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I'm learning abstract algebra and need help with the following problem:

Let $X = \{1, \ldots, 10\}$ and $X_1 = \{2, 3\}, \,\,X_2= \{4, 6, 8\}, \,\,X_3= \{1, 9\}, \,\,X_4= \{10\}, \,\,X_5= \{5, 7\}$. Construct an equivalence relation $\mathcal{R}$ in $X$ such that the quotient set $X/\mathcal{R} = \{X_1, X_2, X_3, X_4; X_5\}$.

Unfortunately I wasn't able to do much with this one. I noticed that the families of the $X_i$'s, $i = 1, \ldots, 5$ constitutes a partition of the set $X$. Indeed, the union of the $X_i$'s gives the set $X$ and they are pairwise disjoint. Therefore there is an equivalence relation $\mathcal{R}$ on $X$. From here, I don't know how to construct $\mathcal{R}$.

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    It seems this exercise is just trying to reinforce the observations that an equivalence relation on a set is the same thing as a partition of the set. So I think the equivalence relation they are looking for is $i\simeq j\Leftrightarrow$ there is a $k$ such that $i,j\in X_k$. – 2017-02-15

2 Answers 2

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There is another way to do this, and for this we'll need a set $Y$ of cardinality $5$. Let's use $Y = \{1,2,3,4,5\}$.

Next, we'll define a function $f:X \to Y$ like so:

$f(1) = 3, f(2) = 1, f(3) = 1, f(4) = 2, f(5) = 5, f(6) = 2, f(7) = 5, f(8) = 2, f(9) = 3, f(10) = 4.$

Note this function is onto.

Define the relation $\mathcal{R}$ on $X$ by:

$a\sim_{\mathcal R}b \iff f(a) = f(b)$

I leave the verification this is a bona-fide equivalence relation on $X$, and gives the desired result, to you.

Because of this, we often say: "onto functions induce a (canonical) equivalence relation." The equivalence classes are the pre-image sets, which are in a nice one-to-one correspondence with the (elements of the) image set.

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Every equivalence relation induces a partition and every partition induces a equivalence relation.

In this case the relation is defined by $x\mathcal{R}y\iff \exists k:x,y\in X_k$

claim:

Let $X$ be a set and $\lbrace X_i\rbrace_{i\in I}$. Then the relation defined above is an equivalence relation

proof:

Let $x\in X$. As $\bigcup_{i\in I}X_i=X$, exists $i_x\in I:x\in X_{i_x}$, so $x,x\in X_{i_x}$ and then $x\mathcal{R}x$

Suppose $x\mathcal{R}y$. Then $\exists j\in I:x,y\in X_j$, so $y,x\in X_j$ and $y\mathcal{R}x$

If $x\mathcal{R}y$ and $y\mathcal{R}z$ exists $j,k\in I:x,y\in X_j,y,z\in X_k$ and as $y\in X_j\cap X_k$ and they belong to a partition, $X_j=X_k$ so $x,z\in X_k$ and $x\mathcal{R}z$

Remember relations are sets, so any other relation that satisfices this is actually the same but described by another way.