Question about proof from a guy with a highschool education

[Fredrik]

I don't understand this.I thought the empty set was generally not an element of sets.

It certainly can be, just like any other set.

The axiom of pairing says that for all sets x,y there's a set z such that $x,y\in z$.

I think that this is equivalent to saying that for all sets x,y, {x,y} is a set. (I think I've seen a proof of that, but I don't want to think about that today).

So for any set x, there's a set that has ∅ and x as elements.

 

[reenmachine]

It certainly can be, just like any other set.

The axiom of pairing says that for all sets x,y there's a set z such that $x,y\in z$.

I think that this is equivalent to saying that for all sets x,y, {x,y} is a set. (I think I've seen a proof of that, but I don't want to think about that today).

So for any set x, there's a set that has ∅ and x as elements.

oh ok I didn't see it that way.Very clear.

Thanks!

It's been a long day , I'll go get some rest.

 

[CompuChip]

BTW , in the book of proof they denote the set $R^2$ with $\{(x,y) : x,y \in R\}$ , which is a notation of the form $\{x : p(x)\}$.It still confuses me a little bit when it is correct or not to use it.

If you prefer more formalism you could write $\{ (x, y) \in R \times R : x, y \in R \}$ or just $R \times R$. Since you are basically restating the definition of $R \times R$ now, you could even insert a trivial condition like $\{ (x, y) \in R \times R : 1 = 1 \}$, or - if you allow p(x, y) to be omitted if it is true for all x and y, $\{ (x, y) \in R \times R \}$.

Formally it is probably never "correct", in practice it is correct when it is clear what you mean and unambiguous, like in this case.

 

[Fredrik]

It's fine to use the notation $\{x:p(x)\}$ in those cases where we have already proved that $\{x:p(x)\}$ is a set.

 

[reenmachine]

Here's an exercise from the book of proof.

$\forall n \in N \ A_n = \{0,1,2,3,...,n\}$

$$\bigcup_{i \in N}A_i = ?$$

$$\bigcap_{i \in N}A_i = ?$$

My answers were:

$$\bigcup_{i \in N}A_i = \{0,1,2,3,...,\infty\}$$

and

$$\bigcap_{i \in N}A_i = \{0\}$$

The book answers are:

$$\bigcup_{i \in N}A_i = \{0\} \cup N$$

and

$$\bigcap_{i \in N}A_i = \{0,1\}$$

I don't understand what they are doing here.It seems the misconception I have comes from the role of 0 in N.In the second one , if $A_0$ , then $1$ isn't going to be an element of that set.In the first one it seems they are saying that $0$ isn't part of $N$ or something like that.

thx!

 

[micromass]

A lot of books don't define 0 as a natural number. So for them

\mathbb{N}=\{1,2,3,4,...\}

The book of proof is following that convention.

Anyway, I want to say something about your first answer. You wrote

\bigcup A_n = \{0,1,2,3,...,\infty\}

The $\infty$ shouldn't be there. Infinity is not a natural number. And infinity is also not in any of the sets $A_n$. So it's not in the union.

Regardless, try to solve these exercises: (I use the convention here that $\mathbb{N}$ does not contain $0$, but I usually don't follow that convention)

\bigcup_{n\in \mathbb{N}} [0,1-(1/n)]
\bigcup_{n\in \mathbb{N}} [0,1-(1/n) )
\bigcap_{n\in \mathbb{N}} [0,1+(1/n)]
\bigcap_{n\in \mathbb{N}} [0,1+(1/n))

 

[reenmachine]

\bigcup_{n\in \mathbb{N}} [0,1-(1/n)]
\bigcup_{n\in \mathbb{N}} [0,1-(1/n) )
\bigcap_{n\in \mathbb{N}} [0,1+(1/n)]
\bigcap_{n\in \mathbb{N}} [0,1+(1/n))

I don't understand the difference between the first two and the last two.I see the brackets changes but it doesn't make sense in my mind.

$$\bigcup_{n\in \mathbb{N}} [0,1-(1/n)] = \{-0,9 , -0,4 , ... , 0,1\}$$

 

[micromass]

I don't understand the difference between the first two and the last two.I see the brackets changes but it doesn't make sense in my mind.

