The axioms of a vector space are satisfied

In summary, this conversation discusses proving the axioms of a vector space for the $\mathbb{F}_2$-vector space $(2^M, +, \cap)$, where $M$ is a non-empty set. The conversation also touches on the elements of $\mathbb{K}$ and the multiplicative unity of $\mathbb{K}$, and the need to prove that $\mathbb{K}$ is a field isomorphic to $\mathbb{F}_2$ in order for the problem statement to be consistent.
  • #1
mathmari
Gold Member
MHB
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Hey! :eek:

We consider the $\mathbb{F}_2$-vector space $(2^M, +, \cap)$, where $M$ is non-empty set and $+ : 2^M\times 2^M \rightarrow 2^M: (X,Y)\mapsto (X\cup Y)\setminus (X\cap Y)$.

I want to show that $(2^M, +, \cap )$ for $\mathbb{K}=\{\emptyset , M\}$ satisfies the axioms of a vector space. I have done the following:
  • Associativity: $(X+Y)+Z=[(X\cup Y)\setminus (X\cap Y)]+Z=([(X\cup Y)\setminus (X\cap Y)]\cup Z)\setminus ([(X\cup Y)\setminus (X\cap Y)]\cap Z)$

    Is this correct so far? (Wondering)

    $ $
  • Existence of neutral element: The empty set.
  • Existence of inverse element: The inverse of an element is the set itself.
  • Commutativity: We have that $X+Y=(X\cup Y)\setminus (X\cap Y)=(Y\cup X)\setminus (Y\cap X)=Y+X$.

Are the last 3 ones correct? (Wondering) Then the next axioms that we have to show are multiplicative axioms (with the intersection). Let $k, k_1,k_2\in \mathbb{K}$.
  • \begin{align*}(X+Y)\cap k&=[(X\cup Y)\setminus (X\cap Y)]\cap k=[(X\cup Y)\cap k]\setminus [(X\cap Y)\cap k]\\ & =[(X\cap k)\cup (Y\cap k)]\setminus [(X\cap k)\cap (Y\cap k)] =X\cap k+Y\cap k\end{align*} here we use that $(A\cap B)\cap C)=(A\cap C)\cap (B\cap C)$, right? (Wondering)

    $ $
  • $X\cap (k_1+k_1)=X\cap [(k_1\cup k_2)\setminus (k_1\cap k_2)]=[X\cap (k_1\cup k_2)]\setminus [X\cap (k_1\cap k_2)]=[(X\cap k_1)\cup (X\cap k_2)]\setminus [(X\cap k_1)\cap (X\cap k_2)]=X\cap k_1+X\cap k_2$.

    Is this correct? (Wondering)

    $ $
  • $X\cap (k_1\cap k_2)=(X\cap k_1)\cap k_2$. So it is satisfied. (Or do we not get that equality immediately? (Wondering) )

    $ $
  • $k\cap 1=k$ where $1$ is in this case $2^M$, or not? (Wondering)
 
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  • #2
mathmari said:
Hey! :eek:

We consider the $\mathbb{F}_2$-vector space $(2^M, +, \cap)$, where $M$ is non-empty set and $+ : 2^M\times 2^M \rightarrow 2^M: (X,Y)\mapsto (X\cup Y)\setminus (X\cap Y)$.

I want to show that $(2^M, +, \cap )$ for $\mathbb{K}=\{\emptyset , M\}$ satisfies the axioms of a vector space. I have done the following:
  • Associativity: $(X+Y)+Z=[(X\cup Y)\setminus (X\cap Y)]+Z=([(X\cup Y)\setminus (X\cap Y)]\cup Z)\setminus ([(X\cup Y)\setminus (X\cap Y)]\cap Z)$

    Is this correct so far? (Wondering)

    $ $
  • Existence of neutral element: The empty set.
  • Existence of inverse element: The inverse of an element is the set itself.
  • Commutativity: We have that $X+Y=(X\cup Y)\setminus (X\cap Y)=(Y\cup X)\setminus (Y\cap X)=Y+X$.

