Question about proof from a guy with a highschool education

[micromass]

tell me if I'm right: If $\{x\in A~\vert ~ x~\text{is a set}\}$ , then this is equal to set A and perfectly acceptable.But if $\{x\in x~\vert ~ x~\text{is a set}\}$ , then this is a contradiction because a set can't be an element of itself?

thanks

According to the ZFC axioms, the following set is well-defined:

\{x\in A~\vert~\varphi(x)\}

where $A$ is a set that does not depend on x, and where $\varphi(x)$ is a property with $x$ as a variable.

You should interpret the set as "the set of all elements $x$ of $A$ such that $\varphi(x)$ is true.

For example:
$\{x\in \{0,1,2\}~\vert~x=1\}$
Here we have $A=\{0,1,2\}$ and $\varphi(x)$ is the formula $x=1$.
The interpretation of the set is: "the set of all elements $x$ of $\{0,1,2\}$ such that $x=1$. Clearly, the set is just equal to $\{1\}$.

$\{x\in \mathbb{R}~\vert~x^2 = 1\}$
Here the set $A=\mathbb{R}$ and $\varphi(x)$ is $x^2 =1$.
This set is the set of all real numbers $x$ such that $x^2 =1$. Clearly, the set is $\{1,-1\}$.

Some stranger examples that are well-defined sets:
$\{x\in \{0,1,2\}~\vert~x\in x\}$
This is the set of all $x$ in $\{0,1,2\}$ with the proporty that $x$ belongs to itself. There are no such $x$ in $\{0,1,2\}$ with that property. So the set is just the empty set.

$\{x\in \mathbb{R}~\vert~1=1\}$
This is a very pathological example, but it is a well-defined set. The set is the set of all real numbers such that 1=1 is true. To check that an element is in the set, we pick a real number $x\in \mathbb{R}$ and we check that $1=1$ is true. But it is always true. So all $x\in \mathbb{R}$ are in the set. So the set equals entire $\mathbb{R}$.
This is very weird since we have a property $\varphi(x)$ that apparently does not depend on $x$. Such sets are well-defined but never show up in practice.

Some counterexamples:
$\{x~\vert~x~\text{is a set}\}$
This is not a good definition of a set since the set $A$ is missing.

$\{x~\vert~x=1\}$
This is not a good definition of a set, since the set $A$ is again missing.
This is a bit strange, since clearly the set that it would define is just $\{1\}$. So if the above is a set, then it would equal $\{1\}$.
We are not saying that $\{1\}$ is not a set (it is a set). But we are saying that $\{x~\vert~x=1\}$ is not a "well-formed".

$\{x\in x~\vert~x=1\}$
This is not a good definition of a set, since the set $A$ is not independent of $x$. We want a set $A$ that has nothing to do with our dummy variable.

So, the only set builder notation that we accept is:
\{x\in A~\vert~\varphi(x)\}
This is the only well-formed formula.

Now, this implies that the follwing is also not a good definition:

$\{x^3\in \mathbb{R}~\vert~x\geq 0\}$

It is clear what I mean with this set. I want to pick all the elements of $\mathbb{R}$ that have the form $x^3$ for some $x\geq 0$. For example, the element 8 is in the set, because we can pick $x=2$.
The element -8 is not in the set. If we pick $x=-2$, then $-8=x^3$, ,but our $x$ does not satisfy $x\geq 0$.

Anyway, this is not a well-formed formula. Our formula is not in the form
\{x\in A~\vert~\varphi(x)\}
The problem is that we have an $x^3$ instead of an $x$.

We can write rewrite our formula as
$\{x\in \mathbb{R}~\vert ~\exists y\in \mathbb{R}: ~y\geq 0~\text{and}~x=y^3\}$
This is a well-formed formula. Indeed, we can take $A=\mathbb{R}$ and $\varphi(x)$ is the formula $\exists y \in \mathbb{R}:~y\geq 0 ~\text{and}~x=y^3$.
The elements of the set are the real numbers $x$ such that a real number $y$ exists such that $y^3 = x$ and $y\geq 0$.

The problem is that the above set is very difficult to write. I'm sure that people find $\{x^3\in \mathbb{R}~\vert~x\geq 0\}$ much easier to read than $\{x\in \mathbb{R}~\vert ~\exists y\in \mathbb{R}: ~y\geq 0~\text{and}~x=y^3\}$. This is why the notation $\{x^3\in \mathbb{R}~\vert~x\geq 0\}$ is used in math texts and is allowed. Allowing this is an abuse of notation since strictly, the notation wouldn't be allowed.

 

[micromass]

I actually don't know if a set can be an element of itself

There is an axiom in ZFC set theory that forbids a set to be an element of itself. This is called the foundation axiom. Other things that are forbidden are sets such that $x=\{x\}$ or $x=\{x,1\}$.

The foundation axiom is however a non-essential part of ZFC. There is very little that would change in entire mathematics if we allowed $x\in x$ to happen. So many people do not accept the foundation axiom.

