Prove that $|\mathcal{F}|=|\mathbb{R}|$.

Question

Let $\mathcal{F} = \Big\{ \mathcal{I}_{a,b}\times\{x\} : a,b,x\in\mathbb{R}\Big\}$, where $\mathcal{I}_{a,b}:=\{\zeta\in\mathbb{R} : a < \zeta <b\}$. Im trying to prove that $|\mathcal{F}|=|\mathbb{R}|$ (that can be denoted by $\mathfrak{c}).$

Please, can someone help with a clue our something. Thanks.

Answer 1

Consider $f: \mathcal{F} \to \mathbb{R}^3$ defined by \begin{align*} \mathcal{I}_{a,b} \times \{x\} \mapsto (a,b,x). \end{align*} It should be easy to show that this is injective, so that $|\mathcal{F}|\leq |\mathbb{R}^3|$. Also notice that for any fixed $a < b$, $\mathcal{A} = \{\mathcal{I}_{a,b} \times \{x\} \,:\, x\in \mathbb{R}\} \subseteq \mathcal{F}$. But there is a bijection from $\mathcal{A}$ to $\mathbb{R}$ via \begin{align*} \mathcal{I}_{a,b} \times \{x\} \mapsto x \end{align*} so $|\mathbb{R}| \leq |\mathcal{F}| \leq |\mathbb{R}^3|$. Now it comes down to arguing that $|\mathbb{R}| = |\mathbb{R}^3|$.

Answer 2

You might try to show that $|\mathcal{F}| = |\mathbb{R}^3|$. Then you need only show that $|\mathbb{R}| = |\mathbb{R}^3|$ (or, more generally, $\mathbb{R}^n$).