As the title says.

I've already proven a bijection between $\mathbb{R}\times\mathbb{R}$ and $\mathbb{R}$ by construction the following bijections $$\mathbb{R}\times\mathbb{R} \sim 2^{\mathbb{N}}\times 2^{\mathbb{N}} \sim 2^{\mathbb{N}\sqcup\mathbb{N}} \sim 2^\mathbb{N} \sim \mathbb{R}$$ Now, let $S\in \mathcal{P}(\mathbb{R}\times\mathbb{R})$, then $S \subseteq \mathbb{R}\times\mathbb{R}$. By the previous bijection, I can map each point in $S$ bijectively to a point in $\mathbb{R}$, the collection of which I call $S'$. Clearly, $S'\subseteq \mathbb{R}$ and so $S' \in \mathcal{P}(\mathbb{R})$.

Because the actual mapping uses an established bijection, this must establish a bijection $\mathcal{P}(\mathbb{R}\times\mathbb{R}) \sim \mathcal{P}(\mathbb{R})$.

Is this a correct argument? If not, which details have I overlooked?


Yes, your argument is correct. If $f:A\rightarrow B$ is a bijection then, $F:\mathcal{P}(A)\rightarrow\mathcal{P}(B)$ such that $F(X)=f[X]$ is a bijection too.

  • 1
    $\begingroup$ Great, thank you. $\endgroup$ – Auclair May 7 '18 at 1:20

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