So whether there are sets that are elements of themselves. The answer is that you can rule out such sets by axiom. Or you can choose not to rule the sets out.
In light of actual mathematics (except set theory), the axiom is not important at all.

 

[Fredrik]

Regarding the definition $R=\{S\in U\,|\,S\notin S\}$...

We state that an element of R is an element of U , and that the property of this element of R is that it's not an element of itself.So to be an element of R , you need NOT to be an element of yourself.

Yes.

But if you take R , which is a set so not an element of itself

I don't know if a set can be an element of itself. This isn't part of the argument anyway.

, then R is an element of itself based on the property of the elements of R.

Yes, the definition of R ensures that if $R\notin R$, then $R\in R$. That is a contradiction. But the proof doesn't end here, because when an assumption leads to a contradiction, we can conclude that the assumption is a false statement. Our assumption was $R\notin R$, so the conclusion is that this is false, i.e. we have $R\in R$.

But if $R\in R$, then the definition of R tells us that $R\notin R$, so we still get a contradiction. This time our assumption was $R\in R$, so we can conclude that this is false, i.e. that $R\notin R$.

So we have proved that $R\in R$ and (its negation) $R\notin R$ are both false. That is the impossible result that forces us to reject the idea that R is a set.

edit: the definition of R is ''everything'' that's an element of U that isn't an element of itself correct? So S isn't an element of R per say , but it represent everything that is an element of U and isn't an element of itself

Yes.

, and everything that could be included in this definition will be part of the set R?

I don't understand this part of the sentence.

It's clear that one of the reasons why you're having difficulties with this proof is that you had not familiarized yourself with the simple examples $\{x\in y\,|\,\varphi(x)\}$ notation before you attacked the "head explode" example of Russell's paradox. Maybe you understand the proof now, but I think it would still be a good idea to do some exercises from the book that involve this notation. The exercises for section 1.1 on page 7 of the online version contain many simple examples.

 

[reenmachine]

According to the ZFC axioms, the following set is well-defined:

\{x\in A~\vert~\varphi(x)\}

where $A$ is a set that does not depend on x, and where $\varphi(x)$ is a property with $x$ as a variable.

You should interpret the set as "the set of all elements $x$ of $A$ such that $\varphi(x)$ is true.

For example:
$\{x\in \{0,1,2\}~\vert~x=1\}$
Here we have $A=\{0,1,2\}$ and $\varphi(x)$ is the formula $x=1$.
The interpretation of the set is: "the set of all elements $x$ of $\{0,1,2\}$ such that $x=1$. Clearly, the set is just equal to $\{1\}$.

$\{x\in \mathbb{R}~\vert~x^2 = 1\}$
Here the set $A=\mathbb{R}$ and $\varphi(x)$ is $x^2 =1$.
This set is the set of all real numbers $x$ such that $x^2 =1$. Clearly, the set is $\{1,-1\}$.

Some stranger examples that are well-defined sets:
$\{x\in \{0,1,2\}~\vert~x\in x\}$
This is the set of all $x$ in $\{0,1,2\}$ with the proporty that $x$ belongs to itself. There are no such $x$ in $\{0,1,2\}$ with that property. So the set is just the empty set.

$\{x\in \mathbb{R}~\vert~1=1\}$
This is a very pathological example, but it is a well-defined set. The set is the set of all real numbers such that 1=1 is true. To check that an element is in the set, we pick a real number $x\in \mathbb{R}$ and we check that $1=1$ is true. But it is always true. So all $x\in \mathbb{R}$ are in the set. So the set equals entire $\mathbb{R}$.
This is very weird since we have a property $\varphi(x)$ that apparently does not depend on $x$. Such sets are well-defined but never show up in practice.

Some counterexamples:
$\{x~\vert~x~\text{is a set}\}$
This is not a good definition of a set since the set $A$ is missing.

$\{x~\vert~x=1\}$
This is not a good definition of a set, since the set $A$ is again missing.
This is a bit strange, since clearly the set that it would define is just $\{1\}$. So if the above is a set, then it would equal $\{1\}$.
We are not saying that $\{1\}$ is not a set (it is a set). But we are saying that $\{x~\vert~x=1\}$ is not a "well-formed".

$\{x\in x~\vert~x=1\}$
This is not a good definition of a set, since the set $A$ is not independent of $x$. We want a set $A$ that has nothing to do with our dummy variable.

So, the only set builder notation that we accept is:
\{x\in A~\vert~\varphi(x)\}
This is the only well-formed formula.

Now, this implies that the follwing is also not a good definition:

$\{x^3\in \mathbb{R}~\vert~x\geq 0\}$

It is clear what I mean with this set. I want to pick all the elements of $\mathbb{R}$ that have the form $x^3$ for some $x\geq 0$. For example, the element 8 is in the set, because we can pick $x=2$.
The element -8 is not in the set. If we pick $x=-2$, then $-8=x^3$, ,but our $x$ does not satisfy $x\geq 0$.

