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I am trying to prove that a connected open set in $\mathbb{R}$ is path-connected. Is the following correct?

Consider an arbitrary open set in $\mathbb{R}$, $\Phi$. Choose any two points $a,b \in \Phi$, where $b>a$. I now prove the existence of a line, contained in $\Phi$, starting at $a$ and terminating at $b$.

Start at $a$. By the openness of $\Phi$, there exists $\epsilon>0$ such that $B_{\epsilon}(a) \subset\Phi$—say, $\epsilon_1$. Now, consider $a+\epsilon_1$. Again by the openness of $\Phi$, there exists some $\epsilon_2>0$ such that $B_{\epsilon_2}(a+\epsilon_1)\subset\Phi$. Note that $B_{\epsilon_1}(a) \cap B_{\epsilon_2}(a+\epsilon_1)$ is non-empty. Now, continue this construction, until a ball of the following type is constructed: $B_{x}(y)$, where $x,y$ are such that $y+x>b$. Note that this final ball is, by construction, guaranteed to be contained in $\Phi$. Also, note that the process will never arrive at a point not contained in $\Phi$, as, by connectedness, there are no gaps between $A$ and $B$.

Now, the open interval ($A$, $B$) is covered by a sequence of intersecting open balls. Further, all these balls are contained within $\Phi$. It follows that points $A$ and $B$ can be connected by a straight line contained in $\Phi$ (by connecting the central points of each of the constructed balls).

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  • $\begingroup$ How do you justify that this final ball indeed exists? $\endgroup$
    – fwd
    Feb 22, 2023 at 14:56
  • $\begingroup$ What if the radii of these balls that you got kept on rapidly decreasing? $\endgroup$ Feb 22, 2023 at 14:57
  • $\begingroup$ This story can not be qualified as a "proof". I think it is more handsome to prove that an open subset $U\subseteq\mathbb R$ that is not path-connected is not connected. $\endgroup$
    – drhab
    Feb 22, 2023 at 15:22
  • $\begingroup$ @drhab In order for it to quality as a proof, must I justify the existence of the final ball? $\endgroup$
    – Charles
    Feb 22, 2023 at 15:25
  • $\begingroup$ At least, and I do not guarantee that this is possible and enough. See the anwer to your question. $\endgroup$
    – drhab
    Feb 22, 2023 at 15:36

2 Answers 2

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As student13 has shown, your proof does not work.

So let $\Phi$ be a connected open set and $a, b \in \Phi$. We claim that the path $u : [0,1] \to \mathbb R, u(t) = a + t(b-a)$, has image in $\Phi$ which shows that $\Phi$ is path-connected.

The case $a = b$ is trivial. In case $a \ne b$ we may w.l.o.g. assume that $a < b$. Assume that there exists $t \in [0,1]$ such that $u(t) \notin \Phi$.

$U = \Phi \cap (-\infty,u(t))$ and $V = \Phi \cap (u(t),\infty)$ are open subsets of $\phi$ such that $a \in U, b \in V$, $U \cup V = \Phi$ and $U \cap V = \emptyset$. This means that $\Phi$ is not connected, a contradiction.

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This proof will not work. To show you why, consider $\Phi = (-1,2)$, $a = 0$ and $b = 1$. Then consider the series of balls, $B_n = B_{\frac{1}{2n}}(\frac{n-1}{2n})$. This series of balls is as you described: every midpoint increases, and is contained in the previous ball, but $b$ will never be contained by any of the balls (the upper boundary of the balls will always remain at $\frac{1}{2}$). Just think of it: if $\Phi$ weren't (path) connected, you could also find such series of balls.

An easier proof is by contradiction. Assume $\Phi$ is connected, but not path connected. That means that there exist $a, b \in \Phi$ and a value $\gamma \in (0,1)$ such that $a + \gamma (b - a) \notin \Phi$. For ease of notation, let's call this number $q$, and we have $a < q < b$.

Now there are many ways to show that $\Phi$ cannot be connected. It kinda depends on what you're allowed to assume about connected sets, because I would call it trivial from here on. But one way to formally show this is via the separated sets definition of connected sets. We define $\Phi_L = \Phi \cap (-\infty, q)$ and $\Phi_R = \Phi \cap (q, \infty)$. Clearly, $\Phi_L \cup \Phi_R = \Phi$ and clearly, $\Phi_L$ and $\Phi_R$ do not overlap each other's closure (i.e. $\Phi_L \cap \overline \Phi_R = \overline \Phi_L \cap \Phi_R = \emptyset$). Therefore, $\Phi_L$ and $\Phi_R$ are separated.

So we have shown that $\Phi$ consists of two separated sets, which is one of the definitions of an unclosed set. Hence a contradiction.

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  • $\begingroup$ "Assume $\Phi$ is connected, but not path connected. That means that there exist $a, b \in \Phi$ and a value $\gamma \in (0,1)$ such that $a + \gamma (b - a) \notin \Phi$." This has to be proved. Being not path connected means that there are $a,b \in \Phi$ and no path in $\Phi$ connects $a,b$. $\endgroup$
    – Paul Frost
    Feb 24, 2023 at 17:14
  • $\begingroup$ @PaulFrost we're talking about the real line. So technically you're right, but it's quite trivial to prove that any continuous path connecting $a$ and $b$ must have the interval $[a, b]$ in its domain. But indeed, your proof is more formal and complete, I'll give it an upvote. $\endgroup$
    – student13
    Feb 25, 2023 at 20:58

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