# If $\{x\}=\{z\}$ then $x=z$?

I think that no. For example, Let $x=\{1,2,3\}$ and $z=\{1,2,3,4\}$. So, $\{\{1,2,3\}\}=\{\{1,2,3,4\}\}$. Yet, $x\neq z$.

Can you explain?

• Your example is incorrect. It is not true that $\{\{1,2,3\}\}=\{\{1,2,3,4\}\}$. – Steve Kass Jun 5 '16 at 20:25
• @SteveKass Why not? – PozcuKushimotoStreet Jun 5 '16 at 20:26
• @Kahler: Because the only element of the set $\{\{1,2,3\}\}$ is the set $\{1,2,3\}$, the only element of the set $\{\{1,2,3,4\}\}$ is the set $\{1,2,3,4\}$, and these elements $\{1,2,3\}$ and $\{1,2,3,4\}$ are not equal. – Brian M. Scott Jun 5 '16 at 20:31
• Why on earth do you think {{1,2,3}} = {{1,2,3,4}}?????? It obviously does not. – fleablood Jun 5 '16 at 21:40
• If $x \ne y$ then {$y$} $\ne$ {$y$}. So contrapositively the statement is true. – fleablood Jun 5 '16 at 21:42

By definition two sets are equal if and only if they have the same members: $x=y$ if and only if $\forall z(z\in x\leftrightarrow z\in y)$. In your example the sets $\{x\}$ and $\{z\}$ are assumed to be equal, so they must have exactly the same elements. The only element of $\{x\}$ is $x$, and the only element of $\{z\}$ is $z$, so it must be the case that $x=z$.

As was noted in the comments, your example is incorrect: if $x=\{1,2,3\}$ and $z=\{1,2,3,4\}$, then $x\ne z$, so $\{x\}\ne\{z\}$.

• @Kahler: You’re welcome. – Brian M. Scott Jun 5 '16 at 20:36

Notice that in order to prove that two set $A$ and $B$ are equal, we need to show: $$A\subseteq B~~~~ \text{and}~~~~ B\subseteq A$$

In this case we have the set $x=\{1,2,3\}$ and $z=\{1,2,3,4\}$.

We can clearly see that $x \subset z$, since every element of $x$ is contained in $z$. In order to show equality, we now also need $z\subset x$.

However, since $4 \in z$, but $4 \not\in x$ we have that $z \not\subset x$.

So $x \neq z$.

I am going to hazard a guess that when the OP says

Let x={1,2,3} and z={1,2,3,4}. So, {{1,2,3}}={{1,2,3,4}}

that he or she is confusing "equal sets" with "equal size sets". It is definitely true that $\{ \{1,2,3 \}\}$ and $\{ \{1,2,3,4\}\}$ are both sets containing a single element, and they are therefore equal in size. But they are not equal sets. "Equal sets" means "the same sets", which means "sets containing the same elements". If you want to indicate that the two sets are the same size, use the notation $\left| \{ \{1,2,3 \} \} \right| = \left| \{ \{ 1,2,3,4 \} \} \right|$.

That is correct. x $\neq z$. $$x=\{1,2,3\}\ and\ z=\{1,2,3,4\}\ then\ x \neq z.$$

But because all of $x$ is in $z$ and $z$ has items $x$ does not have, $x$ $\subseteq$ $z$. Meaning x is a proper subset of $z$.

$x$ consists of $4$ elements, $z$ consists of $3$ elements. Obviously, they cannot be equal.

• The op doesnt think x =z. Yet somehow the op believes {x} = {z} which are two sets with single elements. Yet as the elements are different I haven't got the slightest idea why the op believes these set are equal. – fleablood Jun 5 '16 at 21:48

By extensionality of sets, there is an equivalence

$$\{ x \} = \{ z \} \quad \Longleftrightarrow \quad \forall t: (t \in \{ x \} \Leftrightarrow t \in \{ z \})$$

By definition of set enumeration notation, there is an equivalence

$$a \in \{ b \} \quad \Longleftrightarrow \quad a = b$$

Substituting this into the above equivalence,

$$\{ x \} = \{ z \} \quad \Longleftrightarrow \quad \forall t: (t = x \Leftrightarrow t =z)$$

The right hand side is equivalent to $x=z$. But if you need to see that proved, I will instead prove something simpler: by substituting $t = x$, we get

$$\forall t: (t = x \Leftrightarrow t =z) \implies (x=x \Leftrightarrow x=z) \implies x=z$$

and consequently,

$$\{ x \} = \{ z \} \quad \implies \quad x=z$$

The other direction

$$\{ x \} = \{ z \} \quad \Longleftarrow \quad x=z$$

is proven by substitution, thus

$$\{ x \} = \{ z \} \quad \Longleftrightarrow \quad x=z$$

Your error is that $\{\{1,2,3\}\} \neq \{\{1,2,3,4\}\}$.