Are the last 3 ones correct? (Wondering) Then the next axioms that we have to show are multiplicative axioms (with the intersection). Let $k, k_1,k_2\in \mathbb{K}$.
  • \begin{align*}(X+Y)\cap k&=[(X\cup Y)\setminus (X\cap Y)]\cap k=[(X\cup Y)\cap k]\setminus [(X\cap Y)\cap k]\\ & =[(X\cap k)\cup (Y\cap k)]\setminus [(X\cap k)\cap (Y\cap k)] =X\cap k+Y\cap k\end{align*} here we use that $(A\cap B)\cap C)=(A\cap C)\cap (B\cap C)$, right? (Wondering)

    $ $
  • $X\cap (k_1+k_1)=X\cap [(k_1\cup k_2)\setminus (k_1\cap k_2)]=[X\cap (k_1\cup k_2)]\setminus [X\cap (k_1\cap k_2)]=[(X\cap k_1)\cup (X\cap k_2)]\setminus [(X\cap k_1)\cap (X\cap k_2)]=X\cap k_1+X\cap k_2$.

    Is this correct? (Wondering)

    $ $
  • $X\cap (k_1\cap k_2)=(X\cap k_1)\cap k_2$. So it is satisfied. (Or do we not get that equality immediately? (Wondering) )

    $ $

Hey mathmari!

It is all correct so far. (Nod)

mathmari said:
  • $k\cap 1=k$ where $1$ is in this case $2^M$, or not?

Shouldn't that be $X \cap 1=X$ where $1$ is the multiplicative unity of $\mathbb K$?
What is the multiplicative unity of $\mathbb K$? (Wondering)

Btw, can we assume that $\mathbb K \simeq \mathbb F_2$, or do we need to prove that as well? (Wondering)
 
  • #3
Klaas van Aarsen said:
It is all correct so far. (Nod)

Ah ok!

For the associativity:
\begin{align*}(X+Y)+Z&=[(X\cup Y)\setminus (X\cap Y)]+Z\\ & =([(X\cup Y)\setminus (X\cap Y)]\cup Z)\setminus ([(X\cup Y)\setminus (X\cap Y)]\cap Z) \end{align*}
\begin{align*}X+(Y+Z)&=X+ [(Y\cup Z)\setminus (Y\cap Z)] \\ & = (X\cup [(Y\cup Z)\setminus (Y\cap Z)])\setminus (X\cap [(Y\cup Z)\setminus (Y\cap Z)])\end{align*}
How can we see that these two are equal? Which rules do we have to use here? I got stuck right now. (Wondering)
Klaas van Aarsen said:
Shouldn't that be $X \cap 1=X$ where $1$ is the multiplicative unity of $\mathbb K$?
What is the multiplicative unity of $\mathbb K$? (Wondering)

Oh yes you are right! The elements of $\mathbb{K}$ are $\emptyset$ and $M$, so the multiplicative unity of $\mathbb K$ is $M$ since $X\cap M=X=M\cap X$, right? (Wondering)
Klaas van Aarsen said:
Btw, can we assume that $\mathbb K \simeq \mathbb F_2$, or do we need to prove that as well? (Wondering)

Why does this hold? (Wondering)
 
  • #4
mathmari said:
Ah ok!

For the associativity:
\begin{align*}(X+Y)+Z&=[(X\cup Y)\setminus (X\cap Y)]+Z\\ & =([(X\cup Y)\setminus (X\cap Y)]\cup Z)\setminus ([(X\cup Y)\setminus (X\cap Y)]\cap Z) \end{align*}
\begin{align*}X+(Y+Z)&=X+ [(Y\cup Z)\setminus (Y\cap Z)] \\ & = (X\cup [(Y\cup Z)\setminus (Y\cap Z)])\setminus (X\cap [(Y\cup Z)\setminus (Y\cap Z)])\end{align*}
How can we see that these two are equal? Which rules do we have to use here? I got stuck right now.

We can write $A\setminus B = A\cap B^C$ can't we? (Thinking)

mathmari said:
Oh yes you are right! The elements of $\mathbb{K}$ are $\emptyset$ and $M$, so the multiplicative unity of $\mathbb K$ is $M$ since $X\cap M=X=M\cap X$, right?

(Nod)

mathmari said:
Why does this hold? (Wondering)

Well, the problem statement says that we have an $\mathbb F_2$-vector space, doesn't it?
That means that our scalars must come from the field $\mathbb F_2$.
And then it said that we were looking at a vector space for $\mathbb K=\{\varnothing, M\}$.
That means that $\mathbb K$ must be a field, mustn't it?
And it should be isomorphic to $\mathbb F_2$ for the problem statement to be consistent.

