Question about proof from a guy with a highschool education

[reenmachine]

In retrospective Fredrik , you might have been right that I sometimes confuse $\cap$ with $\cup$ , it's weird because I clearly know which is which but when I'm in the middle of a thinking process I can confuse them out of nowhere.

Like here , for exemple:

$\emptyset \cap A = A$

what the hell was I thinking?

My next move is to do some exercises from the bookofproof , I have to work through some problems if I want to stop making stupid mistakes that I shouldn't be doing.

 

[Fredrik]

In retrospective Fredrik , you might have been right that I sometimes confuse $\cap$ with $\cup$ , it's weird because I clearly know which is which but when I'm in the middle of a thinking process I can confuse them out of nowhere.
...
My next move is to do some exercises from the bookofproof , I have to work through some problems if I want to stop making stupid mistakes that I shouldn't be doing.

You have seen both me and micromass make mistakes in the thread, especially me. It's easy to make mistakes in set theory for some reason. But practice will definitely make you more likely to get through a given problem without a blunder.

One post that made me think that you sometimes confuse unions and intersections is #473. You asked (in words) if
$$\bigcap\mathcal P(\mathbb R)=\mathcal P(\mathbb R).$$ This equality wouldn't be correct even if we flip that cap upside down, but I still thought that maybe you were thinking of unions, since you came up with a "big" set instead of a "small" one as the answer.
\begin{align}
\bigcap\mathcal P(\mathbb R) &=\text{intersection of all elements of }\mathcal P(\mathbb R) =\text{intersection of all subsets of }\mathbb R =\varnothing,\\
\\
\bigcup\mathcal P(\mathbb R) &=\text{union of all elements of }\mathcal P(\mathbb R) =\text{union of all subsets of }\mathbb R =\mathbb R.
\end{align}

 

[reenmachine]

One post that made me think that you sometimes confuse unions and intersections is #473. You asked (in words) if
$$\bigcap\mathcal P(\mathbb R)=\mathcal P(\mathbb R).$$ This equality wouldn't be correct even if we flip that cap upside down, but I still thought that maybe you were thinking of unions, since you came up with a "big" set instead of a "small" one as the answer.
\begin{align}
\bigcap\mathcal P(\mathbb R) &=\text{intersection of all elements of }\mathcal P(\mathbb R) =\text{intersection of all subsets of }\mathbb R =\varnothing,\\
\\
\bigcup\mathcal P(\mathbb R) &=\text{union of all elements of }\mathcal P(\mathbb R) =\text{union of all subsets of }\mathbb R =\mathbb R.
\end{align}

hmm I remember that , I don't think it was related to confusing unions and intersections , but I did confuse them in other posts after anyway

The post you are talking about was probably just a logic mistake.

Sometime you try hard to understand new concepts but you neglect what you already know.

 

[reenmachine]

Exercises (from the book of proof section 1.8):

For each $n \in N$ , let $A_n= \{-2n , 0 , 2n\}$

(a)
$$\bigcup_{i \in N}A_i = ?$$

Answer: $\{x \in Z : 2x\}$

(b)
$$\bigcap_{i \in N}A_i = ?$$

Answer: $\{0\}$

 

[micromass]

Answer: $\{x \in Z : 2x\}$

This is bad notation.

I read it as: the set of all integers $x$ in $\mathbb{Z}$ such that $2x$ is true.
Clearly, this makes no sense.

 

[reenmachine]

Exercises (from the book of proof section 1.8):

(a)

$$\bigcup_{a \in R}\{a\} × [0,1] = ?$$

Here I'm not sure I understand what they are saying with this notation.Is it (a) × [all numbers between 0 and 1] ?

If that's the case:

Answer: \{R\}

 

[micromass]

Exercises (from the book of proof section 1.8):

(a)

$$\bigcup_{a \in R}\{a\} × [0,1] = ?$$

Here I'm not sure I understand what they are saying with this notation.Is it (a) × [all numbers between 0 and 1] ?

If that's the case:

Answer: \{R\}

No, that answer is incorrect. How did you find it??