Anyway, this is not a well-formed formula. Our formula is not in the form
\{x\in A~\vert~\varphi(x)\}
The problem is that we have an $x^3$ instead of an $x$.

We can write rewrite our formula as
$\{x\in \mathbb{R}~\vert ~\exists y\in \mathbb{R}: ~y\geq 0~\text{and}~x=y^3\}$
This is a well-formed formula. Indeed, we can take $A=\mathbb{R}$ and $\varphi(x)$ is the formula $\exists y \in \mathbb{R}:~y\geq 0 ~\text{and}~x=y^3$.
The elements of the set are the real numbers $x$ such that a real number $y$ exists such that $y^3 = x$ and $y\geq 0$.

The problem is that the above set is very difficult to write. I'm sure that people find $\{x^3\in \mathbb{R}~\vert~x\geq 0\}$ much easier to read than $\{x\in \mathbb{R}~\vert ~\exists y\in \mathbb{R}: ~y\geq 0~\text{and}~x=y^3\}$. This is why the notation $\{x^3\in \mathbb{R}~\vert~x\geq 0\}$ is used in math texts and is allowed. Allowing this is an abuse of notation since strictly, the notation wouldn't be allowed.

Very clear thank you!

 

[reenmachine]

The set of even integers: $\{n\in\mathbb Z\,|\,\exists m\in\mathbb Z~~ n=2m\}$. If we write this as $\{n\in\mathbb Z\,|\,\varphi(n)\}$, then $\varphi$ is the property of "being even", and $\varphi(n)$ is the statement "n is even", which in mathematical notation is written as $\exists m\in\mathbb Z~~ n=2m$.

So the fact that any m in Z multiplied by two will give us an even number makes it good to define n?

 

[reenmachine]

Regarding the definition $R=\{S\in U\,|\,S\notin S\}$...


Yes.


I don't know if a set can be an element of itself. This isn't part of the argument anyway.


Yes, the definition of R ensures that if $R\notin R$, then $R\in R$. That is a contradiction. But the proof doesn't end here, because when an assumption leads to a contradiction, we can conclude that the assumption is a false statement. Our assumption was $R\notin R$, so the conclusion is that this is false, i.e. we have $R\in R$.

But if $R\in R$, then the definition of R tells us that $R\notin R$, so we still get a contradiction. This time our assumption was $R\in R$, so we can conclude that this is false, i.e. that $R\notin R$.

So we have proved that $R\in R$ and (its negation) $R\notin R$ are both false. That is the impossible result that forces us to reject the idea that R is a set.


Yes.


I don't understand this part of the sentence.

It's clear that one of the reasons why you're having difficulties with this proof is that you had not familiarized yourself with the simple examples $\{x\in y\,|\,\varphi(x)\}$ notation before you attacked the "head explode" example of Russell's paradox. Maybe you understand the proof now, but I think it would still be a good idea to do some exercises from the book that involve this notation. The exercises for section 1.1 on page 7 of the online version contain many simple examples.

Yes I'm pretty sure I get the proof now.I will follow your advice and do some exercises later today from the book.

 

[Fredrik]

So the fact that any m in Z multiplied by two will give us an even number makes it good to define n?

Define n? I just used the definition of "even" to define the set of even integers. (What I wrote down means "the set of all integers that have the property of being even"). The definition of "even" is this: An integer n is said to be even if there's an integer m such that 2m=n.

 

[Fredrik]

There is an axiom in ZFC set theory that forbids a set to be an element of itself. This is called the foundation axiom. Other things that are forbidden are sets such that $x=\{x\}$ or $x=\{x,1\}$.

The foundation axiom is however a non-essential part of ZFC. There is very little that would change in entire mathematics if we allowed $x\in x$ to happen. So many people do not accept the foundation axiom.

So whether there are sets that are elements of themselves. The answer is that you can rule out such sets by axiom. Or you can choose not to rule the sets out.
In light of actual mathematics (except set theory), the axiom is not important at all.

Thanks for the explanation. According to Wikipedia, the axiom of foundation says that every non-empty set x has an element y that's disjoint from x. This clearly rules out x={x}, but it's not obvious how it rules out the existence of disjoint x,y such that x={x,y}.

 

[reenmachine]

I sincerely apologize but I'm still having trouble using LaTeX and it would take me half a day to write all the exercises questions in the post by copy-pasting each symbol.

I will simply give you the link http://www.people.vcu.edu/~rhammack/BookOfProof/Sets.pdf , the exercises are on page 7 in section (1.1).I will randomly pick a couple of them and give it a shot.

If you find it unacceptable not to see the question here , please tell me so and I'll try to come out with a solution.

A.

Write each of the following sets by listing their elements between braces.

solutions:
3. {-2,-1,0,1,2,3,4,5,6}
7. {3}
8. {-2}
9. {0} ?? (having trouble with the sin/cos factor in this context)
15. {5a,2b} ? (again , not sure I get how to proceed)

B.