The first one asks you to calculate

[0,1-(1/1)]\cup [0,1-(1/2)] \cup [0,1-(1/3)]\cup ...

The second one is

[0,1-(1/1))\cup [0,1-(1/2)) \cup [0,1-(1/3))\cup ...

Do you know the difference between [a,b] and [a,b) (or sometimes written as [a,b[ in Europe)?

 

[micromass]

anyway , \bigcup_{n\in \mathbb{N}} [0,1-(1/n)] = $\{-0,9 , -0,4 , ... , 0,1\}$

Not sure what you mean with this.

 

[reenmachine]

Do you know the difference between [a,b] and [a,b) (or sometimes written as [a,b[ in Europe)?

I have no clue.

 

[micromass]

I have no clue.

[a,b] = \{x\in \mathbb{R}~\vert~a\leq x \leq b\}
[a,b) = \{x\in \mathbb{R}~\vert~a\leq x < b\}
(a,b] = \{x\in \mathbb{R}~\vert~a < x\leq b\}
(a,b) = \{x\in \mathbb{R}~\vert~a < x < b\}

The last one should not be confused with (a,b) as an ordered pair. It's usually clear from context which one is meant.

 

[reenmachine]

not sure what you mean with this.

$0.1 - 1/1 = -0.9$
$0.1 - 1/2 = -0.4$
...
$0.1 - 1/\bar{9} = 0.1?$

I was just trying to write the elements at both extreme.I can't calculate millions of fractions to find all elements.I'm not really sure what I should do here.

 

[Fredrik]

There's no 0.1 in the exercises suggested by micromass. Did you look at the definitions in post #413? For example, $[0,1-1/2]=[0,1/2]=\{x\in\mathbb R:0\leq x\leq 1/2\}$. This is the interval of all real numbers from 0 and 1/2 (including 0 and 1/2). So you're supposed to find the union of infinitely many intervals. Note that we're talking about a sequence of smaller and smaller intervals. Edit: And by "smaller", I meant "bigger".

 

[reenmachine]

There's no 0.1 in the exercises suggested by micromass. Did you look at the definitions in post #413? For example, $[0,1-1/2]=[0,1/2]=\{x\in\mathbb R:0\leq x\leq 1/2\}$. This is the interval of all real numbers from 0 and 1/2 (including 0 and 1/2). So you're supposed to find the union of infinitely many intervals. Note that we're talking about a sequence of smaller and smaller intervals.

that went over my head , I thought it was 0.1 minus something.Rough day at the office for me.

So it's $0 ,$ then $1- 1/n$.

I'm going to the restaurent I'll try them again when I come back.

 

[reenmachine]

\bigcup_{n\in \mathbb{N}} [0,1-(1/n)]
\bigcup_{n\in \mathbb{N}} [0,1-(1/n) )
\bigcap_{n\in \mathbb{N}} [0,1+(1/n)]
\bigcap_{n\in \mathbb{N}} [0,1+(1/n))

\bigcup_{n\in \mathbb{N}} [0,1-(1/n)] = $\{x \in R : 0 ≤ x ≤ 1\}$

\bigcup_{n\in \mathbb{N}} [0,1-(1/n) ) = $\{x \in R : x = 0\}$ ? If n=1 , then it is [0,1-(1/1)) so [0,0) , how do I operate with this? Is this the empty set?

\bigcap_{n\in \mathbb{N}} [0,1+(1/n)] I'm not sure I get it , how could any number intersect if n changes? I apologize for my lack of entrepreneurship on these exercises but I am confused as hell.

What I mean by numbers can't intersect is [0,1+(1/1)] = [0,2] and [0,1+(1/2)] = [0,1.5] and so on...which intersection can we find by changing n?

edit: maybe I get it , is the intersection $\{x \in R : 0 ≤ x ≤ 1\}$ ? So basically the same set as the first.

Hopefully you see my edit before replying if I'm right lol

 

[Fredrik]

First one: Close, but not quite right. There's one little thing you seem to be overlooking.

Second: Nope, that's pretty far off.

