cardinal of a quotient space

Question

Suppose that we have an equivalence relation $R$ defined on an infinite set $X$ and that all equivalence classes are finite. Is it so that the cardinal of the quotient space of $R$ is that of $X$? If so, I would appreciate a proof. If not, a counterexample.

Answer 1

Yes its true. Suppose $\{X_i\}_I$ is a partition of $X$ defined by this equivalence relation (i.e. equivalence classes). $X=\bigcup X_i$. If $Card(X)\ge \aleph_0$ and $Card(X_i)<\aleph_0$ then $Card(X)=Card(\bigcup X_i)\le Card(I)\cdot Card(\aleph_0)\le Card(I)\cdot Card(I)=Card(I)$.

As it was pointed out in comments to get inequality $Card(I)\le Card(X)$ we can define injection from $f:I\rightarrow X$ using axiom of choice. For any $i\in I$ pick (using AC) element $x_i\in X_i$ and set $f(i)=x_i$. This function is injective because $X_i\cap X_j=\emptyset$ for any $i\neq j$.

Now by Cantor–Bernstein–Schroeder theorem we get $Card(X)=Card(I)$.

Answer 2

Assuming the axiom of choice holds, every partition of an infinite set into finite sets must have the same cardinality. The reason is that if $\mathcal A=\{A_i\mid i\in I\}$ is a collection of pairwise disjoint sets, then $$\left|\bigcup\mathcal A\right|=\sup\left\{|A_i|\vphantom{x^2}\mathrel{}\middle|\mathrel{} i\in I\right\}\cdot|I|=\max\left\{\sup\left\{|A_i|\vphantom{x^2}\mathrel{}\middle|\mathrel{} i\in I\right\},|I|\right\}$$

Assuming the $A_i$ are all finite, then we that becomes at most $\aleph_0\cdot|I|=|I|$ since $I$ is infinite.

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Not assuming the axiom of choice, the plot like a slowly cooking gravy, thickens. It is possible in the absence of the axiom of choice to have a set and a partition of that set into strictly more parts than elements; or an incomparable cardinality of parts. Even if the parts themselves are finite!