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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?2012-11-01
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    @Berci yeah it is a hyperbole.2012-11-01
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    But it is not convex.2012-11-01
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    @Berci that is the whole point we are trying to find, create a convex hull for this set.2012-11-01
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    Technically that's a *hyperbola* in English. "Hyperbole" is a non-scientific word that means "exaggeration".2012-11-01

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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 $(1,1-\varepsilon)$?2012-11-01
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    @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.2012-11-01
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    @gt6989b we can also conclude from here that co(X) of a closed set is not necessarily closed.2012-11-01
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    @teodory agree.2012-11-01