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The powerset of $\{a\} = \{\{a\},\emptyset\}$ is different from the powerset of $\{a,\emptyset\} = \{\{a\},\emptyset,\{\emptyset\}\}$. Why is that?

I thought the null set was implicitly within ever single set, because the null set is a subset of all of them? Hence, the two sets I gave should be the same and should have the exact same power set.

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  • $\begingroup$ The empty set is an element of $\{a,\emptyset\}$. That’s not the case for all sets. $\endgroup$ – Steve Kass Feb 7 '18 at 2:54
  • $\begingroup$ Ah I see, I've confused subset with element. Thanks $\endgroup$ – Goldname Feb 7 '18 at 3:05
  • $\begingroup$ Assuming $a\ne\emptyset,$ the set $\{a,\emptyset\}$ has $2$ elements, so its power set will have $4$ elements, not $3.$ The missing element of the power set is $\{a,\emptyset\}$ itself. $\endgroup$ – bof Feb 7 '18 at 3:17
  • $\begingroup$ The power-set of $\{a.\phi\}$ is $\{\;\phi, \{\phi\},\{a\},\{a,\phi\}\;\}.$.. If $a\ne \phi$ then it has $4$ members. $\endgroup$ – DanielWainfleet Feb 7 '18 at 4:37
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I think you have notation confusion here: $\emptyset = \{\}$, so everytime you see that symbol, you should add an empty set, not an empty place, in your understanding of the set.

So $\{a\}$ implies empty space, not an empty set.

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It is the case that $\emptyset \subseteq X$ for every set $X$, but it not the case that $\emptyset \in X$ for every $X$.

In particular, $\{a,\emptyset\} \neq \{a\}$ if $a\neq \emptyset$, as $\emptyset$ is an element of the first, but not the second. Thus, their power sets won't be equal.

Perhaps the issue might be that the word "contains" is overloaded in the English language, and is ultimately ambiguous when used in set theory, sometimes referring to the membership relation, and sometimes to the subset relation.

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  • $\begingroup$ Good point about the word “contains.” To avoid ambiguity, it helps to say things like “$S$ contains $A$ as an element” or “$S$ contains $A$ as a subset.” Or avoid the word “contains” and say “$A$ is a subset of $S$.” or “$A$ is an element of $S$.” $\endgroup$ – Steve Kass Feb 7 '18 at 17:17

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