# How do I prove that there exists a bijection from power set of $D$ to $C'\subseteq C.$

I was proving the Moore's Plane $$(X,\mathscr T)$$ is not normal. Using the theorem. Let $$C=\{(x,y)\in X:y=0\}$$ be a closed and relatively discrete subset of $$X$$. Let $$D=\{(x,y)\in X:x\in \mathbb Q \text{ and } y\in \mathbb Q\}$$ is a dense subset of $$X$$. How do I prove that there exists a bijection from power set of $$D$$ to $$C'\subseteq C.$$

The power set of $$D$$ has size continuum ($$2^{\aleph_0}$$) and so does $$C \simeq \mathbb{R}$$ so they're even bijectively related, they have the same size.

• How do I find a suitable map? – Unknown x Feb 12 at 14:59
• @Unknownx it should be standard knowledge that these sets have a bijection between them. continuum is continuum. It should have been covered in the past during set theory. otherwise search for bijection powerset of countable set and reals, on this site or on Google. There'll be many hits I think. – Henno Brandsma Feb 12 at 15:02
• okay. Thank you very much. :) – Unknown x Feb 12 at 15:04

The set $$D$$ is countable, and so we can think of it as the set $$\mathbb{N}$$ of all natural numbers. Any subset of $$\mathbb{N}$$ can be associated uniquely with a sequence of $$0$$s and $$1$$s; Given $$Y\subset \mathbb{N}$$, define $$f_Y$$:

$$f_Y(x)=\begin{cases} 0, & x\notin Y,\\ 1, & x \in Y. \end{cases}$$

It is easy to check that $$Y = Z$$ if and only if $$f_Y = f_Z$$.

The function that takes a subset $$Y$$ of $$\mathbb{N}$$ and sends it to $$(0.f_Y, 0)\in C$$ satisfies your request.