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I was reading this proof from here:

Theorem 6. An open set A in R n is connected if and only if it is pathconnected. Proof. Since path-connectedness implies connectedness we need to only show that A is path-connected if it is connected. Suppose A is nonempty and connected. Pick p ∈ A. Let U be the set of points in A that can be connected to p by a path in A. Let V = A \ U, so V is the set of points in A that cannot be connected to p by path in A. So A = U ∪ V . We claim that U and V are open. To show that U is open, let q ∈ U and γ be a path connecting p and q. Since A is open, there exists a open ball Br(q) ⊂ A. For each r ∈ Br(q), there is a path λ in A connecting q and r. We can take λ to be the straight line connecting q and r. Thus, Br(q) ⊂ U and U is open. Similarly one can prove that V is open. Thus, either U or V is the empty set by the connectedness of A

But I feel like showing connected implies path connected can be done another way.

Since $A$ is open it can be written as the union of basis elements which are balls.

I can pick some ball containing $p$ so $B_p$, and then I can union all the balls that don't contain $p$, so $\cup B^*=A-{p}$. $A=B_p \cup \cup B^*$, but since $A$ is connected $B_p \cap \cup B^*\neq \emptyset$

I know that $B_p$ is path connected because it is a ball, and since it intersects with $\cup B^*$ I can show A is path connected if $\cup B^*$ is also path connected. The problem is showing that the union of these balls is path connected, which is tricky.

I am certain the union of those balls is path connected, but based on this question, it seems difficult to prove for an arbitrary union; I only see countable examples in the answers.

My question is can the answers for countable sets be extended for arbitrary unions of balls, or is my approach incorrect because there likely are uncountably many balls in this instance?

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  • $\begingroup$ I don't see what your argument is trying to achieve. All you've done is reduced showing that $A$ is path-connected to showing that $A\setminus\{p\}$ is path-connected, but why would you think that is any easier than just showing $A$ is path-connected directly? $\endgroup$ – Eric Wofsey Nov 6 '16 at 23:21
  • $\begingroup$ The difference between your situation and the question you linked to doesn't have to do with uncountability--it has to do with not knowing anything about how the different balls are positioned relative to each other (all you know about them is that their union is somehow connected). On the other hand, the balls in the linked question are arranged in a very special way, so that the $n$th ball always intersects one of the earlier balls. $\endgroup$ – Eric Wofsey Nov 6 '16 at 23:24

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