Write each of the following sets in set-builder notation.

solutions:
17. ${x\in N\,|x is a multiple of 2}$
22. ${x\in N\,|x follows the sequence +3,+5,+7,+9}$ ? Okay I admit I have no clue how to proceed here.
25. ${x\in N\,|1/2x}$

edit: apparently I'm still confused about how to use LaTeX

C.

Find the following cardinalities.

solutions:
31.3
32.1 ?
33.??

I admit I'm struggling quite a bit , some feedback would be of great help for me! No point in continuing this , I'm having too much of a hard time and I feel it's counter-productive until I have further informations.

EDIT: seems I still have massive trouble with latex , this isn't how the notations were suppose to come out This is unfortunate because I feel I spend 50% of my energy trying to actually WRITE what I'm trying to write instead of concentrating on the math.

thanks!

 

[Fredrik]

{ and } are reserved symbols in LaTeX. If you want to display them, you must type \{ and \}. If you want text to be interpreted as text and not as a product of variables, you must write \text{something}. LaTeX Guide. Yes, the first few days of using LaTeX, you may spend as much time on how to write things as on the math, but it's just the first few days.

A

3. That's right.
7. The equation has two solutions in ℝ, and 3 isn't one of them.
8. The equation has three solution in ℝ, and -2 is just one of them.
9. What you need to know is that $\sin 0=0$ and that $\sin(t+2\pi)=\sin t$ for all $t\in\mathbb R$.
15. You seem to have forgotten how to read the notation $\{x\,|\,\varphi(x)\}$. (The author writes a colon where I write a vertical bar). It's read as "the set of all x with property $\varphi$". Can you translate $\{5a+2b\,|\,a,b\in\mathbb Z\}$ to words in this way?

Your answers to 7,8 and 9 suggest that you need to refresh your memory about polynomial equations and the basics of trigonometry. I would guess that Kline covers that. If he doesn't, then you will have to read about it somewhere else.B

17 If N={1,2,...}, your answer means {2,4,6,8,10,...}, so this is incorrect. Hint: 2^n.
22 This one is pretty hard. The sequence is 3+3, 3+3+5, 3+3+5+7, etc. We keep adding larger odd integers to the result of the previous calculation. I don't see a simple formula for it. We can define the sequence in the following way. Define s(0)=3. For each positive integer n, define s(n)=s(n-1)+2n+1. Then we have s(1)=s(0)+2*1+1=3+3=6, s(2)=s(1)+2*2+1=6+5=11, s(3)=s(2)+2*3+1=11+7=18, etc. With this definition of s, we can write the set as $\{s(n)|n\in\mathbb N\}$. Maybe there's a simpler answer that I just don't see.
25 I would recommend that you check your answers by calculating a few members of the set you have just written down. What you wrote would give us {1/2, 1/4, 1/6,...}, which is wrong. Hint: 2^n.

By the way, I think it's slightly more popular these days to define the "natural numbers" as including 0. This author doesn't, and that's fine too. You just need to know that different authors have different conventions.C

31. The question is asking you how many elements the set {{{1},{2,{3,4}},∅}} has.

 

[reenmachine]

{ and } are reserved symbols in LaTeX. If you want to display them, you must type \{ and \}. If you want text to be interpreted as text and not as a product of variables, you must write \text{something}. LaTeX Guide. Yes, the first few days of using LaTeX, you may spend as much time on how to write things as on the math, but it's just the first few days.

A

3. That's right.
7. The equation has two solutions in ℝ, and 3 isn't one of them.
8. The equation has three solution in ℝ, and -2 is just one of them.
9. What you need to know is that $\sin 0=0$ and that $\sin(t+2\pi)=\sin t$ for all $t\in\mathbb R$.
15. You seem to have forgotten how to read the notation $\{x\,|\,\varphi(x)\}$. (The author writes a colon where I write a vertical bar). It's read as "the set of all x with property $\varphi$". Can you translate $\{5a+2b\,|\,a,b\in\mathbb Z\}$ to words in this way?

7.Meant -3.I have no clue how to find the other one.Seems I suffer from memory loss , how do you deal with variables with exposants in algebra to find the value of the variable? This is highly embarrassing :shy:

8.same question as 7.

9.What does t mean? why 2(3.14)?

15.Not sure how to do it.A try: {x : ''there exist'' a,b in Z such that 5a + 2b = x } ?

 

[reenmachine]

Hint: 2^n.

Hint: 2^n.

By the way, I think it's slightly more popular these days to define the "natural numbers" as including 0. This author doesn't, and that's fine too. You just need to know that different authors have different conventions.C

31. The question is asking you how many elements the set {{{1},{2,{3,4}},∅}} has.

Do you mean I can use formulas to the nth - level in notations?