So I think we should prove that $\mathbb K$ is a field with the given addition and multiplication.
Since there is only one field with 2 elements ($\mathbb F_2$), it should suffice if we can identify $0$ and $1$, and verify that $\mathbb K$ has the same addition table and multiplication table as $\mathbb F_2$. (Nerd)
 
  • #5
Klaas van Aarsen said:
We can write $A\setminus B = A\cap B^C$ can't we? (Thinking)

So, we get the following:
\begin{align*}(X+Y)+Z&=[(X\cup Y)\setminus (X\cap Y)]+Z\\ & =([(X\cup Y)\setminus (X\cap Y)]\cup Z)\setminus ([(X\cup Y)\setminus (X\cap Y)]\cap Z) \\ & = ([(X\cup Y)\cap (X\cap Y)^c]\cup Z)\setminus ([(X\cup Y)\cap (X\cap Y)^c]\cap Z) \\ & = ([(X\cup Y)\cup Z]\cap [(X\cap Y)^c\cup Z)\setminus ((X\cup Y)\cap (X\cap Y)^c\cap Z) \\ & = ([X\cup Y\cup Z]\cap [(X\cap Y)^c\cup Z)\setminus ((X\cup Y)\cap (X\cap Y)^c\cap Z) \end{align*}
\begin{align*}X+(Y+Z)&=X+ [(Y\cup Z)\setminus (Y\cap Z)] \\ & = (X\cup [(Y\cup Z)\setminus (Y\cap Z)])\setminus (X\cap [(Y\cup Z)\setminus (Y\cap Z)]) \\ & = (X\cup [(Y\cup Z)\cap (Y\cap Z)^c])\setminus (X\cap [(Y\cup Z)\cap (Y\cap Z)^c]) \\ & = ( [X\cup Y\cup Z]\cap [X\cup(Y\cap Z)^c])\setminus (X\cap (Y\cup Z)\cap (Y\cap Z)^c) \end{align*}
right? How could we continue? (Wondering)
Klaas van Aarsen said:
Well, the problem statement says that we have an $\mathbb F_2$-vector space, doesn't it?
That means that our scalars must come from the field $\mathbb F_2$.
And then it said that we were looking at a vector space for $\mathbb K=\{\varnothing, M\}$.
That means that $\mathbb K$ must be a field, mustn't it?
And it should be isomorphic to $\mathbb F_2$ for the problem statement to be consistent.

So I think we should prove that $\mathbb K$ is a field with the given addition and multiplication.
Since there is only one field with 2 elements ($\mathbb F_2$), it should suffice if we can identify $0$ and $1$, and verify that $\mathbb K$ has the same addition table and multiplication table as $\mathbb F_2$. (Nerd)

Ah ok! (Nerd)
 
  • #6
mathmari said:
So, we get the following:
\begin{align*}(X+Y)+Z&=[(X\cup Y)\setminus (X\cap Y)]+Z\\ & =([(X\cup Y)\setminus (X\cap Y)]\cup Z)\setminus ([(X\cup Y)\setminus (X\cap Y)]\cap Z) \\ & = ([(X\cup Y)\cap (X\cap Y)^c]\cup Z)\setminus ([(X\cup Y)\cap (X\cap Y)^c]\cap Z) \\ & = ([(X\cup Y)\cup Z]\cap [(X\cap Y)^c\cup Z)\setminus ((X\cup Y)\cap (X\cap Y)^c\cap Z) \\ & = ([X\cup Y\cup Z]\cap [(X\cap Y)^c\cup Z)\setminus ((X\cup Y)\cap (X\cap Y)^c\cap Z) \end{align*}
\begin{align*}X+(Y+Z)&=X+ [(Y\cup Z)\setminus (Y\cap Z)] \\ & = (X\cup [(Y\cup Z)\setminus (Y\cap Z)])\setminus (X\cap [(Y\cup Z)\setminus (Y\cap Z)]) \\ & = (X\cup [(Y\cup Z)\cap (Y\cap Z)^c])\setminus (X\cap [(Y\cup Z)\cap (Y\cap Z)^c]) \\ & = ( [X\cup Y\cup Z]\cap [X\cup(Y\cap Z)^c])\setminus (X\cap (Y\cup Z)\cap (Y\cap Z)^c) \end{align*}
right? How could we continue?

