How can a set with one element be equal to a set with two elements

I'm looking into nonstandard analysis, and am in a chapter which introduces the whole load of basic terms they'll use.

One of this is a proof for ordered pairs (Kuratowski definition) by induction. The ordered pairs are defined like this:

\begin{equation} \begin{aligned} (a)_k :&=\{a\} \\ (a,b)_{k} :&= \{\{a\},\{a,b\}\} \\ (a_1,\,...\,,a_n)_k :&= ((a_1,\,...\,,a_{n-1}),a_n) \end{aligned} \end{equation}

The theorem to show is : $(a_1,\,...\,,a_n) = (b_1,\,...\,,b_n) \Rightarrow a_k = b_k \text{ for k = 1, ... , n}$

They do it by induction: Case n = 1 is trivial, and case n = 2 (the part I don't understand) goes like this:

It is $(a_1 , a_2) = (b_1,b_2) $. This is per definition equal to $\{\{a_1\},\{a_1,a_2\}\} = \{\{b_1\},\{b_1,b_2\}\} $.

Now the following cases are possible:
$ \begin{align} \{a_1\} &= \{b_1\} &\text{and}&\quad\quad \{a_1,a_2\} &= \{b_1,b_2\} ,\\ \{a_1\} &= \{b_1,b_2\} &\text{and}& \quad\quad\{b_1\} &= \{a_1,a_2\} \end{align} $

First case seems simple enough, but I don't understand how a set with one element can be equal to a set with two elements. Even worse, they say for both cases follows
$ a_1 = b_1 $ and $ a_2 = b_2$
... but why?


Solution 1:

Hint: the set $\{1, 1\}$ does not have two elements.

Solution 2:

It is possible for $\{x\}$ to equal $\{y,z\}$ if and only if $y=z$ -- because then $\{y,z\}$ is actually a set with one element.