And here I am going backward.I'm now confused about what is an element or not.What the hell is going on

In {{{1},{2,{3,4}},∅}} , you could say 1, 2 ,3 ,4 are elements.Unless this is a powerset , which would make {∅} an element and many more.The multiples brackets are confusing the hell out of me.

This is highly frustrating as it seem I'm regressing instead of progressing today , now I'm confusing subsets and elements again , thought that was behind me

The answer to 31 would be 4?

edit: okay , it doesn't matter how many subsets of a set you have to dig to find an element , if the element is there it's still an element of the big set.

 

[reenmachine]

17 If N={1,2,...}, your answer means {2,4,6,8,10,...}, so this is incorrect. Hint: 2^n.

I'm close to give up

I've tried many things , such as ( x in N : there exist y in N such that x = 2y ) to make sure x is an even integers to operate with them but it still doesn't work out.

EDIT:Think I got it , $\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ , is that it?

This would mean if y=3 , then x = 2^3 = 8 , if y=5 , then x = 2^5 = 32

BOOYAH!

sorry my blood pressure is rising , don't want to sound like I'm pissed but math can do that to me

thanks for the patience

 

[Fredrik]

7.Meant -3.I have no clue how to find the other one.Seems I suffer from memory loss , how do you deal with variables with exposants in algebra to find the value of the variable? This is highly embarrassing :shy:

You understand the formula $(a+b)^2=a^2+b^2+2ab$, right? Suppose that you encounter an equation of the form $x^2+ax+b=0$. There's a clever trick involving that formula that gets x alone on one side of the equality sign:
\begin{align}0 &=x^2+ax+b=x^2+\left(\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2+ax+b =\left(x+\frac a 2\right)^2-\left(\frac{a}{2}\right)^2+b\\
\left(x+\frac a 2\right)^2 &=\left(\frac{a}{2}\right)^2-b\\
x+\frac a 2 &=\pm\sqrt{\left(\frac{a}{2}\right)^2-b}.\\
x &=-\frac a 2 \pm\sqrt{\left(\frac{a}{2}\right)^2-b}.
\end{align} You should memorize both this formula and its derivation. Use the formula to find the solutions.

9.What does t mean?

Did you not see that the sentence ended with "for all $t\in\mathbb R$"? I don't mean to add to your frustration, but if you're still wondering about this after taking the whole sentence into account, then you should go back and read what micromass and I said about dummy variables again.

why 2(3.14)?

Because sin is defined to ensure that this picture tells the truth for all $t\in\mathbb R$.

If you're wondering specifically about the value $2\pi$, then check out this picture, and read the Wikipedia article on "radians" if you need an explanation.

15.Not sure how to do it.A try: {x : ''there exist'' a,b in Z such that 5a + 2b = x } ?

It's "the set of all 5a+2b such that a and b are integers". Edit: I guess I should have looked more closely at what you wrote, because it's the same set as mine. There is however a much simpler way to state the result. Can you think of a set that contains all the sums 5a+2b such that a and b are integers, and doesn't contain anything else?

Do you mean I can use formulas to the nth - level in notations?

I meant e.g. that $2^5=2\cdot 2\cdot 2\cdot 2\cdot 2=32$.

And here I am going backward.I'm now confused about what is an element or not.What the hell is going on

In {{{1},{2,{3,4}},∅}} , you could say 1, 2 ,3 ,4 are elements.Unless this is a powerset , which would make {∅} an element and many more.The multiples brackets are confusing the hell out of me.

This is highly frustrating as it seem I'm regressing instead of progressing today , now I'm confusing subsets and elements again , thought that was behind me

edit: okay , it doesn't matter how many subsets of a set you have to dig to find an element , if the element is there it's still an element of the big set.

That's not it. For example, {x} has only one element, which is denoted by x. This is true even if x={y,z}, so that {x}={{y,z}}. A trick you could use here is to replace matching pairs of curly brackets and the stuff between them with a letter, like x, and then work your way out.

{{{1},{2,{3,4}},∅}} Set a={1} and b={3,4}.
{{a,{2,b},∅}} Set c={2,b}.
{{a,c,∅}} Set d={a,c,∅}.
{d}

This set clearly has only one element. Actually to see this, you only had to make the observation that the expression started with {{ and ended with }}. This made everything between irrelevant. Edit: Oops, not really. See post #258 (written by jbriggs444) for an example that shows that what I just said is incorrect.

EDIT:Think I got it , $\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ , is that it?

This would mean if y=3 , then x = 2^3 = 8 , if y=5 , then x = 2^5 = 32

BOOYAH!

sorry my blood pressure is rising , don't want to sound like I'm pissed but math can do that to me

Yes, that's correct.

There is however a simpler way to write the answer: $\{2^n\,|\,n\in\mathbb N\}$.

 

[reenmachine]

That's not it. For example, {x} has only one element, which is denoted by x. This is true even if x={y,z}, so that {x}={{y,z}}. A trick you could use here is to replace matching pairs of curly brackets and the stuff between them with a letter, like x, and then work your way out.

