Question about proof from a guy with a highschool education

[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!

 

[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.