Show that if $U \subseteq C$ and $V \subseteq C$ are both open and convex sets, then the set $U \cap V \subseteq C$ is open and convex as well.

Question

I think you have to prove that as it is the intersection then both are in open and convex sets seeing as they are on their own. Don't really know how to put this down in notation though.

Is your $C$ supposed to be $\mathbb C$? – MPW Mar 16 '15 at 12:19 — MPW, Mar 16 '15 at 12:19

5xum · Answer 1 · 2015-03-16T13:04:43.653

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Proving it is open: Depending on your definition of open sets, this can be very easy (in topology, any finite intersection of open sets is open) or just easy. If $C$ is a metric space, then take any point $x\in U\cap V$ and find first an $\epsilon_V$ such that $d(x,y)<\epsilon_V$ implies that $y\in V$. Now, do the same with $U$. Taking $\epsilon=\min\{\epsilon_V, \epsilon_U\}$, what can you say about $y$ if $d(x,y)<\epsilon$?
Proving it is convex: Take any two points $x,y$ in $U\cap V$. Now, because $U$ is convex, what can you say about the line between $x$ and $y$? And what does the convexity of $V$ give you?

edited Mar 16 '15 at 13:04

answered Mar 16 '15 at 12:19

5xum

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I think you mean "convex" not "complex" in number 2. – M.G Mar 16 '15 at 12:56
@M.G Thank you. I did. – 5xum Mar 16 '15 at 13:05

Show that if $U \subseteq C$ and $V \subseteq C$ are both open and convex sets, then the set $U \cap V \subseteq C$ is open and convex as well.

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