{{{1},{2,{3,4}},∅}} Set a={1} and b={3,4}.
{{a,{2,b},∅}} Set c={2,b}.
{{a,c,∅}} Set d={a,c,∅}.
{d}

This set clearly has only one element. Actually to see this, you only had to make the observation that the expression started with {{ and ended with }}. This made everything between irrelevant.

Just to comment right away on that part of the post , the worst thing is I actually wrote this but edited it.I knew I had to eliminate all the brackets that were inside brackets (not counting the brackets immediately between the brackets of the set we're searching the cardinality of) but I didn't do it for some reasons.

The answer 1 was the first that got to my mind so this is good news.I noticed I've been getting a little bit emotionnal the last 2 or 3 days.Hopefully I can learn from these frustrations and rise up to the challenge.

Will get back at you about the other part of your post soon enough.

thanks a lot

 

[micromass]

Thanks for the explanation. According to Wikipedia, the axiom of foundation says that every non-empty set x has an element y that's disjoint from x. This clearly rules out x={x}, but it's not obvious how it rules out the existence of disjoint x,y such that x={x,y}.

You should apply the axiom to $\{x\}$. So the axiom tells you that $\{x\}$ has an element disjoint from it. So $x$ and $\{x\}$ are disjoint. But if $x=\{x,y\}$, then $\{x,y\}$ and $\{x\}$ are not disjoint.

Sorry reen, for this off-topic discussion.

 

[Fredrik]

You should apply the axiom to $\{x\}$. So the axiom tells you that $\{x\}$ has an element disjoint from it. So $x$ and $\{x\}$ are disjoint. But if $x=\{x,y\}$, then $\{x,y\}$ and $\{x\}$ are not disjoint.

Ah, of course. So the proof of the proposition that no set in ZFC set theory is an element of itself goes like this: Suppose that there's a set x such that $x\in x$. Then {x} is a set. (There are many ways to see that; one is to use the assumption and the power set axiom). The foundation axiom says that there's a y in {x} that's disjoint from x. Since the only element of {x} is x, this means that x is disjoint from x. This implies that x=∅. This contradicts that $x\in x$.

Sorry reen, for this off-topic discussion.

I suspect that Reenmachine is just relieved to see that I too can fail to find proofs that look pretty simple when you have them in front of you.

 

[jbriggs444]

This set clearly has only one element. Actually to see this, you only had to make the observation that the expression started with {{ and ended with }}. This made everything between irrelevant.

Nit pick.

{{1},{2}} starts with {{ and ends with }} but has two elements.

It is often helpful to decorate expressions like these with white space to make them easier for the reader to parse by eye.

 

[Fredrik]

Agreed. Thanks for the correction. I have added a comment about it to the post where I said that.

 

[reenmachine]

You understand the formula $(a+b)^2=a^2+b^2+2ab$, right?

Well , yeah , I understand it very clearly.

Suppose a=2 and b=3:

The left side is (2+3)(2+3) , the 2×2 and 3×3 on the left side are equal to 2^2 and 3^3 on the right side.Then on the left side you are left with a×b and b×a , which means the same thing , so 2×3 and 3×2 , which is really 2(2×3) denoted by 2ab on the right side.

Suppose that you encounter an equation of the form $x^2+ax+b=0$. There's a clever trick involving that formula that gets x alone on one side of the equality sign:
\begin{align}0 &=x^2+ax+b=x^2+\left(\frac{a}{2}\right)^2-\left(\frac{a}{2}\right)^2+ax+b =\left(x+\frac a 2\right)^2-\left(\frac{a}{2}\right)^2+b\\
\left(x+\frac a 2\right)^2 &=\left(\frac{a}{2}\right)^2-b\\
x+\frac a 2 &=\pm\sqrt{\left(\frac{a}{2}\right)^2-b}.\\
x &=-\frac a 2 \pm\sqrt{\left(\frac{a}{2}\right)^2-b}.
\end{align} You should memorize both this formula and its derivation. Use the formula to find the solutions.

Wow , that's pretty good.But while I understand it on an operating level (like if x=3 , a=4 and b=-21 for exemple) , I would like to understand a little bit more in depth.

Take the way I presented the formula $(a+b)^2=a^2+b^2+2ab$ at the top of the post , I think I made it clear that I connected all the dots as to why this formula was an equation.With this one , not so much.In fact , I guess the key part where I'm not sure what's happening is when ax disappears and (x+a/2)^2 makes his entrance.

Suppose you take my exemple of x=3 , a=4 and b= -21.

This means that 3^2 + 4(3) -21 = 0.

Then the formula introduces (a/2)^2 , but since they both add and substract it it's not that confusing for the moment and it has no immediate impact on the formula.

It still gives 9 + (4/2)^2 - (4/2)^2 +12 - 21 = 9 + 4 - 4 + 12 - 21 = 0.

Then this is where I'm not sure what is happening ''abstractly''.

