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Let $X$ be (Edit: a closed convex subset of ) the unit sphere $Y=\{x\in \ell^2: \|x\|=1\}$ in $\ell^2$ with the great circle (geodesic) metric. (Edit: Suppose the diameter of $X$ is less than $\pi/2$.) Is it true that every decreasing sequence of nonempty closed convex sets in $X$ has a nonempty intersection? (A set $S$ is convex in $X$ if for every $x,y\in S$ the geodesic path between $x,y$ is contained in $S$.)

(I edited my original question.)

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    @Muhammad. If $x\neq -y$, this means $x,y\in S$ implies $\cos(t)x+\sin(t)u \in S$ for $t$ between $0$ and $\cos^{-1}(x,y)$, where $u$ is the unit vector in the direction of $y-(x,y)x$, $(x,y)$ being the inner product of $x,y$.2011-10-31

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No. For example, let $A_n$ be the subset of $X$ consisting of vectors that are zero in the first $n$ co-ordinates.

EDIT: this assumes that when $x$ and $y$ are antipodal, convexity of $S$ containing $x$, $y$ only requires that at least one of the great-circle paths is contained in $S$. If it requires all of them, then the $A_n$ are not convex. t.b. points out in the comments that in this case we can set $A_n$ to consist of all vectors in $X$ that are zero in the first $n$ co-ordinates and non-negative in the remainder.

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    The answer to my e$d$ited question is true; see my post "Geometry of the Hilbert sphere" in Mathoverflow.2011-11-02