Question about proof from a guy with a highschool education

[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

 

[Fredrik]

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?

I would interpret your statement as "there exists a,b such that z=(a,b)", so both variables are the target of a "there exists", and strictly speaking, that makes them dummy variables, and a reference to a or b in the next sentence doesn't make sense. However, since of the abuse of language I described earlier is so common, I would have interpreted a reference to a or b in the next sentence as if you had made two separate statements: "There exists c,d such that z=(c,d)" and "Let a and b be such that z=(a,b)".

It makes sense to change "This implies that z is an element of the form (a,b)" to "This implies that z is an element of the form (a,b) with $a\in\mathbb R$ and $b\in[0,1]$", and I consider this an improvement. But what's really missing from your statement is something that makes it clear that you are assigning values to the variables a and b. (So that it makes sense to refer to them later).

You could e.g. say "define $a\in\mathbb R$ and $b\in[0,1]$ by $z=(a,b)$" or "let a,b be the unique real numbers such that z=(a,b)"

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?

You said that for all real numbers x,y with y in [0,1], we have z=(x,y). This implies that all of the following equalities and infinitely many more are all true: $(1,0)=z,\ (-1/12,0)=z,\ (\pi,1/2)=z$...and they all contradict each other.

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 $z \in (x,y)$.

The last $\in$ should be an equality, but I assume that was just a typo (hmm...repeated at the end of the proof). The statement is fine apart from that. You could say that there's a unique $x\in\mathbb R$ and a unique $y\in[0,1]$ such that z=(x,y) if you want to emphasize that the values of x and y are fully determined by z, but it's not necessary to do that in this proof.

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

The "such that" makes the sentence weird. You can end the sentence after $y\in[0,1]$, and then say "this implies that...". But you don't actually have to mention that $y\in[0,1]$, since you did that earlier.

It looks like what you're trying to do here is what I would say like this: Since $\mathbb R=\bigcup_{a\in R}\{a\}$, this implies that we have $z=(x,y)\in\left(\bigcup_{a\in R}\{a\}\right)\times[0,1]$.

But this right-hand side isn't the one we're interested in. We want to prove that $z\in \bigcup_{a\in R}\left(\{a\}\times[0,1]\right)$.

 

[reenmachine]

I meant z = (x,y) , both times.Sorry about that.Also not sure if you saw the edited version.

 

[reenmachine]

It looks like what you're trying to do here is what I would say like this: Since $\mathbb R=\bigcup_{a\in R}\{a\}$, this implies that we have $z=(x,y)\in\left(\bigcup_{a\in R}\{a\}\right)\times[0,1]$.

But this right-hand side isn't the one we're interested in. We want to prove that $z\in \bigcup_{a\in R}\left(\{a\}\times[0,1]\right)$.

Yes this is pretty much what I tried to do.Seems I confused everything again.

 

[Fredrik]

I think the comment "I'm done editing" was there when I started typing the reply.

 

[reenmachine]

In that case , it should be easier to prove in the sense you only have to prove that $z=(x,y)$ and that $(x,y) \in {a} × [0,1]$

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

?

Is this sufficient? The truth is it looks like much of the same of what I previously attempted.I'm hitting a wall.

 

[Fredrik]

OK, I'll tell you.

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 $z = (x,y)$.

All you need to say after that is that
$$z=(x,y)\in\{x\}\times[0,1]\subseteq\bigcup_{a\in\mathbb R}\{a\}\times[0,1].$$

 

[reenmachine]

OK, I'll tell you.All you need to say after that is that
$$z=(x,y)\in\{x\}\times[0,1]\subseteq\bigcup_{a\in\mathbb R}\{a\}\in[0,1].$$

$$z=(x,y)\in\{x\}\times[0,1]\subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1]?$$

× instead of $\in$ at the end no?

Okay so I had to go through the $(x,y) \in \{x\} × [0,1] \subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1] ...$.