You have (3 + 4/2)^2 - (4/2)^2 -21 = 0.What happened to 12? Why does it work? I thought about it for 15-20 minutes and couldn't figure it out.

The end of the formula is pretty clear and I understand WHY you need to introduce some new elements in the formula in order to isolate the x on an operating level.That I have no problem of understand , it's just this specific case which has me scratching my head as to why it all work out.

Just a minor question about the end of the formula , why the ± sign? And when you have the -a/2 ahead of the ± sign , what difference does it make on -a/2 compared to the rest of the square root?

Anyway , with my exemple of x=3 , a=4 and b= -21 it all works out in the end and 3 is indeed isolated.

thanks!


Then you have

 

[Fredrik]

I think it's much harder to see what's going on when you assign values to the variables and work with the numbers instead.

The statement

For all $a,b\in\mathbb R$, we have $(a+b)^2=a^2+b^2+2ab$.​

is an easy to prove theorem about real numbers. This theorem implies that for all $a,x\in\mathbb R$, we have
$$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$ What I did was to rewrite $x^2+ax+b$ as a sum of five terms, three of which are the ones on the right-hand side of the equality above. So then I could just apply the result above to those three terms.

This trick is called "completing the square".

What happened to 12?

12 is one of the three terms whose sum you rewrote as (3 + 4/2)^2.

Just a minor question about the end of the formula , why the ± sign?

$x=\pm a$ really means $x\in\{-a,a\}$. When you see ± in a formula, you should think of the formula as representing two different formulas, one with a plus sign and one with a minus sign. Each of the formulas is a solution to the equation.

 

[reenmachine]

I think it's much harder to see what's going on when you assign values to the variables and work with the numbers instead.

The statement

For all $a,b\in\mathbb R$, we have $(a+b)^2=a^2+b^2+2ab$.​

is an easy to prove theorem about real numbers. This theorem implies that for all $a,x\in\mathbb R$, we have
$$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$ What I did was to rewrite $x^2+ax+b$ as a sum of five terms, three of which are the ones on the right-hand side of the equality above. So then I could just apply the result above to those three terms.

This trick is called "completing the square".

Sorry I'm a bit confused here , do you mean that the two formulas are related? because I didn't think so.

Also , in $$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$

Shouldn't it be 2ax? If the a/2 is what makes the 2 in 2ax disappear , then why isn't the x divided by 2 also? (I understand in practice it wouldn't make sense if we tried it with numbers but I'm trying to understand the logic).

 

[reenmachine]

I suspect that Reenmachine is just relieved to see that I too can fail to find proofs that look pretty simple when you have them in front of you.

ahahah I don't know if I'm relieved but I feel less lonely

Though I don't mind being the lonely student of many mentors , you certainly won't find that in the school system.Maybe it is my insecurity , but I'm sometimes scared that if I don't ''get it'' fast enough people will get bored so I put pressure on myself to move forward.I think it can be both good and bad , good because I force myself to a higher standard and bad because it can get frustrating when I don't meet that high standard.

In other words , I assume it's funnier to teach someone when you can see his progress (even though I have no clue how to teach so that's just an assumption) , so I want to show progress for the help I'm receiving.

 

[reenmachine]

Sorry reen, for this off-topic discussion.

No problem , I'm honored to have undirectly generated some mathematical discussions

 

[Digitalism]

Shouldn't it be 2ax? If the a/2 is what makes the 2 in 2ax disappear , then why isn't the x divided by 2 also? (I understand in practice it wouldn't make sense if we tried it with numbers but I'm trying to understand the logic).

When you multiply out (x + a/2)^2 you get x^2 + ax/2 + ax/2 + (a/2)^2 the two middle terms add up to ax which shows the equivalence between both sides of the equation you quoted

 

[Fredrik]

Sorry I'm a bit confused here , do you mean that the two formulas are related? because I didn't think so.

Not related? They are essentially the same statement.

Edit: On second thought, I'm not sure what you meant by "the two formulas". The two that I was referring to when I said that they are "essentially the same statement" are the ones in this quote:

The statement

For all $a,b\in\mathbb R$, we have $(a+b)^2=a^2+b^2+2ab$.​

is an easy to prove theorem about real numbers. This theorem implies that for all $a,x\in\mathbb R$, we have
$$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$

Edit 2: OK, maybe I see what you meant. When I said "There's a clever trick involving that formula that gets x alone on one side of the equality sign", it wasn't clear enough that what I meant by "that formula" was $(a+b)^2=a^2+b^2+2ab$.

Also , in $$\left(x+\frac a 2\right)^2=x^2+\left(\frac a 2\right)^2+ax.$$

Shouldn't it be 2ax? If the a/2 is what makes the 2 in 2ax disappear , then why isn't the x divided by 2 also? (I understand in practice it wouldn't make sense if we tried it with numbers but I'm trying to understand the logic).

