Question about proof from a guy with a highschool education

[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 $\{x| x\le 0 and x< 0\}$ 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 $\{x| x\le 0 and x< 0\}$ is precisely {x |x< 0}.

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

$$\{x\in \mathbb{R}~\vert~x\geq 0 ~\text{and}~x<0\}$$

 

[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 $\{x| x\le 0 and x< 0\}$ 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.