Language is non-empty is recursively enumerable (intuitive way) $M$ is some Turing machine, $\left&lt;M\right&gt;$ is the code of the Turing machine.

$L =\{\left&lt;T\right&gt; | L(T) \ne \emptyset\}$

How to see intuitively that $L$ is partially decidable?

We can try running a given $M$ on all strings and accept if $M$ accepts.
However, what if the simulated $M$ gets into an infinite loop?
A language $L$ is partially computable if there is a machine $M$ that on input $x$:


If $x \in L$ then $M$ halts.
If $x \notin L$ then $M$ doesn't halt.


This is equivalent to the other definitions.

To answer your question, what we do is to run the input machine $M$ on all inputs in parallel. If it never accepts on any input, then we never halt, but that's ok.

In order to run $M$ on all inputs in parallel, we use the technique of dovetailing:


We run $M$ for one step on the first input.
We run $M$ for one step on the first two inputs.
We run $M$ for one step on the first three inputs.
And so on.