Bijection Between $\mathbb{Z}_+$ and a Subset of $\mathbb{Z}_+ \times \mathbb{Z}_+$ Let $T=\{(a,b)\mid a,b\in\mathbb Z_+, b\leq a\}.$ Find a bijective function $f: T \to \mathbb{Z}_+$
I have tried to find a function but I can't, how does such function look like?
 A: (1): $(1,1)$
(2):$(2,1)\ \ \ (2,2)$
(3):$(3,1)\ \ \ (3,2) \ \ \ (3,3) $
$\ldots$
($k$):$(k,0)\ \ \ (k,1) \ \ \  (k,2)\ \ \ \ldots \ \ (k,d)\ \ \ldots (k,k)$
$\ldots$
define $f$ by $f(1,1)=0$ , $f(2,1)=1$ , $f(2,2)=2$ , $f(3,1)=3$ $\ldots$ that is
Take $f(k,d)=1+\ldots+(k-1)+(d)-1=k(k-1)/2+d-1$
A: My bijection won't be explicit, but I typically try to place some kind of order on the sets and place them "side by side".
The set $\mathbb{Z}_+$ already has a natural order on it.
We can order $T$ as follows. For each $n \geq 2$, let $T_n$ contains all ordered pairs from $T$ whose coordinates sum to $n$. Order each $T_n$ lexicographically, and then order $T$ by concatenating all these orders.
With these orders in mind, my bijection would therefore be to map $n$ to the $n^\text{th}$ element of $T$. The first few such mappings are
\begin{align*}
1 &\mapsto (1,1)\\
2 &\mapsto (2,1)\\
3 &\mapsto (2,2)\\
4 &\mapsto (3,1)\\
5 &\mapsto (3,2)\\
6 &\mapsto (4,1)
\end{align*}
When choosing the order on $T$, it is important to ensure that any element is reached in a finite amount of time. For example, the order $(1,1) < (2,1) < (3,1) < (4, 1) < \cdots$ is not viable since it takes an infinite amount of time to reach $(2,2)$.