Instead of just giving your answer, I really would appreciate it if you would also give a formal proof of why the answer is correct. For example, what Fredrik did in post 504.

 

[reenmachine]

This is bad notation.

I read it as: the set of all integers $x$ in $\mathbb{Z}$ such that $2x$ is true.
Clearly, this makes no sense.

I know the set I'm looking for is the set of all even integers , positive and negative.Don't remember how to write it.

Let me try it again:

$\{x \in Z : \exists y \in Z \ 2y=x\}$

 

[micromass]

I know the set I'm looking for is the set of all even integers , positive and negative.Don't remember how to write it.

Let me try it again:

$\{x \in Z : \exists y \in Z \ 2y=x\}$

Right.

 

[Fredrik]

The $\times$ in post #511 is a cartesian product.

 

[reenmachine]

No, that answer is incorrect. How did you find it??

Instead of just giving your answer, I really would appreciate it if you would also give a formal proof of why the answer is correct. For example, what Fredrik did in post 504.

I would like to be sure I'm reading the notation correctly at least.

Is it (a) × [all numbers between 0 and 1 , including both of them] ?

 

[micromass]

I would like to be sure I'm reading the notation correctly at least.

Is it (a) × [all numbers between 0 and 1 , including both of them] ?

Yes, it's

\{a\}\times \{x\in \mathbb{R}~\vert~ 0\leq x\leq 1\}

 

[reenmachine]

The $\times$ in post #511 is a cartesian product.

I'm a little bit confused about how you can multiply a number with something like [0,1].I don't think I've ever encountered such a thing.

 

[reenmachine]

Yes, it's

\{a\}\times \{x\in \mathbb{R}~\vert~ 0\leq x\leq 1\}

hmmm ok , let me think about it some more.

 

[micromass]

I'm a little bit confused about how you can multiply a number with something like [0,1].I don't think I've ever encountered such a thing.

We don't multiply a number with [0,1]. We take the cartesian product of the set $\{a\}$ (this is not a number, but a set containing the number) and the set [0,1].

 

[reenmachine]

No, that answer is incorrect. How did you find it??

Instead of just giving your answer, I really would appreciate it if you would also give a formal proof of why the answer is correct. For example, what Fredrik did in post 504.

Okay ,my thought process was that since 1 is included , then all real numbers could be multiplied by 1 and it would give me all real numbers.Since it's the union of all these multiplications , I figured in the end it would be all real numbers present in the set.The other multiplication like (3) × 0,6 for exemple would just give me 1,8 , which is the same as (1,8) × 1.So they are all guaranteed to be there.

 

[micromass]

Okay ,my thought process was that since 1 is included , then all real numbers could be multiplied by 1 and it would give me all real numbers.Since it's the union of all these multiplications , I figured in the end it would be all real numbers present in the set.

But $\times$ isn't multiplication. It's the cartesian product.

 

[reenmachine]

So the cartesian product would look like : $( x \in R , 0 ≤ x ≤ 1)$ ? (an ordered pair)

 

[micromass]

So the cartesian product would look like : $( x \in R , 0 ≤ x ≤ 1)$ ? (an ordered pair)

The cartesian product of which sets would like like that? I don't understand the question.

 

[reenmachine]

How about that? $\{(x,y) \in R : x \in R \ \ , 0 ≤ y ≤ 1 \}$

 

[micromass]

How about that? $\{(x,y) \in R : x \in R \ \ , 0 ≤ y ≤ 1 \}$

Is this supposed to be an answer to the question in 511?

In that case, can you please provide a formal proof of this. Like Fredrik did in 504.

 

[reenmachine]

Is this supposed to be an answer to the question in 511?

In that case, can you please provide a formal proof of this. Like Fredrik did in 504.

Yes , it's suppose to

Okay I will try to prove it.

 

[reenmachine]

(x,y) is an element of the union of all cartesian products of $\{a\} × [0,1]$ with $a$ being an element of the set $R$.

union: (x,y) is an element of at least one cartesian product of $\{a\} × [0,1]$ with $a$ being an element of the set $R$.

