Question about proof from a guy with a highschool education

[micromass]

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.

OK, good!

Try to experiment with the notation a bit, it's the best way to get comfortable with it. So the above answer is (-1,1), for example.

 

[reenmachine]

OK, good!

Try to experiment with the notation a bit, it's the best way to get comfortable with it. So the above answer is (-1,1), for example.

Good!

It pisses me off because I'm having a lot of fun , but I have to leave for school (if I want to ace these high school math exams sooner than later).I'll be back in 2 or 3 hours to try the last two which seems trickier.

thanks a lot for the help , if you have time to take a quick look when I come back I'll try to solve them!

cheers!

 

[reenmachine]

\bigcup_{x\in \mathbb{R}} (x,x+1)
.

screw school it can wait a couple of minutes , I'm giving this one a shot right away.

$\{ y\in R : \exists x \in R \ x < y < (x+1) \}$

 

[micromass]

Good!

It pisses me off because I'm having a lot of fun , but I have to leave for school (if I want to ace these high school math exams sooner than later).I'll be back in 2 or 3 hours to try the last two which seems trickier.

thanks a lot for the help , if you have time to take a quick look when I come back I'll try to solve them!

cheers!

Have fun!

In the meanwhile, let me explain yet another notation. We also have things like this

[a,+\infty) = \{x\in \mathbb{R}~\vert~a\leq x\}
(-\infty, a) = \{x\in \mathbb{R}~\vert x&lt;a\}

These are "half-rays" of real numbers. Try to draw them.
You can also make sense of things like $(-\infty,a]$ and such. But I think it's clear.

Again, the $\infty$ is just a symbol. It is not a real number so it is not included in the set. The bracket ")" should already indicate that we don't include $\infty$. We might include infinity by writing things like $[-\infty, a]$, but this notation is not used because $-\infty$ is not a number, so saying things like "including minus infinity" makes no sense. (actually, in higher mathematics, it does make sense, but I don't want to confuse you now. So just know that every time you encounter $\infty$, it will just be a symbol and not an actual number. This will always be the case until you learn things like analysis).

I actually prefer the notation $[a,\rightarrow )$ instead of $[a,+\infty)$ since the former makes no reference to a non-existent infinity, so it is less confusing. But it is a notation that is rarely used, so I won't use it either.

Now you know this, can you find the following:

\bigcap_{n=1}^{+\infty} [n,+\infty)

\bigcup_{n=1}^{+\infty} (-\infty,n)

 

[micromass]

screw school it can wait a couple of minutes , I'm giving this one a shot right away.

$\{ y\in R : \exists x \in R \ x < y < (x+1) \}$

Perfectly fine. But this set can be written a lot shorter. It is actually a very well-known set. Can you find which one it is? If you don't know, then just read it out loud. Then think which $y\in \mathbb{R}$ satisfy the condition or don't satisfy it.

 

[reenmachine]

Perfectly fine. But this set can be written a lot shorter. It is actually a very well-known set. Can you find which one it is? If you don't know, then just read it out loud. Then think which $y\in \mathbb{R}$ satisfy the condition or don't satisfy it.

Well I would guess it's just the set $R$

 

[micromass]

Well I would guess it's just the set $R$

Right!

 

[reenmachine]

Right!

Kind of funny that to define R we would use R multiple times in it's own definition.

 

[micromass]

Kind of funny that to define R we would use R multiple times in it's own definition.

Well, it's not a definition of $\mathbb{R}$. A definition would not use $\mathbb{R}$ anywhere.

 

[reenmachine]

I'm already back , teacher called in sick so I came back.I'm learning more here than at school anyway.

 

[reenmachine]

Have fun!

In the meanwhile, let me explain yet another notation. We also have things like this

[a,+\infty) = \{x\in \mathbb{R}~\vert~a\leq x\}
(-\infty, a) = \{x\in \mathbb{R}~\vert x&lt;a\}

I'm not sure I understand , in the first one , you have $[a$ and $\infty+)$ , what if $x = 4$ and $a = 2$ , this would satisfy $\{x\in \mathbb{R}~\vert~a\leq x\}$ yet would not satisfy $\infty+)$ since all positive numbers are excluded.

Or does $[a,+\infty)$ automatically qualify $a$ as 0 or a negative number? But if it does , how do you know by simply reading the notation $\{x\in \mathbb{R}~\vert~a\leq x\}$?

EDIT: I think I just understood , $\infty+$ isn't all positive numbers , it's just all numbers going in the positive direction starting from $a$? Does $a$ counts?

