Infinity xkcd style: can a turing machine exist? I recently read this xkcd comic. It's about a guy who simulates a universe by a Turing machine (specifically, Rule 101, a cellular automaton), by laying down infinite rows of rocks, each row corresponding to a 'state' of the universe.
This made me wonder, because my intuition with dealing with infinities always gets a bit muddy. A single row of stones stretching to infinity is perfectly allowed by mathematicians, as far as I know. And just like in the infinite hotel, we can add some more stones to such a row if necessary.
What happens though, when you add multiple rows? At some point, our guy from the comic must have stopped on the first row, and start on the next. However, in an infinite row of stones, there is no last stone - or is there? 
Tldr;
In other words, if we let $t\to \infty$, will we end up with one row with an infinite number of stones, or with an infinite number of rows? 
Edit: I'm not looking into the question whether above machine could describe a universe - let's assume that it can. I'm only wondering whether you can get to the second infinite row if time is infinite. Maybe I should have removed the rest.
 A: If the first row is infinte, then it doesn't matter how much time she has on her hands, she will never finish laying it down, and thus never get to row number two.
However, in the comic, while she says that the desert expands seemingly infinitely, she doesn't say anything about her computer being infinitely large. Just that there's more than enough room for it.
Therefore, I conjecture that the first row isn't infinite, it's just very, very long (a binary representation of our universe, in a format constructed to be easy to calculate with, not compact). Therefore, after a vary long (but finite) time, she gets to row umber two. That row must also take quite some time to lay out, but it's still finite.
As to the intuition about infinities, if the first row were infinite, but there were infinitely many people out there laying rocks, each one responsible for only xkcd (the third panel) many stones or so, the first row would be finished at some point in time.
This is more or less equivalent to the fact that, in Hilbert's hotel, every guest is responsible to move to their own next room. That's very different from the one hotel owner knocking on every door, one by one, asking them to please move over to the next room. In the first scenario, the whole event might be over in an hour or so, depending on how quick people are packing their stuff. In the second one, there is no end. The hotel owner is doomed to forever knock on the next door asking them to please move one room over.
A: If you demand that you finish the first row before starting on the second, then you're right - you'll never get a chance to begin the second (unless you like supertasks, but let's ignore those for the moment). However, you can indeed deal with infinitely many rows by being a bit more clever - this is basically computing in parallel (sometimes called dovetailing). The idea is as follows:


*

*Write a bit of the first row.

*Write a bit of the second row.

*Write a bit more of the first row.

*Write a bit of the third row

*Write a bit more of the second row.

*Write a bit more more of the first row.
.
.
.

*Write a bit of the $n$th row.

*Write a bit more of the $(n-1)$th row.
.
.
.

*Write a bit more more more . . . more more ($n$ times) of the first row.

As you let time go to infinity, each of the infinitely many rows will go to infinity. Going back to Hilbert's hotel, think of this as corresponding to infinitely many buses, each with infinitely many passengers, showing up to the hotel at the same time.