Yes the notation [0,0) is weird, because the "[" suggests that 0 is included, and the ")" suggests that 0 is not included. But the definition gives us a clear answer: $[0,0)=\{x\in\mathbb R:0\leq x<0\}$. At least it's clear to me. You may have to think about it. Can you see it? Is [0,0) equal to {0} or ∅ or perhaps neither? Does our attempt to define [0,0) even make sense?

Third: Look at the definition of "intersection" again. What does the definition say about $$\bigcap_{n\in\mathbb N}\left[0,1-\frac 1 n\right]?$$ Don't try to come up with the final answer, just tell me what the definition says that the expression above (not including the question mark) means.

OK, now I've seen your edit. Nope, that answer is wrong. My advice is still the same.

Edit: Apparently I read the sign wrong in the third exercise.

 

[reenmachine]

First one: Close, but not quite right. There's one little thing you seem to be overlooking.

At least re-assure me that what I am missing isn't 1 divided by n=0 before I start to overheat my brain

 

[micromass]

At least re-assure me that what I am missing isn't 1 divided by n=0 before I start to overheat my brain

That's not what you're missing

 

[reenmachine]

Second: Nope, that's pretty far off.

Yes the notation [0,0) is weird, because the "[" suggests that 0 is included, and the ")" suggests that 0 is not included. But the definition gives us a clear answer: $[0,0)=\{x\in\mathbb R:0\leq x<0\}$. At least it's clear to me. You may have to think about it. Can you see it? Is [0,0) equal to {0} or ∅ or perhaps neither? Does our attempt to define [0,0) even make sense?

$[0,0)=\{x\in\mathbb R:0\leq x<0\}$ This doesn't make sense in my head , in this notation x could be equal to 0 yet it explicitely says that 0 is bigger than x.

Suppose x=0
Then 0 < 0 doesn't make sense

Suppose x=0,00001
Then 0,00001 < 0 doesn't make sense

Suppose x=-1
Then 0 ≤ -1 doesn't make sense

what's left to make sense here?

 

[reenmachine]

That's not what you're missing

I haven't been thinking about it for long but what I see that could be a problem now would be the second ≤ in $\{x \in R : 0 ≤ x ≤ 1\}$

Maybe $\{x \in R : 0 ≤ x < 1\}$ but then again I thought 0,9999999999... = 1.

And that's not to mention your statement that [a,b] = \{x\in \mathbb{R}~\vert~a\leq x \leq b\}

 

[Fredrik]

You need to pick up the habit of first asking yourself what the definition says the given expression means. Do you remember how you should interpret the notation $\{x\in y:P(x)\}$ where P is a property? (To say that P is a property is to say that P(x) is a statement about x). The notation $\{x\in\mathbb R:0\leq x<0\}$ is interpreted the same way. It may help to simply say it out loud. (The notation does make sense).

 

[reenmachine]

Third: Look at the definition of "intersection" again. What does the definition say about $$\bigcap_{n\in\mathbb N}\left[0,1-\frac 1 n\right]?$$ Don't try to come up with the final answer, just tell me what the definition says that the expression above (not including the question mark) means.

It means the intersection (which is the part where elements of 2 or more sets are the same) of $[0,1- 1/n]$ with $n \in N$.

Dont know if you did it on purpose but it's +1/n in micromass exercise.

 

[Fredrik]

I haven't been thinking about it for long but what I see that could be a problem now would be the second ≤ in $\{x \in R : 0 ≤ x ≤ 1\}$

Maybe $\{x \in R : 0 ≤ x < 1\}$

This is the correct answer for the first one. Your answer can be simplified to [0,1). Your observation about 0.999... is correct, but irrelevant here.

1 can't be a member of the union we seek, because it's not a member of any of the sets we're taking a union of.

It means the intersection (which is the part where elements of 2 or more sets are the same) of $[0,1- 1/n]$ with $n \in N$.

An intersection $\bigcap_{i\in I}A_i$ is the set of all x that are elements of all the $A_i$ with $i\in I$. So what I was hoping you would say is that $\bigcap_{n\in\mathbb N}[0,1-1/n]$ is the the set of all real numbers that are elements of all the intervals [0,1-1/n] with $n\in\mathbb N$.

Dont know if you did it on purpose but it's +1/n in micromass exercise.

