Prove that $(a) n < k^+$ iff $n ≤ k$ and $(b) <$ is a transitive relation I wonder if you could check my approach to this question? I am particularly unsure about part (b); it seems like it needs to be shown in a more substantial way. Thank you.
Question: We define the "less than" relation on the natural numbers by $a < b$ iff $a ∈ b$. Prove that
$(a) n < k^+$ iff $n ≤ k$.
$(b) <$ is a transitive relation.
My approach for $(a)$:
(i) Suppose $n = 1$ and $k = 0$. Then
 $n = 0^+ = \{∅\}$ and $k = 0 = ∅$ (Von Neumann ordinals)
(ii) If we want to say $n < k^+$, we want to say that $n ∈ k^+$, which is the same as $n ∈ (k ∪ \{k\}) <$-- (by the Successor function)
(iii) So, in this instance, $n < k^+$ would give us 
$\{∅\} ∈ (∅ ∪ \{∅\} )$
which is the same as 
$\{∅\} ∈ ∅$, which is false.
My approach for (b):
(i) Suppose this time that $n = 0$ and $k = 0$. Then
  $n = 0 = ∅$ and $k = 0 = ∅$
(ii) Therefore $n < k^+$ means 
$0 < 0^+$, or $0 < 1$
i.e.
$∅ < (∅ ∪ \{∅\})$ which is the same as
$∅ ∈ (∅ ∪ \{∅\})$ which is the same as
$∅ ∈ ∅$  
Given that $a < b$ iff $a ∈ b$, then if $<$ is a transitive relation, what is implied is 
$A ∈ B → A ⊆ B$
(From (ii) above) $∅⊆∅$ (trivially true)
Therefore $<$ is a transitive relation.
 A: Your (iii) for (a) is incorrect: you should get $\{\varnothing\in\{\varnothing\}\cup\{\varnothing\}\}$, which is true, not $\{\varnothing\}\in(\varnothing\cup\{\varnothing\})$, which is false. (However, $\varnothing\cup\{\varnothing\}$ is $\{\varnothing\}$, not $\varnothing$.)
To show that $n<k^+$ iff $n\le k$, assume first that $n<k^+$. This means that $n\in k^+$, and since $k^+=k\cup\{k\}$, this means that $n\in k\cup\{k\}$. Thus, either $n\in k$, in which case $n<k$, or $n\in\{k\}$, in which case $n=k$, and it follows that $n\le k$. Conversely, if $n\le k$, then either $n<k$, in which case $n\in k$, or $n=k$, in which case $n\in\{k\}$, and in either case $n\in k\cup\{k\}=k^+$. Note that no induction is required.
I can’t make head or tail of your argument for (b), I’m afraid; as written it simply makes no sense. You should be trying to show that if $n<k$ and $k<m$, then $n<m$, i.e., that if $n\in k$ and $k\in m$, then $n\in m$. It you’ve already shown that von Neumann ordinals are transitive sets with respect to the relation $\in$, this is immediate. If not, then you have a fair bit of work to do.