$R$ is the set of all real numbers.

Since it's $\{a\} × [0,1]$ , the first componant of every ordered pair of this set will be an element of $R$ and the second will be a number between 0 and 1 (with 0 and 1 included because of the [] ).Since all numbers between 0 and 1 are real numbers , then the second componant will also be an element of $R$.

This means that $(x,y) \in R$ , and since $x \in R$ and $0 ≤ y ≤ 1$ , this set will be the set $\{ (x,y) \in R : x \in R \ , 0 ≤ y ≤ 1\}$.

 

[micromass]

(x,y) is an element of the union of all cartesian products of $\{a\} × [0,1]$ with $a$ being an element of the set $R$.

union: (x,y) is an element of at least one cartesian product of $\{a\} × [0,1]$ with $a$ being an element of the set $R$.

$R$ is the set of all real numbers.

Since it's $\{a\} × [0,1]$ , the first componant of every ordered pair of this set will be an element of $R$ and the second will be a number between 0 and 1 (with 0 and 1 included because of the [] ).Since all numbers between 0 and 1 are real numbers , then the second componant will also be an element of $R$.

This means that $\{(x,y) \in R\ : ...\}$ , and since $x \in R$ and $0 ≤ y ≤ 1$ , this set will be the set $\{ (x,y) \in R : x \in R \ , 0 ≤ y ≤ 1\}$.

That only proves

\bigcup_{a\in \mathbb{R}} \{a\}\times [0,1]\subseteq \mathbb{R}\times [0,1]

You need ot prove the other inclusion too.

 

[reenmachine]

That only proves

\bigcup_{a\in \mathbb{R}} \{a\}\times [0,1]\subseteq \mathbb{R}\times [0,1]

You need ot prove the other inclusion too.

How could the first be a subset of the second? It should be the opposite no?

edit:forget it , they are the same aren't they?

 

[reenmachine]

Let assume x is an element of $\bigcup_{a\in \mathbb{R}} \{a\}$ and that y is an element of [0,1].

This implies that $(x,y) \in \bigcup_{a\in \mathbb{R}} \{a\}$ × $[0,1]$.

Now let assume $(x,y) \in \bigcup_{a\in \mathbb{R}} \{a\}$ × $[0,1]$ , this implies that $x \in R$ and that $0 ≤ y ≤ 1$.Therefore it proves that this is the set $\{ (x,y) \in \bigcup_{a\in \mathbb{R}} \{a\} × [0,1]: x \in R \ , 0 ≤ y ≤ 1\}$.

 

[Fredrik]

I think it's almost time to start doing things a more normal way. I don't mind if you keep using the thread this way for the rest of the day (and night), but after that, I would like you to start using the homework forums for questions about textbook problems*.

You can keep using this thread for questions about definitions and theorems in the Book of Proof. When you start with calculus, you should start another thread to discuss concepts from your calculus book.


*) Note that we have specific guidelines about how to ask for homework help.

 

[reenmachine]

Is what I'm missing the fact that the ordered pair will always be [R , 0] , [R , 0,002] , ... , [R , 1]. With R truly meaning all numbers in R?

This would mean : Now let assume $(x,y) \in \bigcup_{a\in \mathbb{R}} \{a\}$ × $[0,1]$ , this implies that $x = R$ and that $0 ≤ y ≤ 1$.Therefore it proves that this is the set $\{ (x,y) \in \bigcup_{a\in \mathbb{R}} \{a\}× [0,1] : x = R \ , 0 ≤ y ≤ 1\}$.

(The union of all elements of R = R.Which is why the x=R was implied)

 

[reenmachine]

I think it's almost time to start doing things a more normal way. I don't mind if you keep using the thread this way for the rest of the day (and night), but after that, I would like you to start using the homework forums for questions about textbook problems*.

You can keep using this thread for questions about definitions and theorems in the Book of Proof. When you start with calculus, you should start another thread to discuss concepts from your calculus book.*) Note that we have specific guidelines about how to ask for homework help.

No problem.I think the thread has been more about questions and definitions than exercises anyway.I only tried exercises here and there to verify if I understood.

