Cardinality of the set of all subsets of $E$ equipotent to $E$

I'm attempting to confirm the adhering to declaration (a workout in Bourbaki is Set Theory ):

If $E$ is a boundless set, the set of parts of $E$ which are equipotent to $E$ is equipotent to $\mathfrak{P}(E)$.

As a tip, there is a reference to a suggestion of guide, which reviews:

Every boundless set $X$ has a dividing $(X_\iota)_{\iota\in I}$ created of countably boundless collections, the index set $I$ being equipotent to $X$.

I do not have any kind of suggestion just how that suggestion could aid.

If $E$ is countable, after that a part of $E$ is equipotent to $E$ iff it is boundless. Yet the set of all limited parts of $E$ is equipotent to $E$. So its enhance in $\mathfrak{P}(E)$ needs to be equipotent to $\mathfrak{P}(E)$ by Cantor is theorem. Therefore the declaration holds true if $E$ is countable. However, I do not see a means to generalise this argument to vast $E$.

I would certainly rejoice for a tiny tip to get me going.

2022-06-07 14:38:38
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Answers: 2

Using the axiom of selection, every boundless set $X$ can be separated right into 2 disjoint collections $X_0\sqcup X_1$, both of which are equinumerous with $X$. (Just well - order $X$, and also take every various other factor in the list.)

Currently, take into consideration ready of the kind $X_0\cup A$ for any kind of $A\subset X_1$. There are $2^X$ several such $A$ and also therefore $2^X$ several such collections, and also each is equinumerous with the initial set $X$. So we've obtained $2^X$ several collections as wanted, and also there can not be greater than this, so this is the specific number.

Incidently, the mentioned response to this inquiry carries out in reality rely on the axiom of selection, given that it is recognized to be regular with $ZF+\neg AC$ that there are boundless Dedekind limited collections, and also these are not equinumerous with any kind of correct parts of themselves. So for such a boundless set $X$, there would certainly be just one part to which it is equinumerous.

2022-06-07 15:08:05

Another strategy would certainly take advantage of the reality that $\kappa_1 + \kappa_2 = \max(\kappa_1,\kappa_2)$ when $\kappa_1,\kappa_2$ are cardinals at the very least among which is boundless. From this it adheres to that, for any kind of part of $S \subset X$, either $S$ or its enhance has cardinality $|X|$. Given that even more uncomplicated cardinal math reveals there are as several corresponding sets of parts as there are parts of $X$, we get the outcome.

Incidentally, JDH is strategy can permit you to show the (a little) more powerful declaration that there are $2^X$ collections $S \subset X$ with both of $S$ and also $X - S$ equipotent to $X$. Merely write $X$ as a disjoint union $X = X_0 \cup X_1 \cup X_2$ with all components equipotent to $X$ and also take into consideration parts of the kind $X_0 \cup A$ where $A \subset X_1$.

An intriguing adhere to - up inquiry could be to see if you can generate $2^X$ collections $S \subset X$ to make sure that $|S|=|X|$, yet $|X-S| < |X|$.

2022-06-07 14:54:59