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What is the coX of {(x,y) $\in$ $\mathbb R^2$ : y = $1\over1+x$, $x \ge 0$ } ?

coX is the convex hull.

I couldn't figure out. coX should be the smallest convex set that contains the set but in this case should it be the hyperbola itself. Thanks for helping.

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Line?? It is a hyperbole. Which line were you thinking about? – Berci Nov 1 '12 at 20:38
@Berci yeah it is a hyperbole. – teodory Nov 1 '12 at 20:39
But it is not convex. – Berci Nov 1 '12 at 20:40
@Berci that is the whole point we are trying to find, create a convex hull for this set. – teodory Nov 1 '12 at 20:44
Technically that's a hyperbola in English. "Hyperbole" is a non-scientific word that means "exaggeration". – Hew Wolff Nov 1 '12 at 21:29

I think $co(X) = \{ (x,y) \in \mathbb{R}^2 : x > 0, \frac{1}{1+x} \leq y < 1 \} \cup {(0,1)}$. The basic idea is, you want to start from the point $(0,1)$ and draw a ray to anywhere on $X$, and the entire ray must be included in the $co(X)$.

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But.. How would $(1,1)$ be in $co(X)$, for example? (I think it is only in the closure). But, what about $(1,1-\varepsilon)$? – Berci Nov 1 '12 at 20:43
@Berci You are right, the entire ray $(x,1)$ for $x>0$ is in the closure, not in the hull itself. Altered the answer (found another bug), please see the change. – gt6989b Nov 1 '12 at 21:02
@gt6989b we can also conclude from here that co(X) of a closed set is not necessarily closed. – teodory Nov 1 '12 at 21:34
@teodory agree. – gt6989b Nov 1 '12 at 21:38

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