# Why does the curly bracket do not equal to the double curly brackets?

$$\{a\} \neq \{\{a\}\}$$

$$\{a\}$$ is the set whose only element is the a (and no others). $$\{\{a\}\}$$ is the set whose only element is the set $$\{a\}$$.

Does this mean the 'element a' is not equal to 'set $$\{a\}$$'?

• Indeed, those are not equal. – StackTD Mar 17 '19 at 14:19
• – JMoravitz Mar 17 '19 at 14:21

## 3 Answers

Even though people sometimes get sloppy about it, $$a$$ and $$\{a\}$$ are not the same object. $$a$$ is the only element of the set $$\{a\}$$.

They are not equal.

Intuitively, $$\{a\}$$ means a set which contains an element $$a$$; while $$\{\{a\}\}$$ means a set that contains a set $$\{a\}$$ as its element.

From ZFC axiom: Every non-empty set $$x$$ contains a member $$y$$ such that $$x$$ and $$y$$ are disjoint sets.

In general: $$\{x\}=\{y\}\iff x=y$$

Then we can conclude that also:$$\{x\}\neq\{y\}\iff x\neq y$$

Applying that in your case we find that the statement $$\{a\}\neq\{\{a\}\}$$ is the same statement as $$a\neq\{a\}$$.

Sidenote:

If also the axiom of regularity is accepted then this statement is true for every $$a$$.

This because on base of that axiom it can be proved that $$a\notin a$$ is true for every $$a$$ while $$a=\{a\}$$ implies that $$a\in a$$.

• From Wikipedia, axiom of extensionality: "what the axiom is really saying is that two sets are equal if and only if they have precisely the same members. The essence of this is: A set is determined uniquely by its members." >> {a} and {{a}} are not same. – Chen Yun Mar 17 '19 at 15:42
• Axiom of regularity: "No set is an element of itself" "We see that there must be an element of {A} which is disjoint from {A}. Since the only element of {A} is A, it must be that A is disjoint from {A}. So, since A ∈ {A}, we cannot have A ∈ A (by the definition of disjoint)." >> I don't get your meaning, I assume as {a} is an element of {{a}}, they are not going to be equal. Is this correct?? – Chen Yun Mar 17 '19 at 15:42
• What you mention ("No set is an element of itself") is not the axiom of regularity itself but is a consequence of the axiom of regularity. The axiom says that $\{A\}$ must have an element $x$ with $x\cap\{A\}=\varnothing$. The only candidate for $x$ is $A$ so that we can conclude that $A\cap\{A\}=\varnothing$. From this it follows that $A\notin A$. If that's what you are saying then I fully agree with you and based on the axiom it has been proved that $a\in a$ cannot be a true statement (as stated in my answer). "I don't get your meaning..." What meaning? – drhab Mar 17 '19 at 16:00
• What I said about the axiom of extensionality in my answer was wrong and I removed it. – drhab Mar 17 '19 at 16:14
• Oh I can't get the false statement "a={a} implies that a∈a". From your comment above "A∩{A}=∅. From this it follows that A∉A". >> No matter what is the element surely it is not equal to the set because they have no intersection of each other, disjoint. Thus we could not find "element of..." within it. – Chen Yun Mar 17 '19 at 17:15