Binary operation on equivalence classes

[PhysicsRock]

Homework Statement

Given a set $M$ and an equivalence relation $\sim$ on $M$, for $a,a^{\prime},b,b^{\prime} \in M$ with $a \sim a^{\prime}$ and $b \sim b^{\prime}$ with a binary operation $*$ such that $a * b \sim a^{\prime} * b^{\prime}$ prove that a binary operation $\overline{*}$ exists that fulfills $[x * y] = [x] \overline{*} [y]$.

Relevant Equations

None else.

So, my approach and solution are as follows:

$$

[x * y] = \{ z \in M : z \sim (x * y) \}

$$

Since we know that $a * b \sim a^{\prime} * b^{\prime}$ we can rewrite $z$ as $x^{\prime} * y^{\prime}$. Plugging this in yields:

$$

[x * y] = \{ x^{\prime}, y^{\prime} \in M : x^{\prime} * y^{\prime} \sim x * y \}

$$

We are also given that if $x^{\prime} * y^{\prime} \sim x * y$ then $x^{\prime} \sim x$ and $y^{\prime} \sim y$. So once again, we can rewrite the expression to obtain:

$$
[x * y] = \{ x^{\prime}, y^{\prime} \in M : x^{\prime} \sim x, y^{\prime} \sim y \}
$$

With $x^{\prime}$ and $y^{\prime}$ now essentially separated, we can break apart the set into two individual ones with

$$
[x*y] = \{ x^{\prime} \in M : x^{\prime} \sim x \} \bigcup \{ y^{\prime} \in M : y^{\prime} \sim y \}
$$

Now one observes that the two sets are the equivalence classes for $x$ and $y$ respectively. So, we can conclude that

$$
[x*y] = [x] \cup [y]
$$

and comparing this to the assignment, we obtain that $\overline{*}$ is equivalent to the union of sets. Hence, there exists a binary operation fulfilling the required demands.

Now, am I correct here or did I screw up at some point?

 

[PeroK]

Homework Statement:: Given a set $M$ and an equivalence relation $\sim$ on $M$, for $a,a^{\prime},b,b^{\prime} \in M$ with $a \sim a^{\prime}$ and $b \sim b^{\prime}$ with a binary operation $*$ such that $a * b \sim a^{\prime} * b^{\prime}$ prove that a binary operation $\overline{*}$ exists that fulfills $[x * y] = [x] \overline{*} [y]$.

I wonder whether something has been lost in translation here. I suspect what you are asked to prove is that the binary operation defined by $[x] \overline{*} [y] = [x * y]$ is well-defined. In the sense that it doesn't matter which members of the equivalence classes $[x]$ and $[y]$ you choose, you get the same result.

$$

[x * y] = \{ z \in M : z \sim (x * y) \}

$$

Since we know that $a * b \sim a^{\prime} * b^{\prime}$ we can rewrite $z$ as $x^{\prime} * y^{\prime}$. Plugging this in yields:

$$

[x * y] = \{ x^{\prime}, y^{\prime} \in M : x^{\prime} * y^{\prime} \sim x * y \}

$$

This doesn't look right. Instead:
$$

[x * y] = \{ z: z \sim x * y \}

$$

$$
[x*y] = \{ x^{\prime} \in M : x^{\prime} \sim x \} \bigcup \{ y^{\prime} \in M : y^{\prime} \sim y \}
$$

I don't understand what you are doing here.

Now one observes that the two sets are the equivalence classes for $x$ and $y$ respectively. So, we can conclude that

$$
[x*y] = [x] \cup [y]
$$

and comparing this to the assignment, we obtain that $\overline{*}$ is equivalent to the union of sets. Hence, there exists a binary operation fulfilling the required demands.

Now, am I correct here or did I screw up at some point?

I don't follow this either.

 

[PhysicsRock]

I wonder whether something has been lost in translation here. I suspect what you are asked to prove is that the binary operation defined by is well-defined. In the sense that it doesn't matter which members of the equivalence classes and you choose, you get the same result.

Yes, you are right. I quoted the statement from my memory and forgot the well-definedness. In that case, is it still recommendable to make use of set notation or should I retreat to use elements?

If it still makes sense, should I explain why I carried out certain steps you said you didn't understand?

 

[PeroK]

Yes, you are right. I quoted the statement from my memory and forgot the well-definedness. In that case, is it still recommendable to make use of set notation or should I retreat to use elements?

If it still makes sense, should I explain why I carried out certain steps you said you didn't understand?

The result should follow quite quickly from what you are given.