Is $\{x\}$ an element of $\{x,\{x,y\}\}$ or not? I'm a math teacher and I have a doubt about this question
$\{x\}\in\{x,\{x,y\}\}$
is true or false? And why? 
I think it's true because $\{x,y\}=\{x\}\cup\{y\}$ then...
But I'm not sure.
Thanks!
 A: You need to understand the difference between $x$ and $\{x\}$. For example $\emptyset$ is just an empty set but $\{\emptyset\}$ is a set containing one element $\emptyset$. Hence the answer to your question is NO as RHS contains $x$ but not $\{x\}$.
A: It is not true since $\{x\}$ is not an element of $\{x, \{x,y\}\}$. 
What is true is that $x \in \{x, \{x,y \}\}$. It is also true that $x\in \{x, \{x,y\}\}$ and that $\{x,y\}\in \{x, \{x,y\}\}$ because $\{x, \{x,y\}\}$ consists of exactly the two elements $x$ and $\{x,y\}$.
So in general you want to only say that something is in a set if exactly that something is on the list of things between the $\{$ and $\}$.
A: By the definition of set enumeration notation,
$$\{ x \} \in \{ x, \{ x, y \} \}$$
if and only if
$$ \{ x \} = x \qquad \text{or} \qquad \{ x \} = \{ x, y \} $$

For the second case,
$$\{ x \} = \{ x, y \} $$
if and only if
$$ \{ x \} \subseteq \{ x, y \} \qquad \text{and} \qquad \{ x \} \supseteq \{ x, y \} $$
if and only if (since the first case is always true we only need to check the second case)
$$ x \in \{ x \} \qquad \text{and} \qquad y \in \{ x \} $$
if and only if (since the first case is always true we only need to check the second case)
$$ y = x$$

Therefore,
$$\{ x \} \in \{ x, \{ x, y \} \}$$
if and only if
$$ x = \{ x \} \qquad \text{or} \qquad y = x $$
...
Depending on your particular version of set theory (e.g. "axiom of foundation"), $x = \{ x \}$ can never be true. In that case, we can follow up with
... if and only if $y = x$
A: *

*"Belongs to" ($\in$) means simply "is an element of". 

*The set $\{x,\{x,y\}\}$ has just two elements, $x$ and $\{x, y\}$, and neither one is $\{x\}$. (To phrase it differently: neither element is equal to $\{x\}$.)*

*So $\{x\}$ does not belong to $\{x,\{x,y\}\}$.


$^*$Footnote: Certainly not for arbitrary $x$ and under standard set theory, as the question intends. 
A: If $x=y$ then $\{x\}=\{x,x\}=\{x,y\}\in\{x,\{x,y\}\}$.
If $x=\{x\}$ then also yes, because $\{x\}=x\in\{x,\{x,y\}\}$.
Otherwise, the answer is no. There is no element in $\{x,\{x,y\}\}$ which is exactly $\{x\}$. It is true that $\{x\}$ is a subset of $\{x,\{x,y\}\}$ and a subset of an element of this set as well. But it is not an element itself.

The general method I often suggest is to use some dummy variable. Set $u=\{x,y\}$ and $v=\{x\}$. Then $u\neq v$ because $y\in u\setminus v$. Then we ask, $v\in\{x,u\}?$ and the answer is quite obvious, $v\neq x$ and $v\neq y$. So $v\notin\{x,u\}$.
