generating a path-connected group from a neighborhood of 1 I am reading the book 'Naive Lie theory'. In it, it is proven that a path-connected group can be generated by a neighborhood of the unity element. The idea is simple and clear. But I cannot overcome some details. 
How is always possible to find $A_i$ and $A_{i+1}$ in the same set $A(t_j ) \mathcal{O}$? 
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 A: I think the point is that the open cover of the path $A(t)$ is a cover, hence the intersection of at least some of the $A(t_j)\mathcal{O}$ must be nonempty.  However, you are right that there is no reason to assume that $A(t_1)\mathcal{O}\cap A(t_2)\mathcal{O}\neq \emptyset$ - the numbering sort of implies it, but this is not actually stated, nor even a necessary consequence.  The usual picture would be of a snaking curve with little balls neatly lined up covering it one by one, but the mere existence of a finite subcover does not necessitate it.
However, if you note carefully, you'll see that the $t_j$ have a different indexing than the points $A_i$; the only claim is that each two consecutive points are in some $A(t_j)\mathcal{O}$, not that the $j$ are in order, or even that there are not many consecutive pairs $A_i/A_{i+1}$ that are all in the same open set. (Maybe we chose our $\mathcal{O}$ particularly well and actually $A\in \mathcal{O}$ already? could happen.)
So I think instead what the author is implying is that first we pick a certain number of points in linear order (via the order on $[0,1]$) which go from $1$ to $A$, and then if for some reason we don't have that each consecutive pair are both in at least one of the open sets of the finite subcover, then we pick more of them to fill in the gaps until we do.  
Explicitly1, take the intersections of the inverse images of the open sets with $[0,1]$.  If we get a set like $\{A_j\}$ for those open sets (which must also be a finite cover), their images will be in the images of those open sets and have the consecutive pairs property in them (which are the original open sets).  So the question reduces to the same question on the unit interval.  
Even if the intersections with the unit interval, which will all be unions of open intervals, are pathological and once again are infinite (try drawing some squiggly open sets on $A(t)$ that don't intersect "nicely") you can ignore all but a finite subcover of those once you get to $[0,1]$.  And now you have a finite set of open intervals on the interval $[0,1]$ which you have to pick points from, every consecutive pair of which are in the same open interval.  You can even get rid of any wholly contained in another one - if they were superfluous on $[0,1]$, they will be on $A(t)$ as well2.  Now just start with $0$ and go to the nearest point which is in an interval around zero and a different one; then repeat, but with a point in the second interval and a third one as well.  Since there are only finitely many such intersections, you will exhaust the supply of these intersections and since you are moving forward, eventually you must come to an interval which contains $1$ as well, in which case you naturally pick $A$ as your last stop.

1I am sure I have made this phenomenally more complicated that it has to be (and hope someone else has a better answer), but as I imagined all sorts of nasty pathological path-connected $\mathcal{O}$ neighborhoods of the identity just in $\mathbb{R}^2$, the more I realized that for me to believe this, we'd have to go back to someplace where things cannot be that complicated, which is the unit interval.  And surely that is the "moral" reason why this works - in essence, the interval is still the interval, even if it's in some bizarre Lie group.
2Obviously the actual open subsets in the Lie group are not superfluous, but ignoring the extras won't take you out of them, it just clarifies where to look.  They will still be there when you go back to the group.
