cardinality of the set of junctions intervals

Question

Let P$=\{(a,b) : a,b \in \mathbb{R}\}$ and Q$=\{$ countable unions of elements of $P\}$. I 'm interested in knowing the cardinality of Q. I denote with $C$ if the cardinal real , then $C \leq \vert Q\vert$. but I wonder if it really will be given equality , appreciate any suggestion

Answer 1

We do indeed have $\vert Q\vert=C$.

To see this, we can argue as follows:

• $Q$ is certainly at most as large as the set $X$ of functions from $\mathbb{N}$ to $P$ (since to each such function $f$ we may assign an element of $Q$, $\bigcup f(n)$, and this assignment is surjective from $X$ onto $Q$).

• Meanwhile, $P$ has cardinality $C^2=C$, so $X$ has cardinality $C^{\aleph_0}$.

• So it is enough to show that $C^{\aleph_0}=C$. To do this, let's look at the set of infinite binary sequences, which has cardinality $C$. To an ${\aleph_0}$-sequence $F=(f_i)$ of infinite binary sequences, we can associate the single infinite binary sequence $g_F$, whose $\langle m, n\rangle$-th bit is the $m$th bit of $f_n$. (Here $\langle\cdot, \cdot\rangle$ is your favorite bijection from $\mathbb{N}^2$ to $\mathbb{N}$.)

• It's now an easy exercise to show that the map $F\mapsto g_F$ is a bijection from the set of infinite sequences of [infinite binary sequences] to the set of infinite binary sequences. So $C^{\aleph_0}=C$.

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The argument that $C^{\aleph_0}=C$ can be more briefly summed up as: $C=2^{\aleph_0}$, so $C^{\aleph_0}=(2^{\aleph_0})^{\aleph_0}=2^{{\aleph_0}\times{\aleph_0}}=2^{\aleph_0}=C$.