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When a set has an empty set as an element, e.g.$ \{\emptyset, a, b \}$. What is the powerset?

Is it: $$ \{ \emptyset, \{ \emptyset \}, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$

Or

$$ \{ \emptyset, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$

Or $$ \{ \{\emptyset\}, \{a\}, \{b\}, \{\emptyset, a\} \{\emptyset, b\}, \{a, b\}, \{\emptyset, a, b\}\}$$

The confusion arises for me because, the powerset of every non-empty set has an empty set. Well the original set already has the empty set. So we don't need a subset with an empty set.

Somehow, the first one seems correct. Yet, I can't seem to accept it.

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    $\begingroup$ The first one: $\;\emptyset\;$ is one of the elements of the given set, besides being a subset of it. $\endgroup$ – DonAntonio Oct 12 '16 at 12:39
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    $\begingroup$ Let $c$ denote $\varnothing$. What is the power set of $\{a,b,c\}$? Now write $\varnothing$ instead of $c$ again. $\endgroup$ – Asaf Karagila Oct 12 '16 at 13:16
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The first one is correct.

This is because $\emptyset$ and $\{\emptyset\}$ are different. The first is an empty set whereas the second is a set whose only element is the empty set.

Both are subsets of the given set. This is because the $\emptyset$ is the subset of every set, and as it happens to be an element of the given set, the set containing it as its element is also its subset.

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If a set $A$ is such that $\emptyset\in A$, its power set must necessarily contain these two sets:

  • $\emptyset$ (like all other power sets), corresponding to selecting nothing from $A$ (not even $\emptyset$, which is something)
  • $\{\emptyset\}$, corresponding to selecting $\emptyset$ only

Therefore only the first of your proposed answers is correct, as you think.

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Your suggestions differ by having $\emptyset$ and/or $\{\emptyset\}$ included or not.

  • We have $\emptyset\in\mathcal P(X)$ because $\emptyset\subseteq X$ (which would hold for any other $X$ as well)
  • We have $\{\emptyset\}\in\mathcal P(X)$ because $\{\emptyset\}\subseteq X$ (which is the case because $\emptyset\in X$ in this specific problem)

Therefore, your first variant is correct (and the other two are incorrect because $\emptyset\ne\{\emptyset\}$).

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    $\begingroup$ Your second bullet is strangely phrased to me; we have $\{\emptyset\}\in\mathcal{P}(X)$ simply because $\emptyset\in X$ in this specific problem. The fact that $\emptyset\subseteq X$, which is true for every set $X$, has nothing to do with it? $\endgroup$ – Inactive - avoiding CoC Oct 12 '16 at 15:47

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