TITLE: Compactness theorem for minimal surfaces
QUESTION [6 upvotes]: I am a bit confused about the statement of Theorem 1.1 in  this paper  by Brian White. For convenience, I will restate it here.

Theorem: Let $\Omega$ be an open subset of a Riemannian $3$-manifold. Let $g_i$ be a sequence of smooth Riemannian metrics on $\Omega$ converging smoothly to a Riemannian metric $g$. Let $M_i \subseteq \Omega$ be a sequence of properly embedded surfaces such that $M_i$ is minimal with respect to $g_i$. Suppose also that the area and the genus of $M_i$ are uniformly bounded on compact subsets of $\Omega$. Then (after passing to a subsequence) the $M_i$ converge to a smooth, properly embedded $g$-minimal surface $M$. For each connected component $\Sigma$ of $M$, either

the convergence to $\Sigma$ is smooth with multiplicity one, or
the convergence is smooth (with some multiplicity $> 1$) away from a
  discrete set $S$.

In the second case, if $\Sigma$ is two-sided, then it must be stable.

My question concerns the last sentence of the statement above. It seems very surprising to me the conclusion about the stability. What if $g$ is a metric with strictly positive Ricci curvature? 
I was thinking about the following 2-dimensional situation. Imagine to have a sequence of metrics $g_i$ on the 2-sphere such that  the limit metric $g$ has positive curvature. Imagine that each $g_i$ carries two closed geodesics which are converging to an unstable geodesic with multiplicity 2. See this poor quality picture: 
The compactness theorem above says that one cannot construct such an example on the $3$ sphere and using minimal surfaces instead of geodesics. Can anyone give me an intuition of what is going on here? 
Thank you!

REPLY [5 votes]: What's going on with the picture you drew is that because the $g_i$ have regions of negative curvature (in the "valleys" where the $M_i$ sit) that become arbitrarily close to the "hilltop" where $M$ sits), one must have that the curvature of $g$ along $M$ is identically zero.  This means that $M$ is actually weakly stable as the constant function $1$ is an eigenfunction of the stability operator (and is the lowest as it doesn't change sign) with eigenvalue $0$.