2
$\begingroup$

enter image description here

Hi so we've started learning about relations and I'm completely lost.. I don't get how to represent the equivalence classes and I'm not even sure the way I've proved the equivalence relation is correct..

Question 1

a.

Note that if for every $a,b ∈ R$ if $a = b = 0 \ $$a*b$ $\not>$ $0$, but in this case it's already part of the relation.

Also note that in order for this inequality to work, $a,b$ must be both negative, or both positive.

Reflexive:

Since $a$ obviously has the same sign as itself, $a * a > 0$.

Therefore for every $a ∈ R/(0) $ $ \ a R a \ $

Symmetric:

By associativity, if $a * b > 0 ➜ b * a > 0$.

Therefore for every $a,b ∈ R/(0) $ if $aRb ➜ bRa$.

Transitive:

Since $a,b,c$ must share the same sign, if $a * b > 0$ and $b * c > 0 ➜ a * c > 0$.

Therefore for every $a,b,c ∈ R/(0) $ if $aRb ∧ bRc ➜ aRc$.

Equivalence classes:

$\left\{R^{+},R^{-},\left\{0\right\}\right\}$

b.

Note that if $8 ∉ XΔY$ then either $8 ∈ X ∧ 8 ∈ Y \ $ or $ \ 8 ∉ X ∧ 8 ∉ X$.

Reflexive:

Since $XΔX$ is the $∅$, then it is obvious $8 ∉ XΔX$.

Therefore for every $X ∈ P(NxN)$ $XRX$.

Symmetric:

Δ is associative, therefore if $8 ∉ XΔY ➜ 8 ∉ YΔX $

Therefore for every $X,Y ∈ P(NxN) \ XRY➜YRX$.

Transitive:

Since $X,Y,Z$ must all contain or all not contain $8$ for the relation to work,

then if $8 ∉ XΔY ∧ 8 ∉ YΔZ ➜ 8 ∉ XΔZ$.

Therefore for every $X,Y,Z ∈ P(NxN) \ $ if $XRY ∧ YRZ ➜ XRZ$.

Equivalence class:

$\left\{\left\{X\ ∈\ \ P(N)\ :\ 8\ ∉\ \ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 8\ ∈\ X\ \right\}\right\}$

c.

Note that we're proven in class equality is both reflexive, symmetric and transitive.

Reflexive: Obviously $X ∩ {1,2} = X ∩ {1,2}$.

Symmetric:

It is also obvious $X ∩ {1,2} = Y ∩ {1,2} ➜ Y ∩ {1,2} = X ∩ {1,2}$.

Transitive:

Again, obviously if $X ∩ {1,2} = Y ∩ {1,2} ∧ Y ∩ {1,2} = Z ∩ {1,2} \ ➜ \ X ∩ {1,2} = Z ∩ {1,2}$.

Equivalence classes:

$\left\{\left\{X\ ∈\ \ P(N)\ :\ 1,2\ ∉\ \ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 1\ ∉\ \ X\ \ ∧\ 2\ ∈\ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 2\ ∉\ \ X\ \ ∧\ 1\ ∈\ X\ \right\},\left\{X\ ∈\ \ P(N)\ :\ 1,2\ ∈\ \ X\ \right\}\right\}$

$\endgroup$
8
  • 2
    $\begingroup$ @ OP, as for your equivalence classes, while true that $[a]_R = \{b~:~aRb\}$, they are not asking for this... that is just the definition. The problem is asking for a ley description of the equivalence classes. In the first problems case, the equivalence classes are: 1) The positive numbers, 2) The negative numbers, and 3) Zero itself. $\endgroup$
    – JMoravitz
    Jan 7 at 13:52
  • 2
    $\begingroup$ For the third... the equivalence classes are: 1) Those sets which contain neither $1$ nor $2$, 2) Those sets which contain both $1$ and $2$, 3) Those sets which contain $1$ but not $2$, and 4) Those sets which contain $2$ but not $1$. The problem is asking for you to find the partition of the domain that this equivalence relation defines. I leave the middle one to you (I honestly haven't even read it, I only skimmed your post) $\endgroup$
    – JMoravitz
    Jan 7 at 13:55
  • 2
    $\begingroup$ As for typesetting, curly braces have special implication in MathJax and so if you want them to be visible you need to escape the characters with slashes.. $\{1,2\}$ yields $\{1,2\}$ $\endgroup$
    – JMoravitz
    Jan 7 at 13:56
  • 1
    $\begingroup$ The notation you are using is incredibly awkward or outright wrong. You are also indeed overlapping things. For question 3, I already gave you the answer. If you insist on writing it with symbols instead of words... the quotient set (i.e. the set of equivalence classes) is perhaps most easily written as $\{\{X\in\mathcal{P}(\Bbb N)~:~1\in X,~2\in X\},~\{X\in\mathcal{P}(\Bbb N)~:~1\in X,~2\notin X\},~\{X\in\mathcal{P}(\Bbb N)~:~1\notin X,~2\in X\},~\{X\in\mathcal{P}(\Bbb N)~:~1\notin X,~2\notin X\}\}$. When describing the equivalence sets explicitly, avoid the subscriptR notation. $\endgroup$
    – JMoravitz
    Jan 7 at 16:05
  • 1
    $\begingroup$ The subscriptR notation you are using does not have the meaning you intend. It is used when you take an individual representative element of the equivalence class and is used as a way of notating the equivalence class it comes from. Again, what appears inside the brackets is an explicit representative element, not a conditional statement like you are doing here. For the first... yes, you could have had the quotient set written as $\{[1]_R,[0]_R,[-1]_R\}$ however at a glance that doesn't tell us as much information as $\{\Bbb R^+,\{0\},\Bbb R^-\}$ does $\endgroup$
    – JMoravitz
    Jan 7 at 16:10
2
$\begingroup$

Instead of 'describing the equivalence classes' it might be more appropriate to rather ask for a set $B$ and a surjective function $f:A\to B$ for each example such that the given equivalence relation on $A$ is just its 'kernel' $\{(a,a'):f(a)=f(a')\}$.
This set $B$ represents the best the equivalence classes.

For 1., the positive numbers are in relation with all each other, and so are the negative numbers, and there's also the $0$, so the function we're looking for is the signum function ${\rm sign}:\Bbb R\to\{-1,0,+1\}$.

For 2. and 3., observe that condition in 2. holds for sets $X,Y$ if and only if either both of them contains $8$ or neither of them, so that this can be translated to the same condition as for 3.: $$X\sim Y\iff X\cap\{8\}=Y\cap\{8\}\,.$$ Generalizing this, for any set $U$ and any subset $A\subseteq U$, the relation $X\sim Y\iff X\cap A=Y\cap A$ is an equivalence relation on $P(U)$, and it is the kernel of the 'restriction' function $$f:P(U)\to P(A)\quad X\mapsto X\cap A\,.$$

So, the equivalence classes of 2. are represented by elements of $P(\{8\})=\big\{\emptyset,\,\{8\}\big\}$ and similarly, those of 3. by $P(\{1,2\})=\big\{\emptyset,\,\{1\},\, \{2\},\, \{1,2\}\big\}$.

$\endgroup$
1
  • $\begingroup$ It's something a lot of students don't realize, but it's surely an useful tool to quickly check if a relation is an equivalence relation. A kernel of a map is also a useful definition in topology and throughout algebra. $\endgroup$
    – Jakobian
    Jan 7 at 18:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.