Powerset of infinite set minus original set equivalent to the poweset I had this idea which I am fairly certain is true but I am not quite certain nor how to go about proving it. Let $\alpha$ be a given infinite set of some cardinality, that is $card(\alpha)=a$ and let $set(\alpha)$ be the set of all singleton sets of $\alpha$. That is if $z\in set(\alpha)$ then $z=\{q\}$ for some $q\in\alpha$
Now my idea is about the cardinality of the difference between the powerset of $\alpha$ and the singleton sets, my idea is that
$$|\mathcal{P}(\alpha)-set(\alpha)|=|\mathcal{P}(\alpha)|$$
I may be wrong but it feels intuitive and I am not quite certain how to go about it.
I know all I need is a bijection between the two given set but I am not certain how to construct it in a rigorous manner that would pass the traditional necessities. I do know on the other hand that $|set(\alpha)|=|\alpha|$ and $|\mathcal{P}(\alpha)|>|\alpha|$ which is what leads me to believe it must be true.
 A: HINT: Recall that for infinite cardinals $\kappa,\lambda$, the following holds: $\kappa+\lambda=\max\{\kappa,\lambda\}$. And note that $\mathcal P(\alpha)=(\mathcal P(\alpha)-\mathrm{set}(\alpha))\cup\mathrm{set}(\alpha)$ is a disjoint union.
(You may wonder, why do we have to use the axiom of choice, in the form of $\kappa+\lambda=\max\{\kappa,\lambda\}$. The answer is that it is consistent that there is a counterexample, where $\mathcal P(\alpha)$ is an infinite Dedekind-finite set, therefore removing elements decreases the cardinality. It suffices, for this proof, however, that for every infinite cardinal $\mathfrak m$, $\frak m+m=m$.)
A: HINT: Clearly $|\wp(\alpha)\setminus\operatorname{set}(\alpha)|\le|\wp(\alpha)|$, so you just need an injection from $\wp(\alpha)$ to $\wp(\alpha)\setminus\operatorname{set}(\alpha)$. One approach is to let $p$ be any object not in $\alpha$, and let $\beta=\alpha\cup\{p\}$.


*

*Show that $|\alpha|=|\beta|$ and hence that $|\wp(\alpha)\setminus\operatorname{set}(\alpha)|=|\wp(\beta)\setminus\operatorname{set}(\beta)|$.

*Now consider the map from $\wp(\alpha)$ to $\wp(\beta)\setminus\operatorname{set}(\beta)$ that takes $\varnothing$ to itself and takes every non-empty $S\subseteq\alpha$ to $S\cup\{p\}$.
