What infinity is greater than the continuum? Show with an example

Solution 1:

The same type of diagonalization argument which is used to show $|\mathbb{R}| > |\mathbb{Z}|$ can be used to show that for any set $X$ the power set of $X$ has strictly greater cardinality than $X$. Thus, if you want a set of cardinality greater than $\mathbb{R}$ try $\mathcal{P}(\mathbb{R}).$

Solution 2:

Since you are not satisfied with the power set being the most natural example, here is one slightly less natural - but still very common (especially in the absence of choice):

Consider all the ordinal numbers that there is an injective function from them into $\mathbb R$, i.e. $A = \{\alpha\mid\alpha\text{ is an ordinal, and }\exists f\colon\alpha\xrightarrow{1-1}\mathbb R\}$.

We claim that $A$ cannot be bijected with $\mathbb R$.

First, a set of ordinals which is downwards-closed (i.e. if $\alpha\in A$ and $\beta<\alpha$, then $\beta\in A$) is an ordinal. There is a simple proof for this fact, simply because a set $x$ is an ordinal if and only if it is transitive and well-ordered by $\in$.

From this we have that $A$ is an ordinal, because if $\alpha\in A$ and $\beta<\alpha$ then the same injective function from $\alpha$ is still injective from $\beta$ when restricted to $\beta$ (recall that $\beta<\alpha$ if and only if $\beta\in\alpha$ if and only if $\beta\subsetneq\alpha$).

Suppose there was a function from $A$ into the reals which is injective, since $A$ is an ordinal we would have $A\in A$, in contradiction to the fact that an ordinal cannot be a member of itself (regardless to whether or not the axiom of foundation holds in the universe).

Alternatively, one could define $A$ as the least ordinal which cannot be injected into $\mathbb R$. This ordinal exists, since $\mathbb R$ is a set, and the ordinals form a proper class; and it would have the exact same properties.

This construction is called Hartogs number.

Solution 3:

Very similar questions have been asked here before, but I am finding it easier to simply answer again than search for the closest precedent.

An example of a set of cardinality greater than the continuum is given in $\S 2.2$ of these notes. In $\S 2.4$ it is shown that there is more than a set's worth of cardinalities of uncountable sets.

Solution 4:

If you don't like the power set, another example is the set of functions $f:\mathbb{R} \rightarrow \mathbb{R}$.