What is the cardinality of the set of all real closed intervals under certain restrictions?

Question

Let $M$ be the set of all real closed intervals: $M=$ { $[a,b] $ | $a \le b, a,b \in \mathbb{R}$}

Define $I_a$ to be all the elements in $M$ that at least one of their endpoints is a rational number, $I_b$ to be all the elements in $M$ that have a rational length, and $I_c$ to be all the elements in $M$ that both of their endpoints are rational numbers.

What is the cardinality of $I_a, I_b$ and $I_c$?

I think that the cardinality of $I_a$ and $I_b$ is $\aleph$, because I can match each real number at least one element from both $I_a$ and $I_b$, and I think that the cardinality of $I_c$ is $\aleph_0$, bacouse I can match every element of $I_c$ to an element in $ \mathbb{Q}^2$, which is a countable set, but I am not sure how to prove those things.

Answer 1

Hint:

Here is half of what you need. Define injection $f : M \to \mathbb{R} \times \mathbb{R}$ as $$f\big([x,y]\big) = \langle x, y+1 \rangle.$$

Obviously,

• $f|_{I_a} \subseteq \mathbb{R} \times \mathbb{R} $,
• $f|_{I_b} \subseteq \mathbb{R} \times \mathbb{R}$,
• $f|_{I_c} \subseteq \mathbb{Q} \times \mathbb{Q}$,

Now all that is left is to simplify these products and find three injections into $I_a$, $I_b$ and $I_c$.

Solution:

First, $\mathbb{R} \times \mathbb{R}$ has the same cardinality as $\mathbb{R}$, as well as $\mathbb{Q} \times \mathbb{Q}$ has the same cardinality as $\mathbb{Q}$. This is why composing $f$ with any bijection $\mathbb{R}\times\mathbb{R}\to\mathbb{R}$ gives you lowerbound on the cardinality of $I_a$ and $I_b$, similarly for $I_c$. Second, use injections $g_a(x) = \big[\lfloor x \rfloor,x+1\big]$, $g_b(x) = \big[x,x+1\big]$, $g_c(x) = \big[\lfloor x \rfloor,x+1\big]$.

I hope this helps $\ddot\smile$