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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 $\mathbb{N}$. A set whose cardinality is strictly equal to the cardinality of $\mathbb{N}$ will be called countably infinite.

Also, $\bigsqcup_{i \in I} X_i = \{ (i,x) \in I \times\bigcup_{i \in I} X_i \mid x \in X_i \}$. 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 $f\colon I\to\mathbb{N}$ and a family $(f_i)_{i \in I}$ of injections $X_i\to\mathbb{N}$.

  • $(2)$ And what if each $X_i$ is countably infinite while $I$ is just countable? That is, there is an injection $f\colon I\to\mathbb{N}$ and a family $(f_i)_{i \in I}$ of injections $X_i\to\mathbb{N}$.

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) $X_i\to\mathbb{N}$ 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 $X_i\to\mathbb{N}$ (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.

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

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  • $\begingroup$ While I agree it's helpful, this seems more of an oversight on the original intention. If the OP edits now, what's the worth of your answer? $\endgroup$ – Asaf Karagila Sep 8 '18 at 7:32
  • $\begingroup$ Thanks! Edited. $\endgroup$ – Jxt921 Sep 8 '18 at 7:50

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