We can use $(A\cap B)^c = A^c\cup B^c$ and $(A\cup B)^c = A^c\cap B^c$ can't we? (Thinking)
 
  • #7
Klaas van Aarsen said:
We can use $(A\cap B)^c = A^c\cup B^c$ and $(A\cup B)^c = A^c\cap B^c$ can't we? (Thinking)

Ah yes!

So, we get:
\begin{align*}(X+Y)+Z&=[(X\cup Y)\setminus (X\cap Y)]+Z\\ & =([(X\cup Y)\setminus (X\cap Y)]\cup Z)\setminus ([(X\cup Y)\setminus (X\cap Y)]\cap Z) \\ & = ([(X\cup Y)\cap (X\cap Y)^c]\cup Z)\setminus ([(X\cup Y)\cap (X\cap Y)^c]\cap Z) \\ & = ([(X\cup Y)\cup Z]\cap [(X\cap Y)^c\cup Z]\setminus ((X\cup Y)\cap (X\cap Y)^c\cap Z) \\ & = ([X\cup Y\cup Z]\cap [(X\cap Y)^c\cup Z])\setminus ((X\cup Y)\cap (X\cap Y)^c\cap Z) \\ & = ([X\cup Y\cup Z]\cap [(X\cap Y)^c\cup Z])\cap((X\cup Y)\cap (X\cap Y)^c\cap Z)^c \\ & = [X\cup Y\cup Z]\cap [(X\cap Y)^c\cup Z]\cap((X\cup Y)\cap (X\cap Y)^c\cap Z)^c \\ & = [X\cup Y\cup Z]\cap [(X\cap Y)^c\cup Z]\cap[(X\cup Y)^c\cup (X\cap Y)\cup Z^c] \\ & = [X\cup Y\cup Z]\cap [X^c\cup Y^c\cup Z]\cap[(X^c\cap Y^c)\cup (X\cap Y)\cup Z^c] \end{align*}
\begin{align*}X+(Y+Z)&=X+ [(Y\cup Z)\setminus (Y\cap Z)] \\ & = (X\cup [(Y\cup Z)\setminus (Y\cap Z)])\setminus (X\cap [(Y\cup Z)\setminus (Y\cap Z)]) \\ & = (X\cup [(Y\cup Z)\cap (Y\cap Z)^c])\setminus (X\cap [(Y\cup Z)\cap (Y\cap Z)^c]) \\ & = ( [X\cup Y\cup Z]\cap [X\cup(Y\cap Z)^c])\setminus (X\cap (Y\cup Z)\cap (Y\cap Z)^c) \\ & = ( [X\cup Y\cup Z]\cap [X\cup(Y\cap Z)^c])\cap (X\cap (Y\cup Z)\cap (Y\cap Z)^c)^c \\ & = ( [X\cup Y\cup Z]\cap [X\cup(Y\cap Z)^c])\cap (X^c\cup (Y\cup Z)^c\cup (Y\cap Z)) \\ & = [X\cup Y\cup Z]\cap [X\cup Y^c\cup Z^c]\cap [X^c\cup (Y^c\cap Z^c)\cup (Y\cap Z)] \end{align*}
Is this correct so far? Which rule do we have to use next? Do we have to simplify the intersections with unions further? (Wondering)
 
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  • #8
Well, we should ultimately find the following Venn diagram for $X+Y+Z$:
\begin{tikzpicture}[fill=red!30!white, scale=0.5]
%preamble \usetikzlibrary{shapes,backgrounds}
\begin{scope}
\clip (210:2) circle (3) (-5,-5) rectangle (5,5);
\clip (330:2) circle (3) (-5,-5) rectangle (5,5);
\fill ( 90:2) circle (3);
\end{scope}
\begin{scope}
\clip ( 90:2) circle (3) (-5,-5) rectangle (5,5);
\clip (330:2) circle (3) (-5,-5) rectangle (5,5);
\fill (210:2) circle (3);
\end{scope}
\begin{scope}
\clip ( 90:2) circle (3) (-5,-5) rectangle (5,5);
\clip (210:2) circle (3) (-5,-5) rectangle (5,5);
\fill (330:2) circle (3);
\end{scope}
\begin{scope}
\clip ( 90:2) circle (3);
\clip (210:2) circle (3);
\fill (330:2) circle (3);
\end{scope}
\node at (90:3) {$X$};
\node at (210:3) {$Y$};
\node at (330:3) {$Z$};
\end{tikzpicture}
It should be $X+Y+Z = (X\cap Y\cap Z) \cup (X\cap Y^c\cap Z^c) \cup (X^c\cap Y\cap Z^c) \cup (X^c\cap Y^c\cap Z)$.
Or alternatively $X+Y+Z = (X\cup Y\cup Z) \cap (X\cup Y^c\cup Z^c) \cap (X^c\cup Y\cup Z^c) \cap (X^c\cup Y^c\cup Z)$.

