I'm a bit struggling in finding the cardinal size of the set $$|P([0,1])\setminus P((0,1))|$$ Where $P$ is the notation for "power set".
I was thinking about starting with the fact that $(0,1)\cup\{0,1\}=[0,1]$. From there, I could point out that for each $S\in P((0,1))$, there are $S\cup\{0\}, S\cup\{1\},S\cup\{1,0\} \in P([0,1])$. Intuitively, this means that $B=P([0,1])$ is bigger than $A=P((0,1))$ times 4 (for each in A, there is that one in B, plus 3 more). I have yet to study the arithmetic of set theory, so claiming that $|P([0,1])\setminus P((0,1))| = 2^\aleph \times 3 = 2^\aleph $ is out scope regarding my calculation.
I don't quite know how to move on from here. Moreover, mathematically speaking, I'm a bit confused on how to properly write these claims. As I said, intuitively it seems logical to me that B is 4 times larger than A, but I want to be precise in me writings.
Any suggestions are greatly appreciated.
Let $A=P([0,1])$, $B=P((0,1))$, $C=A\setminus B$. Then we have the following injective maps: $$ C\to A, s\mapsto s$$ $$ A\to B, s\mapsto \{\,\tfrac{x+1}3\mid x\in s\,\}$$ $$B\to C, s\mapsto \{\,x\cup \{0\} \mid x\in s\,\}$$ From these we conclude that they all have the same cardinality.