Let $X$ be the first uncountable ordinal. In other words $X$ is an uncountable set equipped with a well-ordering relation "$\leq $" such that, for every $x$ in $X$, the set
 $$
 \{y\in X:\ y\leq x\}
 $$
 is countable.
 Let
 $$
 T=\{(x, y)\in X\times X:\ y\leq x \},
 $$
 so that $T$ consist of everything below the diagonal in $X\times X$.


Let us agree to call a subset $G\subseteq T$ a *graph*, provided
 $$
 \big ((x,y\_1)\in G\big ) \ \wedge\ \big ((x,y\_2)\in G\big ) \ \Rightarrow \ y\_1=y\_2.
 $$
 Clearly these are precisely the graphs of $X$-valued functions $f$ defined on subsets of $X$, such that $f(x)\leq x$, for every $x$.



> 
> Question: Is it possible to write $T$ as the union of a countable family of graphs?
> 
> 
> 



Uncountable ordinals and graphs of functions
There's a more general result here:



> 
> Suppose $\mathcal{A}=(A\_i)\_{i\in I}$ is a collection of countable nonempty sets. Then the set $$[\mathcal{A}]=\{(a,i): a\in A\_i\}$$ is the union of countably many graphs of functions.
> 
> 
> 


When phrased this way it's much easier to think about. Simply fix surjections $b\_i:\omega\rightarrow A\_i$, and for $k\in\omega$ let $$f\_k:I\rightarrow\bigcup\mathcal{A}:i\mapsto b\_i(k).$$ Then the union of the graphs of the $f\_k$s is exactly $[\mathcal{A}]$.


(Put another way, $f\_k(i)=b\_i(k)$ - basically, you have $I$-many "columns" of size $\le\omega$, and you view the whole grid as $\omega$-many "rows" of size $I$.)


*Of course, we use the axiom of choice when we pick the $b\_i$s, and in the absence of choice can fail (and in particular the special case of your question can fail without choice). But that's a side point.*