It wasn't my intention to make you guys correct a bunch of exercises for me.

 

[Fredrik]

Let assume x is an element of $\bigcup_{a\in \mathbb{R}} \{a\}$ and that y is an element of [0,1].

Since that union is equal to $\mathbb R$, this is the same thing as assuming that $(x,y)\in\mathbb R\times[0,1]$.

This implies that $(x,y) \in \bigcup_{a\in \mathbb{R}} \{a\}$ × $[0,1]$.

I would interpret the notation on the right here as
$$\bigcup_{a\in \mathbb{R}} \left(\{a\}\times [0,1]\right),$$ not as
$$\left(\bigcup_{a\in \mathbb{R}} \{a\}\right)\times [0,1].$$ The notation is kind of ambiguous though. I don't know if one of these interpretations are "standard". I'm also not sure how you're interpreting it.

You haven't proved that (x,y) is an element of the latter of these two sets, because that statement is your starting assumption.

And you haven't proved that (x,y) is an element of the former of the two sets either, but it's not hard to add a statement that takes care of that.

So let's be really clear here. Let's say that we want to prove that $$\bigcup_{a\in \mathbb{R}} \left(\{a\}\times [0,1]\right)=\mathbb R\times[0,1].$$ The parentheses make this problem unambiguous. This is how I would start:

Let $z\in \bigcup_{a\in \mathbb{R}} \left(\{a\}\times [0,1]\right)$ be arbitrary. Let x be an arbitrary real number such that $z\in\{x\}\times [0,1]$. Let y be an arbitrary element of [0,1] such that $z=(x,y)$. Since $x\in R$ and $y\in[0,1]$, this equality implies that $z\in\mathbb R\times [0,1]$. Edit: You know what, I'm going to rewrite this in an even clearer way. See below in a few minutes.

Can you do the other one, starting like this:

Let $z\in\mathbb R$ be arbitrary.



Edit: Here's the maximal clarity version of the proof above:

Let $z\in \bigcup_{a\in \mathbb{R}} \left(\{a\}\times [0,1]\right)$ be arbitrary. The definition of this notation implies that there's a $b\in\mathbb R$ such that $z\in\{b\}\times[0,1]$. Let $x$ be such a real number. We have $z\in\{x\}\times[0,1]$. The definition of this notation implies that there's a $c\in[0,1]$ such that $z=(x,c)$. Let $y$ be such an element of [0,1]. We have $z\in (x,y)$. Since $x\in\mathbb R$ and $y\in[0,1]$, this implies that $z\in\mathbb R\times[0,1]$.

 

[reenmachine]

Just saw your edit lol Saved by the bell.

 

[reenmachine]

Let $z \in R × [0,1]$ be arbitrary.This implies that there's an arbitrary element $x \in R$ and an arbitrary element $y \in [0,1]$.The notation of the set implies that $(x,y) = z$.Since $y \in [0,1]$ and $x \in R$ , that the union of all elements of $R = R$ , this implies that $z \in \bigcup_{a \in R}\{a\} × [0,1]$.

 

[Fredrik]

I'm done with the edit. My proof at the end is more detailed than the proofs you will find in textbooks. For example, I said "there's a real number b such that..." and then followed up with "let x be such a real number". Textbooks always smash these two statements into one. I think most of the authors choose to emphasize only the "there exists" part by saying "there's a real number x such that..." The problem with this is that when they refer to "x" in the next sentence, it's not really clear that they're referring to an x with the property just discussed, because the x in the "there exists" statement was a dummy variable.

So this way of presenting the proof is an abuse of language, but people use it anyway because it's a pain to have to think of a second variable to use and then make both statements.

What I did in my first version of the proof was to emphasize the assignment of a value to a new variable x instead of the "there exists" part. I often do that by saying something like "let x be a real number such that...". From a strictly logical perspective, this is better because now when the next sentence refers to x, there's no question of what x. But it may still be more difficult to understand, since it's now up to the reader to understand that the previous statement implies the existence of a real number with the necessary property.

