# Proof that a connected open set in $\mathbb{R}$ is path-connected

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).

• How do you justify that this final ball indeed exists?
– fwd
Feb 22, 2023 at 14:56
• What if the radii of these balls that you got kept on rapidly decreasing? Feb 22, 2023 at 14:57
• 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. Feb 22, 2023 at 15:22
• @drhab In order for it to quality as a proof, must I justify the existence of the final ball? Feb 22, 2023 at 15:25
• At least, and I do not guarantee that this is possible and enough. See the anwer to your question. Feb 22, 2023 at 15:36

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.

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.

• "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 $a,b$. Feb 24, 2023 at 17:14
• @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. Feb 25, 2023 at 20:58