Is $\{\varnothing\}\in\{\varnothing,\{\varnothing\}\}$ true?

Can we use the same explanation as in $\{\varnothing\}\subseteq\{\varnothing,\{\varnothing\}\}$? That is, $\{\varnothing,\{\varnothing\}\}$ has two elements $\varnothing$ and $\{\varnothing\}$. The right set $\{\varnothing\}$ is its subset because it contains only one of them. Therefore, true.

With the only difference being that we replace the word subset and instead say that the right set $\{\varnothing\}$ is an element of $\{\varnothing,\{\varnothing\}\}$, therefore true?

Can we say the same thing for $\{\varnothing\}\in\{\varnothing,\{\{\varnothing\}\}\}$, and $\{\{\varnothing\}\}\in\{\varnothing,\{\varnothing\}\}$?

  • $\begingroup$ I'm wondering if we can invoke the axiom of powerset to answer this question in a more generalized manner? {∅, {∅}} is ℘({∅}) and therefore {∅} ∈ {∅, {∅}} is true. Similarly, {∅, {{∅}}} ≠ ℘({∅}) and {∅, {∅}} ≠ ℘({{∅}})? I am not well versed with en.wikipedia.org/wiki/Set_theory#Axiomatic_set_theory to know if a proof through this route exists. $\endgroup$ – Dynamic Stardust Oct 2 '17 at 15:59
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    $\begingroup$ For my own sanity-- er, understanding, ∅ means the empty set, right? i.e. it's equivalent to $\{ \}$. $\{∅\}$ is a set with one element, which is the empty set? $\endgroup$ – Nic Hartley Oct 2 '17 at 18:01
  • $\begingroup$ @QPaysTaxes That's correct. $\endgroup$ – Stefan Mesken Oct 2 '17 at 21:59

Is {∅} ∈ {∅, {∅}} true?


Can we use the same explanation as in {∅} ⊆ {∅, {∅}}?


  • You prove that $$\{\emptyset\}\in \{\emptyset, \{\emptyset\}\}$$ by proving that $\{\emptyset\}$ is an element of $\{\emptyset, \{\emptyset\}\}$. This is fairly obvious, and is the same as proving that $0$ is an element of $\{0,1\}$.
  • You prove that $$\{\emptyset\}\subseteq \{\emptyset, \{\emptyset\}\}$$ by showing that every element of $\{\emptyset\}$ is also an element of $\{\emptyset, \{\emptyset\}\}$. Since $\{\emptyset\}$ has only one element, $\emptyset$, this means you need to prove that $\emptyset$ is an element of $\{\emptyset,\{\emptyset\}\}$.

Can we say the same thing for {∅} ∈ {∅, {{∅}}}, and {{∅}} ∈ {∅, {∅}}?



$A$ is a subset of $B$ (denoted as $A\subseteq B$) if and only if for every $a\in A$, it is true that $a\in B$.

This means that:

  • $\{\emptyset\}\in \{\emptyset, \{\emptyset\}\}$ is true the same way $1\in \{0,1\}$ is true.
  • $\{\emptyset\} \subset \{\emptyset,\{\emptyset\}\}$ is true the same way $\{a\}\subset \{a,b\}$ is true.
  • $\{\{\emptyset\}\} \subset \{\emptyset,\{\emptyset\}\}$ is true, because every element of $\{\{\emptyset\}\}$ (there is only one) is an element of $\{\emptyset,\{\emptyset\}\}$.
  • $\{\{\emptyset\}\}\in \{\emptyset, \{\emptyset\}\}$ is false, because neither of the two elements in $\{\emptyset,\{\emptyset\}\}$ is equal to $\{\{\emptyset\}\}$.

The easiest way to resolve some of your queries is by introducing convenient notation so as not to get lost in the curly braces, such as $0=\emptyset$, $1=\{\emptyset\}$, $2=\{\emptyset,\{\emptyset\}\}=\{0,1\}$, etc. For example, the answer to your last question is negative because it is not true that $\{1\}\in\{0,1\}$, only that $\{1\}\subset\{0,1\}$.

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    $\begingroup$ That is incorrect: If you define 1={∅},the expression is 1∈{0,1} and not {1}∈{0,1}, which is true. $\endgroup$ – Darkhogg Oct 2 '17 at 17:44
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    $\begingroup$ @Darkhogg I think the "last question" mentioned in this answer was $\{\{\emptyset\}\}\in \{\emptyset, \{\emptyset\}\},$ which indeed is false and in which indeed $\{\{\emptyset\}\}=\{1\}$ according to the notation of this answer. I'm not convinced, however, that the extra notation is really going to help someone who isn't yet able to see that $\{\{\emptyset\}\}\not\in\{\emptyset, \{\emptyset\}\}$ in the original notation of the question. $\endgroup$ – David K Oct 2 '17 at 17:59
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    $\begingroup$ That convenient notation may quickly become confusing too when you start asking if $1 \subseteq 2$ is true... $\endgroup$ – Federico Poloni Oct 2 '17 at 20:45
  • $\begingroup$ @FedericoPoloni, obviously if you wish to study the set inclusion relations you have to restore some of the curly braces. It remains true that some of the questions are more transparent using the abbreviated notation. $\endgroup$ – Mikhail Katz Oct 3 '17 at 8:57

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