Monotonicity and Cardinality

Question

Is it true in general that if $B \subseteq A \implies \lvert B \rvert \leq \lvert A \rvert$, where $\lvert \cdot \rvert$ denotes cardinality? My intuition tells me that it would be natural to expect this to be true but it has failed me before.

Answer 1

Yes, this is true by definition of cardinality. Remember that "$\vert B\vert\le\vert A\vert$" means exactly that there is an injection from $B$ to $A$; if $B\subseteq A$, we can take the inclusion map as such an injection.

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I'm being a bit glib here when I talk about "the definition of cardinality." Depending on whether the axiom of choice holds or not, defining cardinality can be easy or a bit tricky. (Either way, the relation $\vert X\vert\le\vert Y\vert$ is easy to define, as I have above - and see the final paragraph, below.)

So let me say a short bit about what "cardinality" means (and by "short", I mean "about twice as long as the rest of this answer"). This may safely be considered tl;dr, but you may also find that it helps clarify matters.

With the axiom of choice, every set is in bijection with an ordinal. We now define the cardinality of a set $X$ to be the least ordinal $\alpha$ such that there is a bijection from $X$ to $\alpha$. Cardinalities are ordered by the ordering on ordinals. If $X\subseteq Y$ and $\alpha,\beta$ are the least ordinals in bijection with $X, Y$ via the maps $g, h$ respectively, it's not hard to show that $\alpha\le\beta$ - letting $i:X\rightarrow Y$ be the inclusion map, we have an injection of ordinals $f:\alpha\rightarrow \beta$ given by $f=h\circ i\circ g^{-1}$. Now we can't have $\alpha>\beta$ since then by Cantor-Shroeder-Bernstein we'd have that $X$ is in bijection with $\beta$, contradicting the assumption on $\alpha$. So we must have $\alpha\le\beta$, since the ordinals are linearly ordered, so $\vert X\vert=\alpha\le\beta=\vert Y\vert$.

Without the axiom of choice, the definition of cardinality becomes more complicated: the ordinals no longer provide "representatives" of each cardinality. Indeed, such a nice choice of representatives may not exist without choice: see this question. We can define the cardinality of a set in a manner similar to Frege:

• Frege identified e.g. "$2$" with "the set of all two-element sets."

• This doesn't work in $ZF$ (the collection of two-element sets is a proper class, not a set), but we can use Scott's trick: we let $\vert X\vert$ be the set of sets which are in bijection with $X$ and have minimal rank amongst such sets, where "rank" refers to the cumulative hierarchy.

Note, however, that $\vert X\vert$ is not a set in bijection with $X$ - rather, it is a set of sets which are in bijection with $X$. Expressions like "$\vert X\vert\le\vert Y\vert$" need to be interpreted accordingly - they now mean "For every $A\in\vert X\vert$ and $B\in\vert Y\vert$, there is an injection $i: A\rightarrow B$."

Note that the picture of comparing cardinalities via injections is more fundamental, at the end of the day, than a specific definition of what "cardinality" is. This is an important point: with some exceptions (e.g. "countable"), relative cardinality tends to be more important mathematically than absolute cardinality.