Vtyjkyuk

First, let me say that when I say countable set I mean a set whose cardinality is less than or equal to the cardinality of $mathbbN$. A set whose cardinality is strictly equal to the cardinality of $mathbbN$ will be called countably infinite.

Also, $bigsqcup_i in I X_i = (i,x) in I timesbigcup_i in I X_i mid x in X_i $. Of course, $|bigcup_i in I X_i| leq |bigsqcup_i in I X_i|$.

Let $(X_i)_i in I$ be an indexed family with the index set $I$.

It is widely known that if $I$ is countable (resp., countably infinite) and $forall i in I, X_i$ is countable (resp., countable infinite), then $bigsqcup_i in I X_i$ is itself countable (resp., countable infinite).

However, what if $I neq varnothing$ and

• $(1)$ $I$ is countably infinite but each $X_i$ is only known to be countable? That is, there is a bijection $fcolon ItomathbbN$ and a family $(f_i)_i in I$ of injections $X_itomathbbN$.




• $(2)$ And what if each $X_i$ is countably infinite while $I$ is just countable? That is, there is an injection $fcolon ItomathbbN$ and a family $(f_i)_i in I$ of injections $X_itomathbbN$.

The intuition tells me that in both cases $(1)$ and $(2)$ the disjoint union $bigsqcup_i in I X_i$ will be countably infinite. However, how to prove it rigorously is another question entirely.

Remark. To construct a family $(f_i)_i in I$ of injections (resp., bijections) $X_itomathbbN$ knowing only that each $X_i$ is countable (resp., countably infinite) we need the axiom schema of replacement to obtain a set of sets of injections $X_itomathbbN$ (each for every $i in I$) and the axiom of choice to obtain a choice function of such set.

P.S. I don't know whether I should tag this as elementary set theory or set theory. Sorry.

Since the empty set is countable, (1) is not necessarily infinite, and neither is (2) if $I=emptyset$.