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I've already asked this a part of another question, but thought it'd be easier to elaborate a bit more on my concern.

Let $x$ be a set. What is the difference between $x$ and $\{x\}$? I get that the latter is a set consisting of a single element - namely $x$, but what is the difference?

For example, we can have $x$ to be the set $\{1\}$, then $\{x\}=\{\{1\}\}$. Aren't those $2$ expressions the same?

Another problem are the brackets - when we have a set, do we always have to surround him with brackets, for instance, can we have $x$ to be the set $2$?

Thanks a lot

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    $\begingroup$ Do you understand the distinction between the emptyset and the set containing the emptyset ($\{\}$ versus $\{\{\}\}$)? $\endgroup$
    – Noah Schweber
    Jan 25, 2018 at 13:36
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    $\begingroup$ Yes, the former contains nothing, while the latter contains the empty set $\endgroup$
    – asdf
    Jan 25, 2018 at 13:36
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    $\begingroup$ Les trivial example: $\mathbb N = \{ 0,1,\ldots \}$ has infinite many elements: all the natural numbers, while $\{ \mathbb N \}$ has only one element: the set of all natural numbers. $\endgroup$
    – Mauro ALLEGRANZA
    Jan 25, 2018 at 13:38
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    $\begingroup$ So then why do you think that $\{1\}=\{\{1\}\}$? That would only be true if $1=\{1\}$, and what's your reason for believing that? $\endgroup$
    – Noah Schweber
    Jan 25, 2018 at 13:38
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    $\begingroup$ You may find John von Neumann's construction of the ordinals interesting. We start by defining $0=\{\}$, then $1=\{0\} = \{\{\}\}$, $2 = \{0, 1\}$, $3 = \{0, 1, 2\}$, etc. $\endgroup$
    – PM 2Ring
    Jan 25, 2018 at 13:42

6 Answers 6

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Think of the brackets as a bag you put things in. Then $\{1\}$ is a bag containing the number $1$. But $\{\{1\}\}$ is a bag containing a bag containing the number $1$. So two bags, one inside the other. These are different. Physically different if you think real paper bags.

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    $\begingroup$ To the OP: note that this is a limited intuition. In particular, it's really more about multisets than sets, since a bag containing two empty bags is different from a bag containing one empty bag. Still, it's quite useful for thinking about this sort of issue at the beginning. $\endgroup$
    – Noah Schweber
    Jan 25, 2018 at 13:39
  • $\begingroup$ Thanks a lot! Really appreciate this intuitive definition $\endgroup$
    – asdf
    Jan 25, 2018 at 13:43
  • $\begingroup$ @NoahSchweber Thanks for the clarification. $\endgroup$
    – Ethan Bolker
    Jan 25, 2018 at 14:57
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    $\begingroup$ $\{\}$ and $\{\{\}\}$ is more fun; one has cardinality 0, the other 1. $\endgroup$
    – Yakk
    Jan 25, 2018 at 16:37
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$$\{1\} $$ is a set whose the unique element is the integer $1$

$$\{\{1\}\} $$ is a set whose the unique element is the set $\{1\} $.

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    $\begingroup$ I see, thanks! But what about the brackets bit - do we have to always put them around a set to indicate that it's actually a set and not, say, some integers? $\endgroup$
    – asdf
    Jan 25, 2018 at 13:38
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    $\begingroup$ @asdf In response I would ask you, what do you consider to be the difference between "a set" and "some integers"? $\endgroup$
    – David Z
    Jan 25, 2018 at 21:13
  • $\begingroup$ @DavidZ it is clear there is a difference between an element and a set. $\endgroup$
    – hamam_Abdallah
    Jan 25, 2018 at 21:45
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    $\begingroup$ It is clear to most of us here, yes, but I'm specifically asking @asdf. $\endgroup$
    – David Z
    Jan 25, 2018 at 21:47
  • $\begingroup$ Yes, that was my problem - I basically made no difference, thank you all $\endgroup$
    – asdf
    Jan 25, 2018 at 21:51
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You are probably getting confused between the name of a set and its description.
When we write $A=\{x\}$, we mean $A$ is a set and inside set $A$, we have an element $x$.

Now if I define another set $B=\{A\}$, then $B$ is a set and inside set $B$, we have an element $A$, which is also a set. In this case, $B$ is a set of sets.

If you want to refer to the latter set, write
its name $B$, or
its description $\{A\}$.

For your last question, YES, we surround the elements of the set by curly braces {}, which also ensures unorderdness and non-repeatability.

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    $\begingroup$ I'd suggest rephrasing to make it clear that $\{B\}$ is a perfectly fine thing to write, it's just an entirely different mathematical object from $B$. $\endgroup$
    – David Z
    Jan 25, 2018 at 18:45
  • $\begingroup$ Thanks for suggestion. I have removed dos and donts. $\endgroup$
    – spkakkar
    Jan 25, 2018 at 19:41
  • $\begingroup$ After the edit, I think this is a really nice way of describing it. $\endgroup$
    – David Z
    Jan 25, 2018 at 21:11
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Well if you have $x=\varnothing$, then $0=\#x\neq \#\{x\}=1$. So clearly both sets are not the same.

Edit: With $\#S$ I refer to the cardinality of a set $S$, i.e. in the finite case the number of elements in $S$.

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    $\begingroup$ Here $\#$ is being used to denote "the cardinality of" a set, i.e. the size of the set in terms of how many elements it contains. $\endgroup$
    – hardmath
    Jan 25, 2018 at 13:41
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    $\begingroup$ @hardmath Thank you, I should have explained the notation. $\endgroup$
    – asdq
    Jan 25, 2018 at 13:42
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    $\begingroup$ @asdq You could still edit your answer to explain the notation. $\endgroup$
    – David Z
    Jan 25, 2018 at 18:44
  • $\begingroup$ @DavidZ Since it is explained right below, I don't see the benefit. $\endgroup$
    – asdq
    Jan 25, 2018 at 22:29
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    $\begingroup$ @asdq On StackExchange comments are temporary. You shouldn't rely on comments sticking around. It's explicitly part of the StackExchange usage guidelines that improvements suggested in comments should be incorporated into the question/answer itself. $\endgroup$
    – R.M.
    Jan 26, 2018 at 0:09
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There is a practical difference when you think about how you might use these sets - namely as a domain of functions. A function that takes a number is not the same as a function that takes a set.

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Perhaps it would be helpful to imagine the difference in concrete terms - say in terms of a computer data structure. Suppose we represent sets using linked lists [disregard for the moment that we cannot represent infinite sets this way]. Then $x$ is (the head pointer of) a sequence of nodes, where each node has a pointer to an element of $x$ and a pointer to the next node in the sequence. Then {$x$} is (the head pointer of) a single node, whose element pointer points to (the head node of) the sequence of nodes representing $x$. Clearer?

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