Is $\emptyset$ equal to $\{\emptyset\}$? I know an emptyset contains no elements. So I feel like they would be equal. Can someone explain how they wouldn't be?
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3The second set contains an element. – Matt Samuel Aug 30 '15 at 00:06
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2See http://math.stackexchange.com/a/65600/120540 – pjs36 Aug 30 '15 at 00:07
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1Two sets are equal if and only if they have the same elements. The former has no elements; the latter has $\varnothing$ as an element. Thus, they are different. – Akiva Weinberger Aug 30 '15 at 05:27
3 Answers
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No they are not equal. $\emptyset$ contains no elements while $\{\emptyset\}$ contains $\emptyset$.
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$\varnothing$ is the same as $\{\}$. It is a set with no members.
$\{\varnothing\}$, on the other hand, is a set with one member.
And $\{\varnothing,\{\varnothing\}\}$ is a set with two members.
You can form an infinite sequence in which each term is $\{\varnothing, \cdots\cdots\}$, where "$\cdots\cdots$" is the previous term. These are the finite von Neumann ordinals.
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The last paragraph is not really true. The constuction happens to produce the von Neumann representations of $1$ and $2$, but in the next step it goes wrong -- it produces ${\varnothing,{\varnothing,{\varnothing}}}$ rather than the correct ${\varnothing,{\varnothing},{\varnothing,{\varnothing}}}$ for $3$. – hmakholm left over Monica Sep 28 '16 at 12:54
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This is a naive answer. To convince yourself that the two sets are not equal see what happens when you try to prove they are equal.
As usual to prove the two sets are equal you need to prove each of them contains each other.
One implication: $x \in \emptyset \implies x \in \{\emptyset\}$ is true vacuously.
But consider the other. If $x \in \{\emptyset\}$ then $x = \emptyset$. Now $x$ has a concrete presence. It is an entity. Therefore $x \in \emptyset$ is false.
Ishfaaq
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