Proof of Natural Numbers using n+1 = n ∪ {n}

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!

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

$$n \ne k \in \mathbb N$$.

$$n = A;$$ 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.

• The set $\{k \in \mathbb N| k < 2\} = \{0,1\}$. – fleablood Feb 6 at 5:03