That was not on purpose. I thought there was a minus there, not a plus. In that case, the first answer you gave for the third one is correct. That answer can be simplified to [0,1].

I'm signing out for tonight...

 

[reenmachine]

This is the correct answer for the first one. Your answer can be simplified to [0,1). Your observation about 0.999... is correct, but irrelevant here.

1 can't be a member of the union we seek, because it's not a member of any of the sets we're taking a union of.

I see , I just thought that 1 - 1/9999999999999... would equal 0.99999999... so would equal 1.

That was not on purpose. I thought there was a minus there, not a plus. In that case, the first answer you gave for the third one is correct. That answer can be simplified to [0,1].

Good , finally something to cheer me up

Still don't have a clue about the 0 ≤ x < 0 thing though.

I'm signing out for tonight...

Yeah that's a good idea.I'm tired myself.

thanks a lot and see you later!

 

[Fredrik]

I guess I have time for one more quick post before I go to bed.

I see , I just thought that 1 - 1/9999999999999... would equal 0.99999999... so would equal 1.

It does, but there's no interval with right endpoint 1-1/99999... among those we're taking a union of. 99999... is not an integer, and every interval we're considering is of the form [0,1-1/n] with n an integer.

Still don't have a clue about the 0 ≤ x < 0 thing though.

I stand by my advice here:

You need to pick up the habit of first asking yourself what the definition says the given expression means. Do you remember how you should interpret the notation $\{x\in y:P(x)\}$ where P is a property? (To say that P is a property is to say that P(x) is a statement about x). The notation $\{x\in\mathbb R:0\leq x<0\}$ is interpreted the same way. It may help to simply say it out loud. (The notation does make sense).

Forget about whether $\{x\in\mathbb R:0\leq x<0\}$ is equal to ∅ or {0} for a moment, and just tell us what the notation means. Say it out loud. Type it in your next post. "The set of..."

 

[reenmachine]

Forget about whether $\{x\in\mathbb R:0\leq x<0\}$ is equal to ∅ or {0} for a moment, and just tell us what the notation means. Say it out loud. Type it in your next post. "The set of..."

the set of all $x$ in $R$ such that 0 is equal or lesser than x and x is lesser than 0

 

[Fredrik]

the set of all $x$ in $R$ such that 0 is equal or lesser than x and x is lesser than 0

Exactly. And how many real numbers are there that have that property (a property that implies that 0<0)?

 

[reenmachine]

Exactly. And how many real numbers are there that have that property (a property that implies that 0<0)?

none

 

[HallsofIvy]

Actually, I disagree with Fredrik (or with what I think he means). "x\le 0" means "x is less than or equal to 0" so the set \{x| x\le 0 and x&lt; 0\} is precisely {x |x< 0}.

 

[micromass]

Actually, I disagree with Fredrik (or with what I think he means). "x\le 0" means "x is less than or equal to 0" so the set \{x| x\le 0 and x&lt; 0\} is precisely {x |x< 0}.

OK, but he's not talking about that set. He's talking about the set

\{x\in \mathbb{R}~\vert~x\geq 0 ~\text{and}~x&lt;0\}

 

[reenmachine]

Actually, I disagree with Fredrik (or with what I think he means). "x\le 0" means "x is less than or equal to 0" so the set \{x| x\le 0 and x&lt; 0\} is precisely {x |x< 0}.

In your exemple you use x ≤ 0 , in our exemple it was 0 ≤ x ...

 

[micromass]

none

So no real numbers have the property. So what is the set $\{x\in \mathbb{R}~\vert~0\leq x<0\}$?

 

[reenmachine]

the empty set? there's no $x \in R$ that has the property the set is describing , the set should therefore be empty.

 

[Fredrik]

Right. We might as well have written $\{x\in\mathbb R:\text{Pigs can fly}\}$. Since there are no real numbers such that pigs can fly, the set is empty.

