If $X$ is an infinite set, then is $\sum_{i=1}^\infty |X|=|X|$?

Question

I used the above to solve a problem, but I am not sure if this is true (and if it is true does it require a proof or is it obvious). I think it is true because $|X| + |X| = |X|$, so it should follow that $\sum_{i=1}^\infty |X|=|X|$. However, I am getting confused because I do not know if there may be different 'rules' for infinite sums of cardinals.

Answer 1

The answer depends on the context. In particular it depends on the Axiom of Choice. Since you put the "cardinals" and "elementary-set-theory" tags on it, I will treat it as a set theoretic question. If instead this question arose from some application (say group theory), then feel free to just assume the Axiom of Choice.

Definitions

In general we have the following definitions for adding cardinalities: if $A$ and $B$ are disjoint sets, then $|A|+|B|=|A\cup B|$. If $\kappa$ is a cardinal number, then we can generalise this by defining the sum $\sum_{\xi<\kappa}|A|$ as the disjoint union of $\kappa$ many sets of size $|A|$.

We can define cardinal multiplication as $|A|\cdot |B|=|A\times B|$.

Although $\infty$ as a symbol is too vague for set theoretic purposes, I guess that with $\sum_{i=1}^\infty$ you mean to say $\sum_{i\in\omega}$, where $\omega=\{0,1,2,3,\dots\}$ is the set of finite ordinals, or natural numbers.

With Axiom of Choice

Under the assumption of the Axiom of Choice every set can be well-ordered, and thus every set has a cardinality equal to a cardinal number. In this case we could show that addition and multiplication of infinite cardinals are quite trivial. In particular, one could show for both $A$ and $B$ infinite sets, that $|A|+|B|=|A|\cdot |B|=\max\{ |A|,|B|\}$, and thus $|X|+|X|=|X|$ for an infinite set $X$ is indeed true under the assumption of the Axiom of Choice.

We can see $\sum_{i\in\omega} |X|$ as the cardinality of the union of $\{X\times\{i\}\mid i\in\omega\}=X\times\omega$. Since $X$ was assumed to be infinite, we then get $|X\times\omega|=\max\{|X|,\omega\}=|X|$.

Note that it does not hold in general that $\sum_{\xi\in\kappa}|X|=|X|$ for any cardinal $\kappa$. To be precise, $\sum_{\xi\in\kappa}|X|=\max\{\kappa,|X|\}$ if $\kappa$ is infinite and $X\neq\varnothing$.

Without Axiom of Choice

Without the Axiom of Choice, it gets very difficult to say what the size of $\sum_{i\in\omega} |X|$ is. For example, without choice, it is possible that the countable union of countable sets is uncountable, and thus we cannot claim that $\sum_{i\in\omega}|X|=|X|$. This also shows that it becomes unreasonable to give any meaning to taking the sum of arbitrary cardinalities, since it does not have a fixed value: the value depends on which sets we take the cardinality of. In other words, even if $|A_i|=|B_i|$ for all $i\in\omega$ and for every $i\neq j$ we have $A_i$ and $A_j$, and $B_i$ and $B_j$, disjoint from each other, it could be that $\left|\bigcup_{i\in\omega} A_i\right|\neq\left|\bigcup_{i\in\omega} B_i\right|$

In fact, even finite sums become problematic: without the Axiom of Choice there could exist Dedekind-finite infinite sets. If $X$ is a Dedekind-finite infinite set, then the cardinality of the disjoint union $|X\sqcup X|$ is still Dedekind-finite, which implies that $|X|+|X|>|X|$.