 

[micromass]

Why do you think all positive numbers are excluded. All $[a,+\infty)$ means is the set I wrote down.

 

[reenmachine]

Why do you think all positive numbers are excluded. All $[a,+\infty)$ means is the set I wrote down.

see edit

 

[micromass]

EDIT: I think I just understood , $\infty+$ isn't all positive numbers , it's just all numbers going in the positive direction starting from $a$? Does $a$ counts?

Correct, $[a,+\infty)$ is all number starting from a (with a included, going in the positive direction). But it's $+\infty$ and not $\infty+$.

If you don't want to include a, then it's $(a,+\infty)$.

Again, $\infty$ is just a symbol. It has no meaning.

 

[reenmachine]

Correct, $[a,+\infty)$ is all number starting from a (with a included, going in the positive direction). But it's $+\infty$ and not $\infty+$.

If you don't want to include a, then it's $(a,+\infty)$.

Again, $\infty$ is just a symbol. It has no meaning.

But in $[a,+\infty)$ , you exclude all larger numbers than $a$ (and $a$) from the set no? Because of the '')''.

So in $\{x \in R : a ≤ x \}$ , this is the set of all larger or equal numbers to $a$.I thought these were the numbers that we excluded.

 

[micromass]

But in $[a,+\infty)$ , you exclude all larger numbers than $a$ (and $a$) from the set no? Because of the '')''.

No. You're interpreting the notation somehow. The ")" in the notation does not mean "excluding" something (unlike in the other [a,b) notation). Here the notation is just defined as

[a,+\infty) = \{x\in \mathbb{R}~\vert~a\leq x\}

The ")" doesn't mean anything specific here.

If you want to understand the notation as "excluding" something,then you can see it as follows. Denote $+\infty$ as something (that is not a real number!) that is somehow larger than all real numbers. So $2<+\infty$ holds and $10000000000<+\infty$ holds. In fact, if $x$ is any real number, then $x<+\infty$ holds. The notation $[a,+\infty)$ just means all real numbers $x$ such that $a\leq x<+\infty$. So we exclude the "thing" $+\infty$. But since $x<+\infty$ holds for any real number $x$, we just write $a\leq x$. So if you want, you can interpret $[a,+\infty)$ as excluding a thing called $+\infty$ that is larger than all real numbers. This interpretation is problematic since it is not clear what $+\infty$ actually is. It is not a real number (by definition), but something else.

 

[reenmachine]

Ok , I think I understand it , but one thing that confuses me is what happens if $[a , +\infty]$ instead of $[a +\infty)$ ?

This is the set of $a$ , every larger real numbers and even the ''thing'' that is larger than real numbers?

 

[micromass]

Ok , I think I understand it , but one thing that confuses me is what happens if $[a , +\infty]$ ?

This is the set of $a$ , any larger real numbers and even the ''thing'' that is larger than real numbers?

Yes, that would be exactly what you describe.
But since the "thing" is not a real number itself, the set $[a,+\infty]$ would not be a subset of the real numbers. So if we only care about real numbers (like usual in calculus), then the set $[a,+\infty]$ is never used.

So although you can give meaning to $[a,+\infty]$, you will never see it used (except if you get to advanced math courses). The only sets you will see used is $[a,+\infty)$.

 

[reenmachine]

Yes, that would be exactly what you describe.
But since the "thing" is not a real number itself, the set $[a,+\infty]$ would not be a subset of the real numbers. So if we only care about real numbers (like usual in calculus), then the set $[a,+\infty]$ is never used.

So although you can give meaning to $[a,+\infty]$, you will never see it used (except if you get to advanced math courses). The only sets you will see used is $[a,+\infty)$.

Yes I intuitively figured this set wouldn't be used that often , if ever.

thanks a lot man!

 

[reenmachine]

\bigcap_{A\in \mathcal{P}(\mathbb{R})} A

Maybe something went over my head as I was thinking about this one , but could this simply be $\mathcal{P}(\mathbb{R})$?

The intersection of all elements of $\mathcal{P}(\mathbb{R})$ is $\mathcal{P}(\mathbb{R})$?

 

[micromass]

That's not correct.

 

[reenmachine]

Hmmm wait , $A \subset R$ , so $A \in \mathcal{P}(\mathbb{R})$.

If $R$ was to be $\{1,2\}$
$\mathcal{P}(\mathbb{R})$ would be $\{ \varnothing , \{1\} , \{2\}, \{1,2\}\}$.

