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In set theory natural numbers are defined by 0 = ∅ and natural number n+1 = n ∪ {n}

I need to prove that for every n ∈ N , n = {k ∈ N | k < n}.

I know that natural numbers

1 = {∅}

2 = {∅,{∅}}

3 = {∅,{∅},{∅,{∅}}}

The reason I'm having issues is that my intuition is not even correct. I know we need to use the definition of natural numbers, but I don't understand how n is equal to k ∈ N, when k < n.

Any help to getting start would be much appreciated!

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  • $\begingroup$ Remember $n = \{\forall k < n\}$, not just some $k \in n$ $\endgroup$ – Robert Lewis Feb 6 at 0:05
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    $\begingroup$ $n$ is equal to the set of $k < n$ $\endgroup$ – J. W. Tanner Feb 6 at 0:06
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    $\begingroup$ You can prove this from the definition of $n+1$ via induction on $n$. $\endgroup$ – Robert Shore Feb 6 at 0:06
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    $\begingroup$ You'll need a definition of what $k<n$ means, too. $\endgroup$ – Henning Makholm Feb 6 at 0:08
  • $\begingroup$ Aren't n,k just Natural numbers? $\endgroup$ – James Pekon Feb 6 at 0:10
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$n \ne k \in \mathbb N$.

$n = A;$ some subset of $\mathbb N$ and $k \in A$.

To get you intuition back in tune:

$0 = \emptyset$.

$1 = \emptyset \cup \{0\} = \{0\} = \{\emptyset\}$

$2 = 1 \cup \{1\} = \{\emptyset\} \cup \{1\} = \{\emptyset, 1\} = \{0, \{\emptyset\}\}$.

$3 = 2 \cup \{2\} = \{0, \{\emptyset\}\}\cup \{2\} = \{0, \{\emptyset\},2\}= \{0, \{\emptyset\},\{0, \{\emptyset\}\}\}$.

And so on.

Or another way of putting it:

$0 = \emptyset$

$1 = 0 \cup \{0\} = \emptyset \cup \{0\} = \{0\}$.

$2 = 1 \cup \{1\} = \{0\}\cup \{1\} = \{0,1\}$.

$3 = 2 \cup \{2\} = \{0,1\} \cup \{2\} = \{0,1,2\}$

$4 = 3 \cup \{3\} = \{0,1,2\}\cup \{3\} = \{0,1,2,3\}$

And so on.

ANd if we were to replace $1$ with $\{\emptyset,\{\emptyset\}\}$ and so on we'd get the stuff that look more familiar.

Anyway if we use induction and

suppose $n = \{0,1,2,3,4,......n-1\}$ then

$n+1 = n \cup \{n\} = \{0,1,2,3,4,...., n-1\} \cup \{n\} = \{0,1,2,3,4,.....,n-1, n\}$

And.... that's it.

Think about and reread this until your intuition works.

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  • $\begingroup$ Okay i think i understand most of what is going on, the one question I have is if k∈n, then in this problem is k the natural number that is within n.. For instance the n=2= {0}∪{1}. Would 0 and 1 be the different k's? $\endgroup$ – James Pekon Feb 6 at 2:24
  • $\begingroup$ The set $\{k \in \mathbb N| k < 2\} = \{0,1\}$. $\endgroup$ – fleablood Feb 6 at 5:03

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