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This question already has an answer here:

Following are two statements from Enderton's book on set theory, I fail to understand that if empty set is a subset of every set then why can't it be a subset of $\{ \{ \emptyset \} \} $

(1) $\emptyset \subseteq A $, ($\emptyset$ is a subset of every set)

(2) $\{ \emptyset \} \nsubseteq \{ \{ \emptyset \} \} $·

$\{ \emptyset \} $ is not a subset of $\{ \{ \emptyset \} \} $ because there is a member of $\{ \emptyset \} $, namely $\emptyset$, that is not a member of $\{ \{ \emptyset \} \} .$

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marked as duplicate by skyking, user91500, Claude Leibovici, Asaf Karagila elementary-set-theory Jan 22 '16 at 8:55

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    $\begingroup$ The empty set is a subset of $\{ \{ \emptyset \} \}$. The empty set is a subset of every set. $\endgroup$ – copper.hat Jan 22 '16 at 7:16
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    $\begingroup$ $\emptyset \neq \{\emptyset\}$. $\endgroup$ – 5xum Jan 22 '16 at 7:27
  • $\begingroup$ $\{\emptyset\}\nsubseteq\{\{\emptyset\}\}$ but $\{\emptyset\}\in\{\{\emptyset\}\}$. $\endgroup$ – marco trevi Jan 22 '16 at 8:24
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$\emptyset$ is a subset of every set indeed. But it is surely not a member of every set. Perhaps you are confusing these two related but different concepts?

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The empty set is indeed a subset of $\{\{\varnothing\}\}$, as it is of every set, but that is not what (2) says. (2) says that the set $\{\varnothing\}$ is a subset of $\{\{\varnothing\}\}$, and $\{\varnothing\}$ clearly not empty: it has one element, the set $\varnothing$.

To see why $\{\varnothing\}\nsubseteq\{\{\varnothing\}\}$, you may find it helpful to give $\varnothing$ a different name temporarily, say $x$. Then the question is whether $\{x\}$ is a subset of $\{\{x\}\}$. What does it mean to say that $\{x\}\subseteq\{\{x\}\}$? It means that every element of $\{x\}$ is also an element of $\{\{x\}\}$. The only element of $\{x\}$ is $x$; is it true that $x\in\{\{x\}\}$? No: the only element of $\{\{x\}\}$ is $\{x\}$, which is not equal to $x$. Thus, $\{x\}\nsubseteq\{\{x\}\}$. This is true no matter what $x$ is. In particular, it’s true when $x=\varnothing$.

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The only member of $\{\phi\}$ is $\phi$. The only member of $\{\{\phi\}\}$ is $\{\phi\}$. It must be clear that $\phi\neq\{\phi\}$

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The empty set is a subset $\{\{\emptyset\}\}$, because is it a subset of every set.

But $\{\emptyset\}$ is not the empty set, but a set containing precisely one element, namely the empty set (it’s non an empty bag, but a bag containing an empty bag). So that $\{\emptyset\} \subsetneq \{\{\emptyset\}\}$ does not contradict the empty set being a subset of every set.

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The empty set is a subset of every set. In particular, $\emptyset$ is a subset of $\{\{\emptyset\}\}.$

On the other hand, $\{\emptyset\}$ is not a subset of $\{\{\emptyset\}\}.$ So what? $\{\emptyset\}\ne\emptyset.$

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The empty set is a subset of every set, irregardless of what Dilbert says.

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    $\begingroup$ Who is Dilbert? $\endgroup$ – 5xum Jan 22 '16 at 7:25
  • $\begingroup$ @5xum: A satirical comic about large corporate management in the US (ironically written by a manager in a large US corporation). $\endgroup$ – copper.hat Jan 22 '16 at 7:30

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