So I think we need to be a bit smart on how to get there. (Sweating)
 
  • #9
Klaas van Aarsen said:
So I think we need to be a bit smart on how to get there. (Sweating)

But how? (Wondering)
 
  • #10
mathmari said:
But how? (Wondering)

Let's try to write everything as a disjunction of conjunctions.
For starters we can write:
$$A+B=(A\cup B)\setminus (A\cap B) = (A\cap B^c) \cup (A^c \cap B) \tag 1$$
So:
$$(X+Y)+Z= ((X+Y)\cap Z^c) \cup ((X+Y)^c \cap Z) \tag 2$$
The first half of (2) expands into a disjunction as:
$$((X+Y)\cap Z^c) = [(X\cap Y^c) \cup (X^c \cap Y)]\cap Z^c
= (X\cap Y^c\cap Z^c) \cup (X^c \cap Y\cap Z^c)$$
The second half of (2) expands as:
$$\begin{align*}((X+Y)^c \cap Z) &= [(X\cap Y^c) \cup (X^c \cap Y)]^c \cap Z
= (X\cap Y^c)^c \cap (X^c \cap Y)^c \cap Z
= (X^c\cup Y) \cap (X \cup Y^c) \cap Z \\
&= [((X^c\cup Y) \cap X) \cup ((X^c\cup Y) \cap Y^c)] \cap Z \\
&= [(Y \cap X) \cup (X^c \cap Y^c)] \cap Z \\
&= (X\cap Y \cap Z) \cup (X^c \cap Y^c \cap Z) \end{align*}
$$
Substitute in (2) to find:
$$(X+Y)+Z = (X\cap Y^c\cap Z^c) \cup (X^c \cap Y\cap Z^c) \cup (X\cap Y \cap Z) \cup (X^c \cap Y^c \cap Z) \tag 3$$
(Whew)
 
  • #11
Associativity can indeed be proven by brute force, reducing both sides to disjunction of conjunctions of sets or their complements. The result of simplifying $A_1+\dots+A_n$ should be the disjunction of conjunctions where each conjunction has all $A_i$'s and among them an odd number of $A_i$'s not under complement. So for $n=3$ each conjunction must have either zero or two complements, thus one or three sets without complements. I also prefer omitting $\cap$ and writing $\bar{A}$ for the complement of $A$.
 

FAQ: The axioms of a vector space are satisfied

What are the axioms of a vector space?

The axioms of a vector space are a set of rules that must be satisfied for a set of objects to be considered a vector space. These axioms include closure under vector addition and scalar multiplication, associative and commutative properties, and the existence of an additive identity and inverse for each vector.

Why is it important for the axioms of a vector space to be satisfied?

If the axioms of a vector space are not satisfied, then the set of objects in question cannot be considered a true vector space. This means that the set cannot be used for mathematical operations and cannot be studied using vector space properties and theorems.

How do we know if the axioms of a vector space are satisfied?

The axioms of a vector space can be checked by verifying that each axiom is satisfied for all vectors in the set. This can be done through mathematical proofs or by testing specific examples.

Can the axioms of a vector space be satisfied for different types of objects?

Yes, the axioms of a vector space can be satisfied for a variety of different objects, as long as they satisfy the properties of closure under addition and scalar multiplication, and possess an additive identity and inverse. For example, we can have vector spaces of matrices, polynomials, or functions.

What happens if one of the axioms of a vector space is not satisfied?

If one of the axioms of a vector space is not satisfied, then the set of objects in question cannot be considered a vector space. This means that the objects in the set cannot be studied using vector space properties and theorems, and mathematical operations involving these objects may not be valid.

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