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I need to prove that any convex subset of $R^k$ is connected. I have seen the proof in Rudin's book and on numerous websites but they all use some prior results. I want to do it without using results where one establishes something for some other question and then uses the result to prove this.

Andrés E. Caicedo
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Silver moon
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    Do you know about path connectedness? – dannum Oct 03 '14 at 03:18
  • Sorry I haven't been taught that. Is there any other proof? – Silver moon Oct 03 '14 at 03:20
  • What do you know about connectedness? –  Oct 03 '14 at 03:30
  • A set $E$ is connected if $E$ is not a union of two nonempty separated sets – Silver moon Oct 03 '14 at 03:32
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    It's much easier to prove that if a set is convex in $\mathbb R^k$, it is path connected, and since all path connected spaces are connected, you are done. The proof of each of these smaller pieces isn't too demanding. – dannum Oct 03 '14 at 03:36
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    Suppose $C$ is a non-empty convex set and that $C=U\cup V$ where $U,V$ are two disjoint nonempty open sets. Take two elements in $C$: $a\in U$, $b\in V$. Since $C$ is convex, define $\alpha:[0,1]\to C$ by $\alpha(t)=a(1-t)+tb$. Since $\alpha$ is continuous and connectedness is preserved by such functions, $\alpha([0,1])=P$ is connected, but $P\cap U$ and $P\cap V$ are two disjoint nonempty open sets whose union is $P$. This is a contradiction. –  Oct 03 '14 at 03:53
  • @Zircht- Can you please explain this? Since $α$ is continuous and connectedness is preserved by such functions, $α([0,1])=P$ is connected, but $P∩U$ and $P∩V$ are two disjoint nonempty open sets whose union is $P$. This is a contradiction – Silver moon Oct 03 '14 at 04:13
  • @Martin What Zircht means is that $P$, the image of $[0,1]$ by $\alpha$, is connected because $[0,1]$ is. But $P$ starts in $U$ because $\alpha(0)=a$ and ends in $V$ for $b=\alpha(1)$. So there are points in $C$ that are neither in $U$ neither in $V$ and this contradicts $C=U \cap V$. – Marc Bogaerts Oct 03 '14 at 07:16
  • Anyone can explan why $P \cap U$ is open? – user May 28 '15 at 03:15

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First a preliminary lemma:

Lemma: If $\mathcal{F}$ is a family of connected sets such that $\bigcap \mathcal{F} \neq \emptyset$ then $\bigcup \mathcal{F}$ is connected, for, if $\bigcup \mathcal{F}$ is not connected, then it can be written as the union of two disjoint open subsets $X$ and $Y$. Let $x \in \bigcap{F}$. Suppose that $x \in X$, so if $y \in Y$ then there is $F \in \mathcal{F}$ such that $y\in F$ so $F = (A\cap F) \cup (B\cap F)$, wich is a contradiction, since $A$ is open in $\bigcup \mathcal{F}$, $A \cap F$ is open in $F$ and $B \cap F$ is open in $F$.

Now, the problem:

Let $X\subset M$ be a non-empty convex subset and let $a \in X$. Let $L_v = \{x \in X \ |\ x = y + \lambda \ v$ for some $\lambda \in \mathbb{R} \}$. As $X$ is convex, $$X = \bigcup_{\|v\|=1} L_v$$

It remains to prove that each $L_v$ is connected, but since $L_v$ is homeomorphic to some real interval (because $X$ is convex!), $L_v$ is also connected. And since $\{a\} \subset \bigcap_{\|v\|=1} L_v $, lemma aplies and thus $X$ is connected.

Jonas Gomes
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