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I need to show that in $R^n$ : Convex implies Connected

I try to prove the contrapositive. Disconnected sets cannot be convex. A is a disconnected set. Assume A is convex. $A = B \cup C$ where B and C are non empty disjoint open sets. As they are non-empty I can pick $b \in B$ and $c \in C$. As B is open, there exists $x>0$ such that $B(b,x) \subseteq B$ (open ball contained in B). As C is open, there exists $y>0$ such that $B(c, y) \subseteq B$ (open ball contained in C). We consider $$ t = \frac{x}{x+y}$$ $t \in (0,1)$. By convexity of A, $m=tb+(1-t)c \in A$. As B and C are disjoint there are two possible cases.

Case 1: $m \in C$ and $m \notin B$ Case 2: $m \in B$ and $m \notin C$ I feel like both should lead to a contradiction but I'm not sure how to proceed. I don't think B and C are necessarily convex but I probably need to use that and the fact that we're operating in an Euclidean space but can't think of anything. Thank you so much in advance!!

P.S. I cannot use path-connectedness at this point in the textbook so I cannot rely on the other post about this. Sorry!

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Shivansh Raman
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    convex => path-connected => connected, a path from $x$ to $y$ being given by $t\mapsto ty+(1-t)x$ – Jakobian Sep 02 '24 at 14:26
  • Thanks!! I'm sorry but they haven't taught us path-connectedness in this real analysis textbook I'm working through so I was hoping to use another way – Shivansh Raman Sep 02 '24 at 14:29
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    Suppose a convex set $A$ can be written as union of two non-empty open sets $U_1, U_2$, pick $x\in U_1, y\in U_2$. Continuity of the path above $p:t\mapsto ty+(1-t)x$ implies that $[0, 1] = p^{-1}[A] = p^{-1}[U_1]\cup p^{-1}[U_2]$ is a disjoint union of two non-empty open sets. This contradicts that $[0, 1]$ is connected @ShivanshRaman – Jakobian Sep 02 '24 at 14:33
  • @Jakobian Why not an official answer? – Paul Frost Sep 02 '24 at 23:47
  • @PaulFrost why would I be answering a duplicate/FAQ – Jakobian Sep 02 '24 at 23:52
  • @Jakobian Then why not close it as a duplicate? – Paul Frost Sep 03 '24 at 08:06
  • You claim "$A = B \cup C$ where $B$ and $C$ are nonempty disjoint sets", but this is incorrect. You only have that $A \subseteq B \cup C$. – MartianInvader Sep 03 '24 at 16:43
  • @PaulFrost sure, it's just bothersome to search for a duplicate candidate, and I'm lazy – Jakobian Sep 03 '24 at 17:17
  • @MartianInvader It's techhnically not incorrect, it depends where are $B, C$ open. If they are open in $A$ then everything is fine. – Jakobian Sep 03 '24 at 17:18
  • @Jakobian Sure, but that's clearly not what was meant, given the follow-up logic about the existence of balls contained in $B$ and $C$. – MartianInvader Sep 03 '24 at 20:21
  • @MartianInvader It might be that the author didn't mean that, but the balls still don't raise concern as we could be talking about the balls relative to the space $A$. – Jakobian Sep 03 '24 at 21:41

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