Let $\{M\_i\}$ be a sequence of 2-dimensional orientable closed surfaces of genus $g$ with smooth Riemannian metrics with the Gauss curvature at least $-1$ and diameter at most $D$. By the Gromov compactness theorem, one can choose a subsequence converging in the Gromov-Hausdorff sense to a compact Alexandrov space with curvature at least $-1$ and Hausdorff dimension 0,1,or 2. One can show (see below) that if $g\geq 2$ then the limit space cannot be a point, thus the dimension of the limit space is at least 1 (while for $g=0,1$ it can be 0). 


Let us assume that the limit space has dimension 1. Then it is either circle or segment. 


**Whether these both possibilities (circle and segment) can be obtained in the limit?** 


**ADDED:** It is not hard to see that one can get segment for $g=0$ and circle for $g=1$. I suspect (but cannot prove) that for $g\geq 2$ and $g=0$ one *cannot* get circle in the limit. In fact I do not even know whether in the case $g\geq 2$ a 2-dimensional limit is the only possibility.


**UPDATE:** Based on the answer by Igor Belegradek, let me summarize the situation. Let $\{M\_i\}$ be a sequence of genus $g$ orientable closed surfaces with Riemannian metrics with Gauss curvature at least -1 which converges in the Gromov-Hausdorff sense to an Alexandrov space $X$. 


1) If $g=0$ then $X$ is either a point, or a segment, or $X$ is homeomorphic to $S^2$ (by Perelman stability theorem), and all the three cases are possible.


2) If $g=1$ then $X$ is either a point, or a circle $S^1$, or homeomorphic to the 2-torus, and all the three cases are possible.


3) If $g\geq 2$ then $\dim X=2$ and hence $X$ is homeomorphic to an orientable genus $g$ closed surface.


**ADDED:** Let me add a proof that if $g\geq 2$ then a point cannot be the limiting space. Indeed otherwise we would have $d\_i:=diam(M\_i)\to 0$. Let us divide the metric of $M\_i$ by $d\_i$ and denote the new metric space by $N\_i$. Then the sectional curvature of $N\_i$ is at least $-d\_i^2$ and diameter 1.


By the Gauss-Bonnet $$4\pi(1-g)=\int\_{N\_i}K\geq -d\_i^2vol(N\_i).$$
By the Bishop inequality $vol(N\_i)$ is bounded from above. Hence the right hand side in the above inequality tends to 0. Hence $1-g\geq 0$ which is a contradiction.



Gromov-Hausdorff limits of 2-dimensional Riemannian surfaces
As mentioned in the comments, if the limit is a circle, then by the Yamaguchi fibration theorem, $M\_i$ fibers over the circle, and hence it is a torus (or Klein bottle in the non-orientable case).


If the limit is a segment, then $M\_i$ is $S^2$ for all large $i$. One (somewhat heavy handed) way to see this is to apply Corollary 0.4 of Shioya-Yamaguchi's [paper](http://projecteuclid.org/euclid.jdg/1090347524). 
Indeed, the product of $M\_i$ and a unit circle, collapses to the cylinder. Hence in Corollary 0.4 we have $g=0$ and $k=2$. Hence for large $i$ the fundamental group of $M\_i\times S^1$ is a free product of $\mathbb Z$ and finitely many finite cyclic groups. Such a group cannot have a center unless all the finite cyclic groups are trivial. The circle factor of $M\_i\times S^1$ is central in the fundamental group, so $\pi\_1(M\_i\times S^1)=\mathbb Z$. Hence $M\_i$ is a sphere (for large $i$). 


This argument uses orientability of $M\_i$ because Shioya-Yamaguchi only deal with orientable manifolds.


