If A = {1,2} , then p(A) = {∅,{1},{2},{1,2}}

If A = {1,2} and B = {4,5} , then A × B = {(1,4),(1,5),(2,4),(2,5)}

If R = {(1,5),(2,5)} , then R' is {(1,4),(2,4)}.The domain of the relation is (1,2).The codomain is (4,5) and the range is (5)? Still having a little bit of trouble here with codomain vs range for some reasons.

R^-1 = {(5,1),(5,2)}

 

[CompuChip]

∀ x and y ∈ N , we define f(x) = (f(y) = y-5) + 3x

If x=2 and y=3 , then f(2) = 4
If x=13 and y=4 , then f(13) = 38

That doesn't make sense. You can define f(x, y) = y - 5 + 3x.
But you can't define f to be two different things (y - 5 or 3x, depending on what you call your variable). Also, I suggest including the y between the brackets to indicate that the function value also depends on y. In terms of relations, you could say this is a subset of N x N x N.

Let set B = {1,2}

∀ x ∈ N , we define f(x) = x + B

If x=5 , the f(5) = 8

Here my problem is how can I express that x would add all elements of B? (x+1+2 in this example)

If B is fixed, you could write f(x) = x + 3 :P
Or you could use sum notation:
f(x) = x + \sum_{b \in B} b

If you want to define such a function for all possible subsets of N, you can include that in your notation. A common way of putting it is f_B(x), to indicate that f depends on B, but not really as a variable but more as an arbitrary - but fixed - value.

Still having a little bit of trouble here with codomain vs range for some reasons.

Yes, the codomain is what you specify in the definition, and guarantees that the function will take values in that set. However, it is not required to cover all the values in the set. In your example, 4 has no pre-image. The range is the subset of the codomain that is actually "used", i.e. if f: A \to B is a function then A is its domain, B is the codomain and the range is
\{ y \in B \mid \exists x \in A : f(x) = y \}
which may or may not equal B (i.e. it's a subset of B).

 

[Fredrik]

This seems to be a good time to point out a pecularity with the definition of a function from X into Y as a special kind of subset of X×Y. Suppose that f is a function from X into Y according to the definition you have studied, and that Y is a proper subset of some set Z. Then f is also a subset of X×Z, and it satisfies both conditions in the definition of a function from X into Z. So Z has an equal right to be called the codomain of f.

With this definition of "function", the range is simply the smallest set that can be called the codomain of f.

Note however that there's an alternative (and not equivalent) definition of "function" that ensures that each function has exactly one codomain. Check out my post #122 again. Definition 1 is equivalent to the definition you have studied. Definition 2 is the alternative I'm talking about. These definitions are two different ways to turn the idea of "a rule that associates exactly one element of Y with each element of X" into a precise mathematical concept.

It rarely matters which definition of function we're using. It only affects the precise choice of words we should use in certain statements. For example, I said earlier that a function $f:X\to Y$ is said to be surjective if f(X)=Y. But if we're using definition 1, every function from X into Y is also a function from X into Z, where Z is any set such that $Y\subseteq Z$. So the statement "f is surjective" is ambiguous. We can't tell if it means f(X)=Y or f(X)=Z. This is not a problem if we use definition 2, which ensures that every function has a unique codomain. If we use definition 1, we should (ideally) use the phrase "f is surjective onto Y" instead of "f is surjective".

 

[reenmachine]

Damn , wrote a long response but the update deleted it one way or another

 

[reenmachine]

That doesn't make sense. You can define f(x, y) = y - 5 + 3x.
But you can't define f to be two different things (y - 5 or 3x, depending on what you call your variable). Also, I suggest including the y between the brackets to indicate that the function value also depends on y. In terms of relations, you could say this is a subset of N x N x N.

So f has to be the exact same function as long as you operate in the same ''universe''?

My point was that the 2nd f was just a second different function that was part of another function's mechanism.

If B is fixed, you could write f(x) = x + 3 :P
Or you could use sum notation:
f(x) = x + \sum_{b \in B} b

Very nice! Thank you! Should be useful if you don't know how many elements are in B.

If you want to define such a function for all possible subsets of N, you can include that in your notation. A common way of putting it is f_B(x), to indicate that f depends on B, but not really as a variable but more as an arbitrary - but fixed - value.

Not sure I understand the logic behind this.

Yes, the codomain is what you specify in the definition, and guarantees that the function will take values in that set. However, it is not required to cover all the values in the set. In your example, 4 has no pre-image. The range is the subset of the codomain that is actually "used", i.e. if f: A \to B is a function then A is its domain, B is the codomain and the range is
\{ y \in B \mid \exists x \in A : f(x) = y \}
which may or may not equal B (i.e. it's a subset of B).

Just to get used to what is a new symbol and concept for me , can you explain more in depth what is the role of ∃ here? I know it means ''there exist'' , but I would like to understand his role better in this specific exemple.