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A textbook on Elementary Set theory shows an example which says that:-

given a set $A=\{a,\{b\},\{c\}\}$ find out if the statements are correct - $$\\ a).\ \ \ \{a,\{b\}\}\in P(A) \\ b). \ \ \ \{a,\{b\}\} \subseteq P(A)$$

Definition : Power set of any set $A$ is the set of all subsets of $A$, including the empty set and $A$ itself, i.e. $$\\ P(A) = \{\phi, \{a\},\{\{b\}\}, \{\{c\}\}, {\{a,\{b\}\}}, {\{a,\{c\}}\}, \{\{b\},\{c\}\}, \{a,\{\{b\},\{c\}\}\}\}$$

From the definition it is clear that the option $a$ is obviously true but the textbook claims that the second option $b$ is incorrect and here I am stuck.

According to the definition $a, \{b\}\notin P(A)$, since $P(A)$ contains the set that consists of the elements $a, \{b\}$ and the set $\{a, \{b\}\}$ that contains the elements $a, \{b\}$ should be the subset of $P(A)$ and that is why the statement $\{a,\{b\}\}\subseteq P(A)$, should be true. But somehow it is incorrect.

Would anyone kindly like to mention where am I wrong? Any help is highly appreciated.

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  • $\begingroup$ I have found a similar video tutorial youtube.com/watch?v=y8arakl4WNM which also supports the claim of the book. $\endgroup$ – vbm Aug 3 '19 at 4:08
  • $\begingroup$ What do you mean when you write $a, \{b\}\nsubseteq P(A)$? $\endgroup$ – Taroccoesbrocco Aug 3 '19 at 4:11
  • $\begingroup$ I mean as elements they are not in $P(A)$ $\endgroup$ – vbm Aug 3 '19 at 4:14
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    $\begingroup$ So, you should write $a, \{b\} \notin P(A)$, and not: $a, \{b\}\nsubseteq P(A)$ . $\endgroup$ – Taroccoesbrocco Aug 3 '19 at 4:15
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Your textbook is right, $\{a,\{b\}\} \not\subseteq \mathcal{P}(A)$. Indeed, $a$ is an element of $\{a,\{b\}\}$ but it is not an element of $\mathcal{P}(A)$ (see the list of the elements of $\mathcal{P}(A)$ that you correctly gave), thus it is not true that every element of $\{a,\{b\}\}$ is an element of $\mathcal{P}(A)$ (in general, given two sets $B$ and $C$, $B \subseteq C$ means that every element of $B$ is an element of $C$).

So, when you say that "the set $\{𝑎,\{𝑏\}\}$ [...] should be the subset of $\mathcal{𝑃}(𝐴)$", you are wrong. You can say that $\{𝑎,\{𝑏\}\}$ is a subset of $A$.

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  • $\begingroup$ I got it. Thank you for your kind attention to my problem. $\endgroup$ – vbm Aug 3 '19 at 4:24

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