I know that "THE" Universal set does not exists, that all sets exists relative to other sets. That because of the Russel Paradox we cannot naively create a set only by arbitrary properties of its elements, we must take the elements that sattisfies the properties from another well known set.
The thing is that I'm proving something and at some point I got to something that looks like $A\cap(B\cup B^c)$.
Can I say that $(B\cup B^c) = \Omega$ with $\Omega$ the universal set? ($B^c$ is the complement of $B$)
If that is the case then I can simply do
\begin{align}
&A\cap(B\cup B^c)\\
\iff& A\cap \Omega\\
\iff& A
\end{align}
But I'm afraid from the many proofs that the Universal set does not exist.
Nonetheless I believe that what does not exist is the Absolute Universal set. What I mean is that for a set (or class?) of sets, there is a Relative Universal Set that contains all possible sets derived from them…
I just want to know how to justify the above manipulation…
P.S. I'm proving that the cardinality of a Power set is 2 to the power of its set cardinality.
Edit:
I'm in a part of the proof where I want to prove:
if $\ a \in x \ \text{and } y = x \backslash \{a\} \implies x = y \cup \{a\}$
\begin{align*}
& \ y \cup \{a\} \\
\iff \ &(x \backslash \{a\}) \cup \{a\} \\
\iff \ &(x \cap \{a\}^c)\cup\{a\} \ \text{<— Here I'm using a definition for set difference} \\
\iff \ &(x \cup \{a\}) \cap (\{a\}^c \cup \{a\}) \\
\iff \ &(x \cup \{a\}) \cap \Omega \ \text{ < —–I used $\Omega$ as Universal set}\\
\iff \ &x \cup \{a\}\\
\iff \ & x \ ( \text{because } a \in x \iff \{a\} \subseteq x )\\
\\
\therefore\ & x = y \cup \{a\}
\end{align*}
Best Answer
As people have commented, one problem is that complementation is not actually a set operation, only relative complement (i.e. set difference). Otherwise we'd need to have $\emptyset^c=\Omega,$ but $\Omega$ is not a set.
So the definition of $x\setminus y$ is not $x\cap y^c$ since $y^c$ isn't a thing. The definition is $x\setminus y:=\{z\in x: z\notin y\}.$
To show $x=(x\setminus\{a\})\cup\{a\},$ show that $x\subseteq(x\setminus\{a\})\cup\{a\}$ and $(x\setminus\{a\})\cup\{a\}\subseteq x.$
For the first direction, assume $z\in x.$ If $z\in \{a\}$, then $z\in(x\setminus\{a\})\cup\{a\}$ by the defintion of $\cup$. If not, then $z\in x$ and $z\notin\{a\},$ which by definition means $z\in x\setminus\{a\},$ so again, $z\in(x\setminus\{a\})\cup\{a\}.$
For the other direction, if $z\in(x\setminus\{a\})\cup\{a\}$ then either $z\in x\setminus\{a\},$ in which case $z\in x,$ or $z\in \{a\},$ in which case $z= a\in x$.
Note you can still use absolute complementation in situations where you are dealing only with a collection of subsets of a given set (which can be called a "universal set" for the given context and “absolute complementation” just means difference with this set). I suspect that may be the case here, but I'm not sure where this fits in to the proof of $|P(A)|= 2^{|A|},$ which I would recommend proving by establishing a one-to-one correspondence between a subset $B\subseteq A$ and its indicator function $1_B: A\to 2$ where $1_B(x) = 1$ if $x\in B,$ else $0,$ with an inverse corresponence of $f:A\to 2$ to the subset $B_f = \{x\in A: f(x)=1\}.$