 

[reenmachine]

Right. We might as well have written $\{x\in\mathbb R:\text{Pigs can fly}\}$. Since there are no real numbers such that pigs can fly, the set is empty.

good , now it finally make sense

 

[reenmachine]

\bigcap_{n\in \mathbb{N}} [0,1+(1/n))

To conclude since I finally solved the three other exercises ,

\bigcap_{n\in \mathbb{N}} [0,1+(1/n)) = $\{x \in R : 0 ≤ x ≤ 1\}$

 

[micromass]

To conclude since I finally solved the three other exercises ,

\bigcap_{n\in \mathbb{N}} [0,1+(1/n)) = $\{x \in R : 0 ≤ x ≤ 1\}$

It's correct, but let me make sure you get this. Can you explain my why $1$ is included in

\bigcap_{n\in \mathbb{N}} [0,1+\frac{1}{n})

but not in

\bigcup_{n\in \mathbb{N}} [0,1-\frac{1}{n})

?

 

[reenmachine]

It's correct, but let me make sure you get this. Can you explain my why $1$ is included in

\bigcap_{n\in \mathbb{N}} [0,1+\frac{1}{n})

but not in

\bigcup_{n\in \mathbb{N}} [0,1-\frac{1}{n})

?

Yes I can.In the first one , we have 0, 1 + something , so the right sided result will always be bigger than 1 , even by a microscopic margin , and THAT is the number that is excluded from the ).1 is always there and therefore it should be included in the intersection.

In the second one , we have 0, 1 - something , so the right sided result will always be smaller than 1 , even by a microscopic margin , and THAT is the number that is excluded from the ) , so if a smaller number than 1 is excluded , then so is 1 (in the context we're speaking of).

edit: btw , I guess even if you put a bigcap instead of a bigcup there on the second one , 1 is still excluded.

 

[micromass]

Yes I can.In the first one , we have 0, 1 + something , so the right sided result will always be bigger than 1 , even by a microscopic margin , and THAT is the number that is excluded from the ).1 is always there and therefore it should be included in the intersection.

In the second one , we have 0, 1 - something , so the right sided result will always be smaller than 1 , even by a microscopic margin , and THAT is the number that is excluded from the ) , so if a smaller number than 1 is excluded , then so is 1.

Right. This was one of the more tricky exercises in set theory and everybody I know fails to solve this right when they first encounter it. Understanding that very exercise is (to me) a big step in understanding sets. So good job.

Let's see if you can tackle somewhat related exercises (note again that $\mathbb{N}$ does not include $0$ here and that $(a,b)$ denotes an interval and not an ordered pair).

\bigcap_{n\in \mathbb{N}} (-\frac{1}{n},\frac{1}{n})
\bigcup_{n\in \mathbb{N}} (-\frac{1}{n}, \frac{1}{n})
\bigcup_{x\in \mathbb{R}} (x,x+1)
\bigcap_{A\in \mathcal{P}(\mathbb{R})} A

I'm curious how you will solve these.

 

[micromass]

Also, the following notation is synonymous:

\bigcap_{n\in \mathbb{N}}

and

\bigcap_{n=1}^{+\infty}

But please don't make mistakes when reading that second notation. The symbol $\infty$ is not a number here, it is only a symbol.
I realize that you could read it as follows

\bigcap_{n=1}^{+\infty} \{n\} = \{1\} \cap \{2\} \cap \{3\} ... \cap \{+\infty\}

But this is incorrect. We don't add the $\{+\infty\}$ in the end (although the notation suggests it). In fact, in elementary math such as this, we don't work with $+\infty$ at all. It is just a symbol without meaning.

The notation $\bigcap_{n=1}^{+\infty}$ is way more popular than the $\bigcap_{n\in \mathbb{N}}$ (even though they mean exactly the same thing). One reason is to avoid the ambiguity whether to include $0$ in $\mathbb{N}$ or not. If an author writes $\bigcap_{n\in \mathbb{N}}$ then this is very ambiguous. But if he writes $\bigcap_{n=1}^{+\infty}$, then it is clear he does not include $0$. If he wants to include $0$, then he writes $\bigcap_{n=0}^{+\infty}$.
The notation is also very flexible, for example, things like

\bigcap_{n=12}^{+\infty}\{n\} = \{12\} \cap \{13\} \cap \{14\} ...

make sense.

 

[reenmachine]

Right. This was one of the more tricky exercises in set theory and everybody I know fails to solve this right when they first encounter it. Understanding that very exercise is (to me) a big step in understanding sets. So good job.

Thank you!

