Is $(\mathbb{N\times N\times N}, \leq_{lex})$ isomorphic to $(\mathbb{N\times N}, \leq_{lex})$?

Question

Is $(\mathbb{N\times N\times N}, \leq_{lex})$ isomorphic to $(\mathbb{N \times N}, \leq_{lex})$? Prove it.

$\leq_{lex}$ - lexicographical order

I think that $(\mathbb{N\times N\times N}, \leq_{\text{lex}})$ is isomorphic to $(\mathbb{N\times N}, \leq_{\text{lex}})$. I defined function $f(l,m,n)=l, \frac{(m+n)(m+n+1)}{2}+m$.

I want to show that if $l_1, m_1, n_1 \leq l_2, m_2, n_2$, then $f(l_1, m_1, n_1) \leq f(l_2, m_2, n_2)$, and in this case that is always true (should I show it somehow or is it enough just to state it?).

Can you tell me if my reasoning is correct? If not, where am I making the mistake?

Answer 1

$\varphi: (m,n) \mapsto \dfrac{(m+n)(m+n+1)}{2}$ is kind of the standard bijection between $\Bbb N \times \Bbb N$ and $\Bbb N$, so your proposed function is indeed a bijection.

However, the homomorphism property is not satisfied, as $(0,0,1) \le (0,1,0)$ but $f(0,0,1) = (0,2) \not\le (0,1) = f(0,1,0)$.