Why a unit set is not the same as its element? $\{x\} \ne x$?

I was solving a list of exercises from homework and I answered one question wrong because I thought that a singleton set and its element were the same thing...

I found only a few results from a Google search:

May anyone please explain this to me?

• Just as an example: $\{x\}$ has exactly one element, namely $x$. However, how many elements $x$ has depends on $x$. Mar 23, 2014 at 15:16
• @Git Gud: Oh, sorry, the problem is that I often receive more than a good answer, I normally just UP vote them, and let the votes of other user decide which answer is the best (by receiving more votes than the others). But I will try to use the accept answer feature from now on. Mar 23, 2014 at 15:20
• @JoãoRimu That's a sensible approach. Do note, however, that I didn't tell you to accept any answer or anything, it's your choice. But I do recommend you accept some answers. Mar 23, 2014 at 15:21

First of all, it can't always be the same. $\{\varnothing\}\neq\varnothing$. On the right-hand side, we have the empty set which has no elements but on the left-hand side we have a set which does include an element.
More generally, one of the axioms of modern set theory allows us to prove that no set is an element of itself. In particular $\{x\}\notin\{x\}$. Since the only element of $\{x\}$ is $x$, when we say $\{x\}\notin\{x\}$ we actually say that $x\neq\{x\}$.
If a singleton set and its element were the same thing, then the set $\{\{1,2,3\}\}$, which has just one member, would be the same thing as the set $\{1,2,3\}$, which has three members. Suppose one wants to enumerate the partitions of the set $\{1,2,3\}$. They are: \begin{align} & \{\{1,2,3\}\} \\[6pt] & \{\{1,2\},\ \ \{3\}\} \\[6pt] & \{\{1,3\},\ \ \{2\}\} \\[6pt] & \{\{2,3\},\ \ \{1\}\} \\[6pt] & \{\{1\},\ \ \{2\}, \ \ \{3\}\} \end{align}