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Jul 23, 2015 at 12:45 comment added Eric Wofsey Given any point $y$ in $V$, if you draw a line segment from $y$ to $x$, and then extend the line segment a little bit past $x$, you will enter the set $W\cdot U$. Thus $x$ lies on the line segment joining $y$ to some point in $W\cdot U$. Since $W\cdot U\leq C$ and $C$ is convex, $y\in C$ would imply $x\in C$.
Jul 23, 2015 at 12:41 comment added Rafał Gruszczyński Eric, I have some second thoughts about the proof. Could you please explain to me the transitions in the following passage: "If $y\in V\cdot C$, then the line from $y$ to $x$ extends into $W\cdot U$, so by convexity $C$ would have to contain $x$". I am not sure if I have understood it properly.
Jul 17, 2015 at 9:10 history bounty ended Rafał Gruszczyński
Jul 17, 2015 at 9:10 vote accept Rafał Gruszczyński
Jul 17, 2015 at 9:10 comment added Rafał Gruszczyński Ah, OK - I misinterpreted <i>dense in</i> property. Thank you very, very much!
Jul 16, 2015 at 6:58 comment added Eric Wofsey That's just the definition of $D+E$: it is the set of points which have a neighborhood in which $D\cup E$ is dense.
Jul 15, 2015 at 22:13 comment added Rafał Gruszczyński Eric, where does existence of an open ball $U$ in which $D\cup E$ is dense come from?
Jul 15, 2015 at 13:38 history answered Eric Wofsey CC BY-SA 3.0