Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

The classical definition of n-tuple $(x_i)_{i < n}$ starts at $n=2$. In this case $$(x_0,x_1) := \{\{x_0\},\{x_0,x_1\}\}$$(1).

For $2<n=k+1$, $(x_i)_{i < n}:=((x_i)_{i < k},x_k)=\{\{(x_i)_{i < k}\},\{(x_i)_{i < k},x_k\}\}$(2).

Hence it well defined on $2 \le n \in \mathbb N$.

One would tend to extend this definition to every natural number.

In the case $n=1$, if we let $(x_0)=x_0$, then (2) can be extended to $1 \le n \in \mathbb N$, so $(x_0,x_1)=\{\{(x_0)\},\{(x_0),x_1\}\}=\{\{x_0\},\{x_0,x_1\}\}$, which also compatible with (1).

However, if we go on do this to $n=0$, we will get a trouble. That is, $()=?$ If $()=\emptyset$, then $(x_0)=((),x_0)=\{\{()\},\{(),x_0\}\}=\{\{\emptyset\},\{\emptyset,x_0\}\}$, but it contradict to $(x_0)=x_0$ as defined. Similar, if we let $()$ remains blank, then $(x_0)=((),x_0)=\{\{()\},\{(),x_0\}\}=\{\{\},\{ ,x_0\}\}=\{\emptyset,\{x_0\}\}$, which also contradict to $(x_0)=x_0$.

So my question: Is there any methodology to climb over this barrier?

share|improve this question
    
You want $()=\emptyset$. The problem you have is $(x_0)=x_0$, this is what causes the contradiction. Simply define $(x_0)=\{\{\emptyset\},\{\emptyset,x_0\}\}$. See this question: math.stackexchange.com/questions/964092/… –  EthanAlvaree Oct 14 at 21:54

2 Answers 2

up vote 2 down vote accepted

There is no way to define a $0$-tuple to that $1$-tuples are ordered pairs (without contradicting Foundation): Suppose that there were such a definition of a $0$-tuple $\langle\;\rangle$ so that $$a = \langle a \rangle = \langle \langle \; \rangle , a \rangle = \{ \{ \langle \; \rangle \} , \{ \langle \; \rangle , a \} \}$$ then $a \in \{ \langle \; \rangle , a \} \in a$.

share|improve this answer
    
See this question: math.stackexchange.com/questions/964092/… –  EthanAlvaree Oct 14 at 21:54

There are ordered pairs, and there are tuples. Often we start with ordered pairs, then define an $n$-tuple as a function from $n$ to $X$. This is a good definition, because it carries over to the transfinite case as well. Whereas inductively extending the Kuratowski definition of ordered pairs will not carry over.

Note that for $n=0$ the tuple would be empty. Since we index from $0$ to $n-1$, in this case we simply don't index.

The empty tuple, if so, is $\varnothing$, which is exactly a function from $0$ into $X$.

share|improve this answer
    
Well, if memory serves me right, the function from $n$, even $\alpha$, any transfinite ordinal, into $X$ is called a sequence over $X$. However the definition of sequence bases on the definition of function, which bases on the definition of relation, which is also bases on the definition of tuple... Thus the Kuratowski definition seems cannot be ignored if we don't want to cause a circular definition. –  Popopo Mar 23 '13 at 1:26

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.