These elements would originally comes from $R$ , but the problem is that they weren't elements but subsets.So that's why I'm hesitant before going with the answer ''$R$''.

 

[micromass]

$\{ \varnothing , \{1\} , \{2\}, \{1,2\} , \{1,\varnothing\} , \{2,\varnothing\} , \{1,2,\varnothing\}\}$.

This is not $\mathcal{P}(\{1,2\})$

Anyway, $\mathbb{R}$ is the wrong answer.

 

[reenmachine]

This is not $\mathcal{P}(\{1,2\})$

I was in the moon.

$\{ \varnothing , \{1\} , \{2\}, \{1,2\}\}$

there you go

 

[micromass]

OK, so let $E= \{1,2\}$. Can you find

\bigcap_{A\in \mathbb{P}(E)} A

This is the same as

\emptyset \cap \{1\} \cap \{2\} \cap \{1,2\}

 

[Fredrik]

I don't mean to interfere with the discussion, which seems to be going well. Just a little observation: You seem to sometimes be confusing unions with intersections. Intersections give us something smaller, unions something bigger: $$A\cap B\subseteq A\subseteq A\cup B.$$

 

[reenmachine]

OK, so let $E= \{1,2\}$. Can you find

\bigcap_{A\in \mathbb{P}(E)} A

This is the same as

\emptyset \cap \{1\} \cap \{2\} \cap \{1,2\}

1 and 2?

I admit I'm a bit confused

 

[micromass]

1 and 2?

I really don't understand how you got this.

 

[reenmachine]

Can the empty side intersect with anything anyway? In this case this would be the empty set?

How could nothing intersect with something.

 

[micromass]

Can the empty side intersect with anything anyway?

Why not??

What is $\emptyset \cap A$?? Apply the definition of the intersection.

In this case this would be the empty set?

Yes, but you should really try to understand why your previous answers were wrong and this is right.

 

[micromass]

How could nothing intersect with something.

Always go back to the definition. What is the definition of intersection?

 

[reenmachine]

I don't mean to interfere with the discussion, which seems to be going well. Just a little observation: You seem to sometimes be confusing unions with intersections. Intersections give us something smaller, unions something bigger: $$A\cap B\subseteq A\subseteq A\cup B.$$

If I confuse both of them it's really a concentration mistake.Those symbols are now installed in my brain.

 

[reenmachine]

Why not??

What is $\emptyset \cap A$?? Apply the definition of the intersection.

Yes it's true , $\emptyset \cap A = A$

I notice it sometimes take a couple of minutes before concepts I didn't directly touch in the last 3 days or so comes back to my mind clearly.

 

[micromass]

Yes it's true , $\emptyset \cap A = A$

No. Apply the definition of intersection. What does $\cap$ mean?

 

[reenmachine]

The definition of intersection is something like : the part where some elements of A are also elements of B and some elements of B are also elements of A and no other elements are there.

?

 

[reenmachine]

$\{2\} \cap \{1,2\} = \{2\}$ ?

 

[micromass]

There is a very formal definition of intersection and it is as follows:

$x\in A\cap B$ if and only if $x\in A$ and $x\in B$.

If you want, you can also write this in set builder notation:

A\cap B = \{x\in A~\vert~x\in B\}

 

[micromass]

$\{2\} \cap \{1,2\} = \{2\}$ ?

Yes.

 

[reenmachine]

I don't mean to interfere with the discussion, which seems to be going well. Just a little observation: You seem to sometimes be confusing unions with intersections. Intersections give us something smaller, unions something bigger: $$A\cap B\subseteq A\subseteq A\cup B.$$

btw , you don't ''interfere'' with any conversations , your feedbacks are always greatly appreciated!

 

[reenmachine]

There is a very formal definition of intersection and it is as follows:

$x\in A\cap B$ if and only if $x\in A$ and $x\in B$.

If you want, you can also write this in set builder notation:

A\cap B = \{x\in A~\vert~x\in B\}

yes it's clear now.

thanks!

 

[reenmachine]

Basically , it's unreasonable to think that there would be some common ground for all subsets of $R$ , and therefore this is the empty set.

In a set that represent many and many supposed intersected parts , such as $A \cap B \cap C$ , if any of the sets doesn't intersect with any other sets , then this is the empty set?

 

[micromass]

Basically , it's unreasonable to think that there would be some common ground for all subsets of $R$ , and therefore this is the empty set.

