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?

  • 7
    $\begingroup$ The actual answer is so simple, I think if I didn't ask, I'd never have figured it out ... - Thank you both! $\endgroup$
    – Sudix
    Aug 6, 2017 at 23:11
  • 4
    $\begingroup$ In the proposition that is to be proven, the thesis has equality between the elements of the tuples. Therefore, the cases should also have equality between elements: $a_1=b_1$ and $\{a_1,a_2\}=\{b_1,b_2\}$ for the first case. $a_1=\{b_1,b_2\}$ and $b_1=\{a_1,a_2\}$ for the second. $\endgroup$
    – Marja
    Aug 6, 2017 at 23:11
  • $\begingroup$ Never mind. It is in the definition where the braces are missing. It should be $(a,b)=\{\{a\},\{a,b\}\}$. $\endgroup$
    – Marja
    Aug 6, 2017 at 23:21
  • 6
    $\begingroup$ It seems a bit odd that a book on nonstandard analysis would spend time building up such set theoretic basics. That would be like having a book on public key cryptography start with a chapter that builds up the natural numbers from the Peano postulates. $\endgroup$ Aug 7, 2017 at 11:52
  • 1
    $\begingroup$ It's a key detail, used to build the super structure. It's a means of creating an order (a first/second), so this super structure can express relations. $\endgroup$
    – Sudix
    Aug 8, 2017 at 16:08

2 Answers 2


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


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.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .