Easy proof of $\mathcal{P}(\mathbb{Q})$ is uncountable [Big list]

Question

I'm looking for a easy proof of uncountability of $\mathcal P(\mathbb Q)$. I'll contribute with this:

Let $\mathcal{P}(A)$ denote the power set of $A$, since $\mathbb{N}\subset\mathbb{Q}\Rightarrow\mathcal{P}(\mathbb{N})\subset\mathcal{P}(\mathbb Q)$ let be $f:\mathcal{P}(\mathbb{N})\to\mathcal{P}(\mathbb Q)$ such $A\mapsto A$. then $f$ is injective and since $\mathcal{P}(\mathbb{N})$ is uncountable, then $\mathcal{P}(\mathbb Q)$ is uncountable

Answer 1

Define $f : \mathbb{R} \to \mathcal{P}(\mathbb{Q})$ as $f(x) = (-\infty, x) \cap \mathbb{Q}$. It is injective, so $|\mathbb{R}| \leqslant |\mathcal{P}(\mathbb{Q})|.$

Answer 2

For each real number $x$ pick a strictly increasing sequence $x_n$ of rationals converging to $x$.

Then $$f(x)=\{ x_n| n \}$$ is a one to one function from $\mathbb R$ to $P(Q)$. As $\mathbb R$ is uncountable....

Answer 3

Follows directly by Cantor's Theorem.

(Note that you asked a proof that $\mathcal{P}(\mathbb{Q})$ is uncountable, not that its cardinality is the same of $\mathbb{R}$)

Answer 4

By transfinite induction, I will show that every countable ordinal $\alpha$ there is a set of rationals $S_\alpha\subseteq (0,1)$ which has order type $\alpha$. Then we will have $\omega_1$ clearly distinct sets of rationals.

For $S_0$ we can (indeed, we have to) take an empty set. For set $S_{\alpha+1}$ we can take $\frac{1}{2}S_\alpha\cup\{\frac{1}{2}\}$, where $\frac{1}{2}S_\alpha=\{\frac{1}{2}x:x\in S_\alpha\}$.

Now if $\alpha$ is sum of countably many ordinals $\beta_i$ (some of which could be 0) then we could take $S_\alpha=\frac{1}{2}S_{\beta_1}\cup\frac{1}{4}S_{\beta_2}+\frac{1}{2}\cup\frac{1}{8}S_{\beta_3}+\frac{1}{4}\cup...$ where $aS_\beta+b=\{ax+b:x\in S_\beta\}$.