OK, but can you mathematically prove that

\bigcap_{A\in \mathcal{P}(\mathbb{R})} A = \emptyset

Just saying that it's unreasonable doesn't quite cut it.

 

[reenmachine]

OK, but can you mathematically prove that

\bigcap_{A\in \mathcal{P}(\mathbb{R})} A = \emptyset

Just saying that it's unreasonable doesn't quite cut it.

I know , I was trying to simplify it to a everyday-conversation level.

Just with $\{1\}$ and $\{2\}$ you already know this is the empty set.

Is this a (dis)proof by contradiction?

$\{1\},\{2\} \in P(R) : \{1\} ≠ \{2\}\}$

 

[micromass]

I know , I was trying to simplify it to a everyday-conversation level.

Yes, but it's very important for me that you understand this. And if you give "everyday-conversation" proofs, then it's hard for me to see whether you understand it or not. I really prefer rigorous mathematical proofs.

Is this a (dis)proof by contradiction?

That could work.

 

[reenmachine]

Maybe you didn't see in the edit

Proof that this is the empty set:

$\{\{1\},\{2\} \in P(R) : \{1\} ≠ \{2\}\}$

Proof that $\{1\} ≠ \{2\}$

$\{1 \in \{1\} : 1 \not \in \{2\}\}$

I don't know how to prove that 1 ≠ 2 as far as elements are concerned.

 

[micromass]

What does it mean by definition that

x\in \bigcap_{A\in \mathcal{P}(\mathbb{R})}A

?

 

[reenmachine]

What does it mean by definition that

x\in \bigcap_{A\in \mathcal{P}(\mathbb{R})}A

?

it means that x is an element of the set of the intersection of all the elements of P(R).

 

[Fredrik]

it means that x is an element of [strike]the set of[/strike] the intersection of all the elements of P(R).

Right. And if you use the definition of "intersection" to explain what that means...

You seem to be making it a bit more complicated than it needs to be.

 

[Fredrik]

I'm going to explain this one. I'm going to use spoiler tags so that you won't see the answers until you put your mouse pointer over them. I suggest that you keep thinking about each one for a while before you peek.

$$x\in\bigcap_{A\in\mathcal P(\mathbb R)} A.$$ This set can also be written as $\bigcap\mathcal P(\mathbb R)$ by the way.

Use the definition of the notation:

Spoiler

x is an element of the intersection of all the elements of P(ℝ).

Use the definition of the term "intersection":

Spoiler

x is an element of every element of P(ℝ)

Use the definition of the powerset:

Spoiler

x is an element of every subset of ℝ.

Think of the most useful consequence of that:

Spoiler

x is an element of ∅

Summary and conclusion:

Spoiler

This is impossible. No set is an element of ∅. But the statement above is a logical consequence of the statement that x is an element of the intersection I LaTeXed at the beginning of this post. So that statement must be false.

Since this argument holds for all sets x, we have proved that no set is a member of that intersection. So that intersection must be empty.

I think the simplest way to think here is this: "We're looking for the intersection of all the subsets of ℝ. Since one of those subsets is ∅ and we never get something bigger when we take an intersection, the intersection must be ∅".

 

[reenmachine]

I'm going to explain this one. I'm going to use spoiler tags so that you won't see the answers until you put your mouse pointer over them. I suggest that you keep thinking about each one for a while before you peek.

$$x\in\bigcap_{A\in\mathcal P(\mathbb R)} A.$$ This set can also be written as $\bigcap\mathcal P(\mathbb R)$ by the way.

Use the definition of the notation:

Spoiler

x is an element of the intersection of all the elements of P(ℝ).

Use the definition of the term "intersection":

Spoiler

x is an element of every element of P(ℝ)

Use the definition of the powerset:

Spoiler

x is an element of every subset of ℝ.

Think of the most useful consequence of that:

Spoiler

x is an element of ∅

Summary and conclusion:

Spoiler

This is impossible. No set is an element of ∅. But the statement above is a logical consequence of the statement that x is an element of the intersection I LaTeXed at the beginning of this post. So that statement must be false.

Since this argument holds for all sets x, we have proved that no set is a member of that intersection. So that intersection must be empty.

I think the simplest way to think here is this: "We're looking for the intersection of all the subsets of ℝ. Since one of those subsets is ∅ and we never get something bigger when we take an intersection, the intersection must be ∅".

My apologies for signing off out of nowhere , I had unexpected visitors.

I just did the exercise of thinking about each statement without using the spoilers and I understand pretty clearly.Don't know why I struggled earlier.

Thanks a lot!