# Sequence of rationals converges to an irrational

I'm doing some set theory and I have three questions. I'm not quite sure how to prove them though.

(a) Show that for any irrational number $x$ there is a sequence of rational numbers $\{r_{i}\}$ such that $r_{i}\rightarrow x$ as $i\rightarrow\infty$

(b) Use (a) to show that there is an injection of the irrationals $\mathbb{R}$ \ $\mathbb{Q}$ into $P(\mathbb{Q})$, the power set of $\mathbb{Q}$.

(c) Show that the cardinality of $P(\mathbb{Q})$ is $c$.

Any help with this will be greatly appreciated!

• Hint: Think of an irrational in decimal notation. Then, look at the sequence of consisting of the first $i$ decimal places. Mar 8, 2015 at 14:21
• for (a), see this Mar 8, 2015 at 14:22
• Sorry. I didn't realise.
– user221122
Mar 11, 2015 at 4:44

a) Given $x\in\mathbb{R}\setminus\mathbb{Q}$, consider $a_n=\frac{\lfloor nx\rfloor}{n}$. For sure, $a_n\in\mathbb{Q}$.
On the other hand, since $r-\lfloor r\rfloor\in[0,1)$,
$$\left|x-a_n\right|\leq\frac{1}{n}\;\Longrightarrow\,\{a_n\}_{n\geq 1}\to x.$$ b) Given a), any irrational number $x$ can be associated with an increasing sequence of rational numbers converging towards $x$.
c) Since $\mathbb{Q}$ and $\mathbb{N}$ have the same cardinality (by Cantor-Bernstein theorem, for instance), $P(\mathbb{Q})$ and $P(\mathbb{N})=2^{\mathbb{N}}$ have the same cardinality, where $2^{\mathbb{N}}$ is the set of functions from $\mathbb{N}$ to $\{0,1\}$. By considering the binary representation of any number in $[0,1)$, then using the Cantor-Bernstein theorem again, we have that the cardinality of $2^{\mathbb{N}}$ equals the cardinality of $[0,1)$, that equals the cardinality of $(0,1)$ (by Hilbert's paradox, for instance), that equals the cardinality of $\mathbb{R}$ since the map $\varphi:(0,1)\to\mathbb{R}$ given by: $$\varphi(x) = -\cot(\pi x)$$ is bijective.