So one thing I could have done is to add a comment after my "let x..." statement. For example: (The definition of "union" implies that such an x exists).

 

[micromass]

This implies that there's an arbitrary element $x \in R$ and an arbitrary element $y \in [0,1]$.The notation of the set implies that $(x,y) = z$.

Well, $x$ and $y$ aren't arbitrary anymore. They are completely determined by $z$. So $z=(x,y)$ for a unique $x$ and $y$.

 

[reenmachine]

Well, $x$ and $y$ aren't arbitrary anymore. They are completely determined by $z$. So $z=(x,y)$ for a unique $x$ and $y$.

I'm not sure I understand , what am I suppose to say then? Just leave the word ''arbitrary'' out?

Should I say , R × [0,1] implies that z = (x,y)?

 

[micromass]

I'm not sure I understand , what am I suppose to say then? Just leave the word ''arbitrary'' out?

Yeah, you shouldn't say that $x$ and $y$ are arbitrary, since they are completely determined. Only $z$ is arbitrary.

 

[reenmachine]

Yeah, you shouldn't say that $x$ and $y$ are arbitrary, since they are completely determined. Only $z$ is arbitrary.

Corrected version:

Let $z \in R × [0,1]$ be arbitrary.This implies that there's an element $x \in R$ and an element $y \in [0,1]$.The notation of the set implies that $z = (x,y)$.Since $y \in [0,1]$ and $x \in R$ , that the union of all elements of $R = R$ , this implies that $z \in \bigcup_{a \in R}\{a\} × [0,1]$.

 

[reenmachine]

I'm done with the edit. My proof at the end is more detailed than the proofs you will find in textbooks. For example, I said "there's a real number b such that..." and then followed up with "let x be such a real number". Textbooks always smash these two statements into one. I think most of the authors choose to emphasize only the "there exists" part by saying "there's a real number x such that..." The problem with this is that when they refer to "x" in the next sentence, it's not really clear that they're referring to an x with the property just discussed, because the x in the "there exists" statement was a dummy variable.

So this way of presenting the proof is an abuse of language, but people use it anyway because it's a pain to have to think of a second variable to use and then make both statements.

What I did in my first version of the proof was to emphasize the assignment of a value to a new variable x instead of the "there exists" part. I often do that by saying something like "let x be a real number such that...". From a strictly logical perspective, this is better because now when the next sentence refers to x, there's no question of what x. But it may still be more difficult to understand, since it's now up to the reader to understand that the previous statement implies the existence of a real number with the necessary property.

So one thing I could have done is to add a comment after my "let x..." statement. For example: (The definition of "union" implies that such an x exists).

Hmmm I see , I was wondering why you were doing that.I do ''get it'' , but it's harder to do than it looks when you read it and approve

thanks man!

 

[reenmachine]

I would interpret the notation on the right here as
$$\bigcup_{a\in \mathbb{R}} \left(\{a\}\times [0,1]\right),$$ not as
$$\left(\bigcup_{a\in \mathbb{R}} \{a\}\right)\times [0,1].$$ The notation is kind of ambiguous though. I don't know if one of these interpretations are "standard". I'm also not sure how you're interpreting it.

To be honest at the beginning I was wondering which one it was.

In the second one , does it mean that each ordered pair would look like [all real numbers , 1]?

 

[Fredrik]

I had started typing this when my brother interrupted me with a phone call. I see that you are already working on these issues with micromass, but I might as well post it since I had finished this post before I saw that.

Let $z \in R × [0,1]$ be arbitrary.

So far so good.

This implies that there's an arbitrary element $x \in R$ and an arbitrary element $y \in [0,1]$.

Either say "this implies that there's an $x\in\mathbb R$ such that..." or "let x be an arbitrary real number such that...". Don't say "implies that there's an arbitrary $x\in\mathbb R$..."

More importantly, what you're saying here is just that ℝ and [0,1] are non-empty sets. They are, but, we didn't need to use the statement $z \in\mathbb R × [0,1]$ to see that.

The notation of the set implies that $(x,y) = z$.

