I am trying to understand the significance of curly brackets in set theory.

Let $A = \{a,b,c,d\}$.

I understand $A$ is a set that includes the objects $a,b,c,d$. However, what does $\{\{a\}\}$ mean? Why use two curly brackets on either side instead of just $\{a\}$ ?


How does $\{a\}$ differ from $\{\{a\}\}$? Well, Omnomnomnom and vadim123 have pointed out the key differences, but perhaps a small elaboration will help.

Is $a\in\{a\}$? Yes. Is $a\in\{\{a\}\}$? Well, $a\neq\{a\}$ so the answer is a resounding no. Ah, but do you know what it means for a set to be a subset of another set? If $A$ is a subset of $B$, then this is often denoted by $A\subseteq B$ and may be thought of as "if $x$ is $A$, then $x$ is in $B$"; symbolically, $A\subseteq B$ may be represented as $x\in A\to x\in B$. This illustrates some key points:

  • $a\in \{a\}$
  • $\{a\}\in\{\{a\}\}$
  • $\color{blue}{\{a\}\subseteq \{a\}}$
  • $\color{blue}{\{a\}\not\in\{a\}}$

Pay special attention to the last two points highlighted in blue.

  • $\begingroup$ Also of note is that in ZFC a can not possibly even be possibly be {a}. $\endgroup$ – PyRulez May 6 '15 at 9:11

$\{\{a\}\}$ means the set containing the set containing $a$.

Think of sets like bags. $a$ is an apple. $\{a\}$ is an apple in a bag. $\{\{a\}\}$ is an apple in a bag, that is in another bag.


The set $\{\{a\}\}$ is the set whose only object is the set $\{a\}$. $\{a\}$, in turn, is a set whose only object is $a$.

It might help to consider the set $$ \{a,\{a\},\{b\}\} $$ whose objects are $a$, the set $\{a\}$, and the set $\{b\}$. Note that this set is quite different from $\{a,b\}$.

  • $\begingroup$ Ah, ok gotcha. So then in that case, if A = {a,b,c,d}, then {{a}} would not be a subset of P(A), correct? $\endgroup$ – Omar N May 5 '15 at 18:31
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    $\begingroup$ @OmarN $\{\{a\}\}$ is a subset of $P(A)$ because each of its elements is an element of $P(A)$ (that is, each element is a subset of $A$). Indeed the only element $\{a\}$ is a subset of $\{a,b,c,d\}$ because each element of the former(there is only $a$) is an element of the latter. $\endgroup$ – Hagen von Eitzen May 5 '15 at 18:34
  • $\begingroup$ Actually, $\{\{a\}\}$ would be a subset of $P(A)$ since its only element is also an element of $P(A)$. $\endgroup$ – Omnomnomnom May 5 '15 at 18:37
  • $\begingroup$ @HagenvonEitzen thanks for the comment; I had misread the question $\endgroup$ – Omnomnomnom May 5 '15 at 18:37
  • $\begingroup$ Thank you for the clarification. Would it also follow to say that $\emptyset$ is not a subset of P(A), but {$\emptyset$} is a subset of P(A)? $\endgroup$ – Omar N May 5 '15 at 18:39

If $\{a\}$ is the set whose only element is $a$; denote that set by $x$, then $\{x\}$ is the set whose unique element is $x$, so its unique element is $\{a\}$, which in other words mean that it is the set $\{\{a\}\}$.


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