Don't replace the variables with numbers. Just use the theorem

For all $a,b\in\mathbb R$, we have $(a+b)^2=a^2+b^2+2ab$.​

to evaluate
$$\left(x+\frac a 2\right)^2,$$
or perhaps even better, forget about the theorem and just use that multiplication is distributive over addition. Don't replace the variables with numbers.

Maybe it is my insecurity , but I'm sometimes scared that if I don't ''get it'' fast enough people will get bored so I put pressure on myself to move forward.I think it can be both good and bad , good because I force myself to a higher standard and bad because it can get frustrating when I don't meet that high standard.

In other words , I assume it's funnier to teach someone when you can see his progress (even though I have no clue how to teach so that's just an assumption) , so I want to show progress for the help I'm receiving.

You clearly have made some progress, and I think that the rate at which you make progress will increase significantly once you have reached the level where you understand the basics. We just have to get you there.

I am getting a little bit tired, but I can certainly handle one more bunch of exercises, like the bunch we're discussing now. After that, I think you should start using the homework forum for exercises. Keep using this thread for discussions about definitions and theorems about sets, relations and functions.

 

[Fredrik]

I should also have said that this is a fun thread and a fun experiment. These are interesting topics, and it's also interesting to me to see how this works out. These things are usually taught to people with a lot more experience of math than you have, so I had no idea if you would find them easy or hard. That's why this is a bit of an "experiment".

I think you would have found it much easier to understand things like set-builder notation, dummy variables, and the formal definitions of relations and functions if you had had more experience with basic algebra, 1st and 2nd degree polynomial equations, the informal description of a function from X into Y as "a rule that associates exactly one element of Y with each element of X", plotting graphs of functions, etc.

I guess there's a reason why these things aren't taught at the high school level. I still think this is doable. You just have to be prepared to do a lot of silly-looking exercises, and maybe also to take some time to refresh your memory about some high school stuff that you don't know as well as you will need to. You will also have to be willing to try and try again when you don't see the solution to a problem. Stubbornness pays off. Think of what you did to find the expression $\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ for one of the problems discussed above. When you finally had it in front of you, you knew that it was right before I had confirmed it. This is a good way to learn stuff. The downside is that it takes a lot of time.

 

[reenmachine]

You clearly have made some progress, and I think that the rate at which you make progress will increase significantly once you have reached the level where you understand the basics. We just have to get you there.

I am getting a little bit tired, but I can certainly handle one more bunch of exercises, like the bunch we're discussing now. After that, I think you should start using the homework forum for exercises. Keep using this thread for discussions about definitions and theorems about sets, relations and functions.

I think I've also had a rough week on a mental energy level compared to the two other weeks I've spend in this thread.Do not worry , I won't come here with countless exercises everyday , I might use one here and there just to ensure that I understood correctly by presenting something I've worked on , but I will try to keep this thread for definitions and concepts.

 

[reenmachine]

I should also have said that this is a fun thread and a fun experiment. These are interesting topics, and it's also interesting to me to see how this works out. These things are usually taught to people with a lot more experience of math than you have, so I had no idea if you would find them easy or hard. That's why this is a bit of an "experiment".

This is an experiment for me as well.I'm well aware that I lack some basic stuff that would make my life easier , but at this point I'm in so I have to fight without these basics.If I do end up understanding most of what is in the book of proof , I might become a weird case in the sense that I'll know some deep math all the while lacking some basics which would normally be illogical in a student that followed a normal path.

I think you would have found it much easier to understand things like set-builder notation, dummy variables, and the formal definitions of relations and functions if you had had more experience with basic algebra, 1st and 2nd degree polynomial equations, the informal description of a function from X into Y as "a rule that associates exactly one element of Y with each element of X", plotting graphs of functions, etc.

I guess there's a reason why these things aren't taught at the high school level. I still think this is doable. You just have to be prepared to do a lot of silly-looking exercises, and maybe also to take some time to refresh your memory about some high school stuff that you don't know as well as you will need to. You will also have to be willing to try and try again when you don't see the solution to a problem. Stubbornness pays off. Think of what you did to find the expression $\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ for one of the problems discussed above. When you finally had it in front of you, you knew that it was right before I had confirmed it. This is a good way to learn stuff. The downside is that it takes a lot of time.

I do have an history of self-learning something successfully (at least I think) , the english language! Three years ago , I wasn't capable of writing a single sentence nevermind have conversions about advanced mathematics.At the beginning it was painful , it was hard to keep reading and trying to have discussions when I struggled to understand half of what was said , but eventually it got easier and one day without realizing it I could think in english in my head.

Thanks a lot Fredrik , your feedback is always greatly appreciated and can't thank you enough for the time you've spend helping me.I know I'm repeating myself , but I want you to know that I'm sincerely grateful.

 

[reenmachine]

$\{x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$.

About this , would it be better to write $\{∀x\in\mathbb N\,|\,\exists y\in\mathbb N~~ x=2^y\}$ ?