Let's see if you can tackle somewhat related exercises (note again that $\mathbb{N}$ does not include $0$ here and that $(a,b)$ denotes an interval and not an ordered pair).

\bigcap_{n\in \mathbb{N}} (-\frac{1}{n},\frac{1}{n})

I'm curious how you will solve these.

I know the intersection is 0 here , not sure how to write it.Let me give it a try:

$\{x \in R : 0 ≤ x ≤ 0 \}$ My thought process to use this notation is that if you choose that 0 is less than on the first symbol , then the second symbol (statement) will be false either way.So you have no choice but to choose ''is equal to'' for both.

 

[reenmachine]

\bigcup_{n\in \mathbb{N}} (-\frac{1}{n}, \frac{1}{n})

$\{x \in R : -1 ≤ x ≤ 1\}$

edit:corrected

 

[micromass]

Thank you!



I know the intersection is 0 here , not sure how to write it.Let me give it a try:

$\{x \in R : 0 ≤ x ≤ 0 \}$ My thought process to use this notation is that if you choose that 0 is less than on the first symbol , then the second symbol (statement) will be false either way.So you have no choice but to choose ''is equal to'' for both.

What about $\{0\}$?

 

[reenmachine]

What about $\{0\}$?

I can just write it like this?

 

[micromass]

$\{x \in R : x < 0 > x\}$

No, that notation is illegal. You can only "compose" inequality if they point in the same direction. So things like $0\leq x\leq 1$ and $0<x\leq 2$ are perfectly allowed. Even $0<x=2$ is ok. But if the arrows point in opposite directions then it's not allowed. So $x<0>x$ is not ok and $x>0<x$ is also not ok.

Can you rewrite this set by using "and" or "or"?

 

[micromass]

I can just write it like this?

Sure, the equality holds

\{x\in \mathbb{R}~\vert~0\leq x\leq 0\} = \{0\}

Prove this equality if you're not convinced.

 

[reenmachine]

No, that notation is illegal. You can only "compose" inequality if they point in the same direction. So things like $0\leq x\leq 1$ and $0<x\leq 2$ are perfectly allowed. Even $0<x=2$ is ok. But if the arrows point in opposite directions then it's not allowed. So $x<0>x$ is not ok and $x>0<x$ is also not ok.

Can you rewrite this set by using "and" or "or"?

Think I did a mistake anyway , 0 should be included here.

$\{x \in R : -1 ≤ x ≤ 1\}$

 

[micromass]

$\{x \in R : -1 ≤ x ≤ 1\}$

edit:corrected

Can you explain why you include $1$ and $-1$?

 

[reenmachine]

Also, the following notation is synonymous:

\bigcap_{n\in \mathbb{N}}

and

\bigcap_{n=1}^{+\infty}

But please don't make mistakes when reading that second notation. The symbol $\infty$ is not a number here, it is only a symbol.
I realize that you could read it as follows

\bigcap_{n=1}^{+\infty} \{n\} = \{1\} \cap \{2\} \cap \{3\} ... \cap \{+\infty\}

But this is incorrect. We don't add the $\{+\infty\}$ in the end (although the notation suggests it). In fact, in elementary math such as this, we don't work with $+\infty$ at all. It is just a symbol without meaning.

The notation $\bigcap_{n=1}^{+\infty}$ is way more popular than the $\bigcap_{n\in \mathbb{N}}$ (even though they mean exactly the same thing). One reason is to avoid the ambiguity whether to include $0$ in $\mathbb{N}$ or not. If an author writes $\bigcap_{n\in \mathbb{N}}$ then this is very ambiguous. But if he writes $\bigcap_{n=1}^{+\infty}$, then it is clear he does not include $0$. If he wants to include $0$, then he writes $\bigcap_{n=0}^{+\infty}$.
The notation is also very flexible, for example, things like

\bigcap_{n=12}^{+\infty}\{n\} = \{12\} \cap \{13\} \cap \{14\} ...

make sense.

Ok that's good to know thank you!

 

[reenmachine]

Can you explain why you include $1$ and $-1$?

I shouldn't have included them since the notation was between ( ) and not [ ].

so $\{x \in R -1 < x < 1 \}$

I'm still getting used to these notations , never heard of such a concept before.