You said that x,y,z were all arbitrary, so this doesn't follow from what you said.

 

[Fredrik]

In the second one , does it mean that each ordered pair would look like [all real numbers , 1]?

I don't see what that would mean. The union in parentheses is equal to $\mathbb R$, so the set is
$$\mathbb R\times [1,0] =\{(x,y)\in \mathbb R\times \mathbb R:0\leq y\leq 1\}.$$

 

[reenmachine]

I don't see what that would mean. The union in parentheses is equal to $\mathbb R$, so the set is
$$\mathbb R\times [1,0] =\{(x,y)\in X\times Y:0\leq y\leq 1\}.$$

I know what you mean , but wasn't aware it was possible to say $(x,y)\in X\times Y$.This doesn't give any indication of what X is.

 

[reenmachine]

So is this version correct?

Let $z \in R × [0,1]$ be arbitrary.This implies that there's an element $x \in R$ and an element $y \in [0,1]$.The notation of the set implies that $z = (x,y)$.Since $y \in [0,1]$ and $x \in R$ , that the union of all elements of $R = R$ , this implies that $z \in \bigcup_{a \in R}\{a\} × [0,1]$.

thanks!

 

[Fredrik]

I know what you mean , but wasn't aware it was possible to say $(x,y)\in X\times Y$.This doesn't give any indication of what X is.

Sorry, I don't know why I typed X and Y. I meant $(x,y)\in\mathbb R\times\mathbb R$. But even if we do, what we get is just a more complicated way of saying $\mathbb R\times[0,1]$.

 

[Fredrik]

Let $z \in R × [0,1]$ be arbitrary.This implies that there's an element $x \in R$ and an element $y \in [0,1]$. The notation of the set implies that $z = (x,y)$

The second sentence tells us nothing other than that $\mathbb R\times[0,1]$ is non-empty. The statement $z \in\mathbb R × [0,1]$ doesn't say anything about x or y.

 

[reenmachine]

Let $z \in R × [0,1]$ be abitrary.This implies that z is an element of the form (a,b).For all $x\in R$ and $0≤y≤1$ , $z=(x,y)$ and $(x,y) \in R × [0,1]$.Since $x \in R$ , it implies that $x \in \bigcup_{a\in R}\{a\}$ and since $0≤y≤1$ , it implies that $y\in[1,0]$.Since $z=(x,y)$ for all x and y , it implies that $z \in \bigcup_{a\in R}\{a\} × [0,1]$.hmmm , I'm not sure at all about that one , it is pretty hard to connect the dots for some reasons.

 

[Fredrik]

Let $z \in R × [0,1]$ be abitrary.This implies that z is an element of the form (a,b)

This is true, but after this you never mentioned a and b again.

For all $x\in R$ and $0≤y≤1$ , $z=(x,y)$

Here you're saying that z is equal to infinitely many things.

It is pretty difficult to do these proofs until you get used to them. One tip is to leave the proof for a while, and then read what you have written. If you don't understand your own argument then, no one else will either.

 

[reenmachine]

Ok , I'm going to take a dinner break and try to come up with something after.

thanks man!

 

[reenmachine]

This is true, but after this you never mentioned a and b again.

If I were to say that a is in R and b is in [0,1] , would that make sense or would it be irrelevant because a and b are dummy variables?

Here you're saying that z is equal to infinitely many things.

I'm not sure I understand this , doesn't it say that z=(x,y) and that those x and y are the x in R and y in [0,1] I was talking about?

 

[reenmachine]

Let $z \in R × [0,1]$ be arbitrary.This implies that there's an $x \in R$ and a $y \in [0,1]$ such that $z \in (x,y)$.This implies that $x \in \bigcup_{a\in R}\{a\}$ and that $y \in [0,1]$ such that $(x,y) \in \bigcup_{a \in R}\{a\} × [0,1]$.Since $z \in (x,y)$ , this proves that $z \in \bigcup_{a \in R}\{a\} × [0,1]$

Here I have a feeling I'm missing something between $z \in (x,y)$ and the conclusion , but I just can't find it.

editk I am done editing