# Does there exist a subset of $\mathbb{R}$ of cardinality $2^{\aleph_0}$ that has no perfect subset?

Assuming the axiom of choice, is there a way to construct a subset of the reals of cardinality $2^{\aleph_0}$ that has no perfect subset?

-
A set with such a property is called a Bernstein set. –  Damian Sobota Dec 6 '11 at 2:42

Yes: just build it one point at a time. There are only $2^\omega$ perfect subsets, so we can enumerate them as $\{P_\xi:\xi<2^\omega\}$. Assume that at stage $\eta<2^\omega$ we’ve picked points $x_\xi,y_\xi\in P_\xi$ for each $\xi<\eta$ in such a way that the points $x_\xi$, $\xi<\eta$, and $y_\xi$, $\xi<\eta$, are all distinct. $|P_\eta|=2^\omega$, and we’ve chosen fewer than $2^\omega$ points so far, so we may choose distinct

$$x_\eta,y_\eta\in\{x_\xi:\xi<\eta\}\cup\{y_\xi:\xi<\eta\}$$

and continue the construction. Finally, let $X=\{x_\xi:\xi<2^\omega\}$ and $Y=\{y_\xi:\xi<2^\omega\}$. Clearly $|X|=|Y|=2^\omega$, $X\cap Y=\varnothing$, and both $X$ and $Y$ have non-empty intersection with every perfect set (since $x_\xi\in X\cap P_\xi$ and $y_\xi\in Y\cap P_\xi$), so neither $X$ nor $Y$ can contain a perfect subset.

Added: A more sophisticated version of this construction can be used to partition $\mathbb{R}$ into $2^\omega$ sets of cardinality $2^\omega$, none of which contains a perfect subset. Let $\{\langle\alpha_\xi,\beta_\xi\rangle:\xi<2^\omega\}$ be an enumeration of $2^\omega\times 2^\omega$, and let $\preceq$ be a well-ordering of $\mathbb{R}$ in type $2^\omega$.

Suppose that at stage $\eta$ we’ve chosen distinct $x_\xi\in\mathbb{R}$ such that $x_\xi\in P_{\alpha_\xi}$ for each $\xi<\eta$; as before, $P_{\alpha_\eta}\setminus \{x_\xi:\xi<\eta\}\ne\varnothing$, and we let $x_\eta$ be the $\preceq$-least element of $P_{\alpha_\eta}\setminus \{x_\xi:\xi<\eta\}$. Clearly the construction goes through to $2^\omega$.

Now for each $\eta<2^\omega$ let $X_\eta=\{x_\xi:\beta_\xi=\eta\}$; clearly $|X_\eta|=2^\omega$, and the sets $X_\eta$, $\eta<2^\omega$ are pairwise disjoint. Choosing the $\preceq$-least candidate at each stage ensures that $\{X_\eta:\eta<2^\omega\}$ is a partition of $\mathbb{R}$. Finally, for any $\eta,\gamma<2^\omega$ there is a $\xi<2^\omega$ such that $\alpha_\xi=\gamma$ and $\beta_\xi=\eta$, and by construction $x_\xi\in X_\eta\cap P_\gamma\ne\varnothing$; thus, each $X_\eta$ meets each perfect set.

-
Interesting. Thank you for providing two types of constructions! –  Jens Wallet Dec 6 '11 at 3:42