But why is it necessary to say that $(x,y) \in \{x\} × [0,1]$ , isn't implied in $(x,y) \in R × [0,1]$ when it was already said that $z=(x,y)$ and that $z \in R × [0,1]$ ?

 

[reenmachine]

I guess my next move is to sleep on it and see what happens tomorrow

This has been one of my roughest day in this thread.

 

[Fredrik]

× instead of $\in$ at the end no?

Yes. (I have edited that now).

Okay so I had to go through the $(x,y) \in \{x\} × [0,1] \subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1] ...$.

But why is it necessary to say that $(x,y) \in \{x\} × [0,1]$ , isn't implied in $(x,y) \in R × [0,1]$ when it was already said that $z=(x,y)$ and that $z \in R × [0,1]$ ?

Actually, what you said in #560 is fine. I just think it's clearer with the extra step. I mean, it follows immediately from the definition of $\times$ that $(x,y)\in\{x\}\times[0,1]$, and now the definition of "union" implies that $(x,y)\in\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$.

 

[reenmachine]

Yes. (I have edited that now).Actually, what you said in #560 is fine. I just think it's clearer with the extra step. I mean, it follows immediately from the definition of $\times$ that $(x,y)\in\{x\}\times[0,1]$, and now the definition of "union" implies that $(x,y)\in\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$.

This was the problem , I wasn't sure what was missing.I would lie if I told you I'm 100% confidant in my understanding of everything we've discussed today.But I'll get there eventually.

Not sure that it was the easiest exemple to try either.

thank you infinitely for the patience!

 

[reenmachine]

By the way , it's been a month now since this thread has been created!

I am extremely satisfied with the help I've received.This has been an incredibly positive experience for me to the point where I deeply regret not doing it years ago.

Learning some completely unknown mathematics has been tough and remains tough , especially with my poor background , but I'm impressed by the quality of the helpers helping me getting through these concepts despite my lack of knowledge.

A special thank to you Fredrik , who has been the most active helper all along , you truly deserve to be called a mentor! I think I should also name micromass who has been extremely helpful to me.

thank you very much to everybody who helped me in the thread!

 

[reenmachine]

When we said $(x,y)\in\{x\}\times[0,1]$ yesterday , what does $\{x\}$ truly mean here? I know it's suppose to be any real number , but what difference does it make to put an $x$ out there instead of $R$ or $a$ like in the union notation? Is it because we want to pinpoint that $x$ is there as $\{x\}$ confirming both the position of $x$ and the position of $y$ in the formula ? Isn't $(x,y)$ enough? I'm trying to understand the importance of this statement.Is it because we want to pass through $\{x\} × [0,1]$ to reach $\subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1]$? If so , why is that crucial? Why not $\{a\}$ instead of $\{x\}$?

 

[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.

I think this was the most problematic aspect of how I handled the whole thing yesterday.

I think I even confused which notations I was trying to prove in the middle of my whole thought process.Here we want to prove the first one correct?

So by saying that $(x,y)\in\{x\}\times[0,1]$ , we're saying that $z \in (G)$ and that $z \in \bigcup (G)$ if $z = (x,y)$ and $(G)=\{\{x\}×[0,1]\}$.Is that a correct way of viewing things? Because in my mind , if that's correct , it's much clearer now than yesterday.

 

[reenmachine]

A very simple thing seems to confuse me at this very moment , what's the difference between a set $A$ and $\bigcup A$?

Suppose $A=\{ (0,1) , (1,2)\}$.

