Let $U$ be the universal set, $A,B\subseteq U$.

Corrected due to my mistakes detected in comments. This works only for finite sets.

Prove these statements:

$\mathcal P(A\cup B)\;\supseteq\; \mathcal P(A)\cup \mathcal P(B)$

$\mathcal P(A\cap B)\;=\; \mathcal P(A)\cap \mathcal P(B)$

explanation by cardinality: $$|\mathcal P(A\cup B)|\;=\;2\;^{|A|\;+\;|B|\;-\;|A\cap B|}$$ $$\;\;\;\;|\mathcal P(A)\cup \mathcal P(B)|\;=\;2\;^{|A|}+2\;^{|B|}-2\;^{|A\cap B|}$$ $$|\mathcal P(A\cup B)|\;\geq\;|\mathcal P(A)\cup \mathcal P(B)|$$ $$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;|\mathcal P(A\cap B)|\;=\;|\mathcal P(A)\cap \mathcal P(B)|\;=\;2\;^{|A\cap B|}$$

Is this correct?

  • 3
    $\begingroup$ The fact that two sets have the same cardinality does not mean that the two sets are the same set; consider e.g. $\mathbb N, \mathbb Z$. $\endgroup$ – Mauro ALLEGRANZA Nov 12 '19 at 8:20
  • $\begingroup$ @MauroALLEGRANZA, these are the power sets of union (intersection) of sets and union (intersection) of power sets of different sets . $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 8:25
  • $\begingroup$ @MauroALLEGRANZA this works with $\mathbb N$ and $\mathbb Z$ under the argument . $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 8:26
  • $\begingroup$ So, since $|\Bbb{N}| = |\Bbb{Z}|$, you would conclude that $\mathcal P(\Bbb{N}) = \mathcal P(\Bbb{Z})$? $\endgroup$ – J.-E. Pin Nov 12 '19 at 8:37
  • $\begingroup$ @J.-E.Pin, no, I would not, but there are both sets under operations on both sides, therefore, the question is, bearing in mind there are always common elements, which of the sets contains more than those common. I didn't compare power sets of different sets of the same cardinality. $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 8:46

Your proof isn't correct : if $X$ is an infinite set, you can always find a proper subset of $X$ which has the same cardinality as $X$.

All you need to use is the definition of the power set : $$X \in \mathcal{P}(S) \Longleftrightarrow X \subseteq S$$

proof of $\mathcal P(A\cup B)\;\supseteq\; \mathcal P(A)\cup \mathcal P(B)$ :

Let $X \in \mathcal P(A)\cup \mathcal P(B)$,
If $X \in > \mathcal{P}(A)$, then $X \subseteq A$, hence $X \subseteq A \cup B$ whence $X \in \mathcal{P}(A\cup B)$.
Likewise, if $X\in > \mathcal{P}(B)$, then $X \in \mathcal{P}(A\cup B)$.

proof of $\mathcal P(A\cap B)\;=\; \mathcal P(A)\cap \mathcal P(B)$:

  • "$\subseteq$" : Let $X\in \mathcal{P}(A\cap B)$ ie $X \subseteq A \cap B$. Then $X\subseteq A$ (ie $X \in \mathcal{P}(A)$) and $X\subseteq B$ (ie $X \in \mathcal{P}(B)$). Hence, $X \in \mathcal{P}(A) \cap \mathcal{P}(B)$.
  • "$\supseteq$" : Let $X \in \mathcal{P}(A) \cap \mathcal{P}(B)$, that is to say $X \in \mathcal{P}(A)$ (ie $X \subseteq A$) and $X \in \mathcal{P}(B)$ (ie $X \subseteq B$). We hence have $X\subseteq A \cap B$, in other words $X \in \mathcal{P}(A\cap B)$.
  • $\begingroup$ But if I have a universal set, it must have a higher cardinality than $A, B$. If I take infinite sets, then, as far as I'm concerned, $\mathbb R$ should be the universal set. But thanks for the correction. I wasn't precise and I made a mistake. Better now, than on exam. $\endgroup$ – Praskovya2.718281828 Nov 12 '19 at 9:18

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