The reason I'm asking is I'm not sure of the difference between

$$\bigcup_{a\in \mathbb{R}} \left(\{a\}\times [0,1]\right),$$
and

$\{a\} × [0,1]$ with $a \in R$

 

[Fredrik]

When we said $(x,y)\in\{x\}\times[0,1]$ yesterday , what does $\{x\}$ truly mean here? I know it's suppose to be any real number , but what difference does it make to put an $x$ out there instead of $R$ or $a$ like in the union notation? Is it because we want to pinpoint that $x$ is there as $\{x\}$ confirming both the position of $x$ and the position of $y$ in the formula ? Isn't $(x,y)$ enough? I'm trying to understand the importance of this statement.Is it because we want to pass through $\{x\} × [0,1]$ to reach $\subseteq\bigcup_{a\in\mathbb R}\{a\} ×[0,1]$? If so , why is that crucial? Why not $\{a\}$ instead of $\{x\}$?

x and y are the unique real numbers such that z=(x,y). {x} is the singleton set with x as its only element. I'm not sure where you would like to put an $a$ instead of $x$.

We want to prove that every element of $\mathbb R\times [0,1]$ is an element of $\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$. So we start by saying "Let $z\in\mathbb R\times[0,1]$ be arbitrary", and set out to prove that z is an element of the set on the right-hand side.

We define x and y by z=(x,y). Since $z\in\mathbb R\times[0,1]$, this ensures that $x\in\mathbb R$ and $y\in[0,1]$. The three statements $z=(x,y)$, $x\in\mathbb R$ and $y\in[0,1]$, together imply that $z\in\{x\}\times[0,1]$. Since $x\in\mathbb R$ and $y\in[0,1]$, this result implies that $z\in\bigcup_{a\in\mathbb R}\left(\{a\}\times[0,1]\right)$.

Another way of saying that last bit: Since x is a real number and z is an element of $\{x\}\times[0,1]$, z is an element of one of the infinitely many sets $\{a\}\times[0,1]$ with $a\in\mathbb R$, and by definition of "union", that implies that z is an element of the union of those sets.

 

[Fredrik]

A very simple thing seems to confuse me at this very moment , what's the difference between a set $A$ and $\bigcup A$?

Suppose $A=\{ (0,1) , (1,2)\}$.

You're making it complicated by choosing a set with ordered pairs as elements, but maybe you meant { when you wrote (? I will consider the example with { instead.

If $A=\{\{0,1\},\{1,2\}\}$, then $\bigcup A=\{0,1\}\cup\{1,2\}=\{0,1,2\}$.

 

[reenmachine]

You're making it complicated by choosing a set with ordered pairs as elements, but maybe you meant { when you wrote (? I will consider the example with { instead.

If $A=\{\{0,1\},\{1,2\}\}$, then $\bigcup A=\{0,1\}\cup\{1,2\}=\{0,1,2\}$.

Okay , but what if $A = \{1,2\}$? Is it the same set?

 

[micromass]

A very simple thing seems to confuse me at this very moment , what's the difference between a set $A$ and $\bigcup A$?

Suppose $A=\{ (0,1) , (1,2)\}$.

Everything in mathematics is a set, so $(0,1)$ and $(1,2)$ are sets. So $(0,1)\cup (1,2)$ makes sense. So the answer is

\bigcup A = (0,1)\cup (1,2)

However, while it does make sense, it is something you never really want to do. If you ever encounter a situation where you would have to evaluate $(0,1)\cup (1,2)$, then you made a mistake somewhere.

That said, I absolutely hate the notation $\bigcup$. I think it's a very bad notation. But it's used quite a lot.

 

[micromass]

Okay , but what if $A = \{1,2\}$? Is it the same set?

In that case, you have

\bigcup A = 1\cup 2

Again, this makes perfect sense since $1$ and $2$ are sets. But it is not something you ever want to do.

 

[reenmachine]

In that case, you have

$$\bigcup A = 1\cup 2$$

Again, this makes perfect sense since $1$ and $2$ are sets. But it is not something you ever want to do.

So if $1=\{u,y\}$ and $2=\{z,x\}$ , then

$$\bigcup A = 1\cup 2 = \{u,y,z,x\} ?$$

Basically it's just not good to reach the level where you can't split sets into elements.