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This is a question that has been asked before , but now I came up with another proof so I would like to know if I am doing anything wrong.

Let $A$ be an open connected supspace of $\mathbb{R}^2$. Consider point $x \in A$. Let $B$ be a subset of $A$ such that it contains all the points with a path to $x$. Since $A$ is supspace of $\mathbb{R}^2$ and $B$ a subset of $A$, it follows $B$ belongs to the subspace topology so it is also open. Any point has a path to itself. So $B$ is not empty. Then $B\neq \phi$. Moreover we know $B \cap (A-B) = \phi$ and $B \cup (A-B) = A$. By similar reasoning as before, $A-B$ is also in the subspace topology so it is open. Since $A$ is connected, then it cannot have a separation. This implies $A-B=\phi$ and so $A=B$. We said $B$ is the set of all points that have a path to an arbitrary point $x \in A$. It follows $A$ is path connected.

Is this a valid proof?

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    $\begingroup$ You've assumed $B$ is open, or possibly closed. It could be neither, in which case $B$ and $A \setminus B$ would not form a separation of $A$. $\endgroup$ Commented Oct 9, 2020 at 5:39
  • $\begingroup$ I am not sure where I have made that assumption. I say $B$ is a subset of $A$ such that it contains all the points with a path to $x$. I don't see if I have said it is open or closed. Could you elaborate please? $\endgroup$ Commented Oct 9, 2020 at 5:43
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    $\begingroup$ Separations have to be open sets. It's part of the definition of a disconnected set. And you are arguing that "Since $A$ is connected, then it cannot have a separation." $\endgroup$ Commented Oct 9, 2020 at 5:45
  • $\begingroup$ Note that you have not used the definition of $B$. $\endgroup$ Commented Oct 9, 2020 at 5:49
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    $\begingroup$ The implication $B\subseteq A \implies B\in \tau_A$ is only true when $A$ is discrete (+connected means $A$ is a singleton, which isn't open in $\mathbb{R}^2$, so this is a contradiction). You actually need to use the metric topology on $\mathbb{R}^2$ and the assumption $A$ is open to show $B$ is open in $A$. $\endgroup$ Commented Oct 9, 2020 at 6:23

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Look at the chain-characterisation of connectedness I gave here, which is close to your idea. Take a cover of $O$ by open balls that stay inside $O$. This can be done obviously as $O$ is open in $\Bbb R^2$. If $x,y \in O$ we can apply the charcterisation to this cover and have a finite sequence of open balls (with some centre and radius) $B_1, B_2, \ldots B_n$ such that $x \in B_1$, $y \in B_2$ and $B_i \cap B_{i+1} \neq \emptyset$ for $i=1,\ldots, n-1$.

Now we can form a polygonal path (all straight line segments, as all open balls are convex, these all lie within a ball) from $x$ to some intersection point of $B_1 $ and $B_2$, next to some intersection of $B_2$ and $B_3$, and so on till the some last intersection point to $y$. So then we have a nice polygonal path from $x$ to $y$ entirely inside $O$. So $O$ is path-connected.

In general we show in this way that a locally path-connected connected space is path-connected.

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  • $\begingroup$ Now we can form a polygonal path --- Thus, open connected subsets of the plane are actually polygonally connected (also called $L$-sets). A professor I once had for a complex analysis class, after presenting essentially this proof (same proof as in our textbook), made the off-hand remark that we've actually shown something stronger, namely polygonal connectedness (not mentioned in our text). This was an epiphany moment for me, (continued) $\endgroup$ Commented Oct 9, 2020 at 23:16
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    $\begingroup$ as I began to realize how one can sometimes obtain an unexpected stronger result from a proof of a theorem, and thus a proof was not necessarily just a "legal justification" for the theorem statement. That professor, by the way, wound up being my Ph.D. advisor. $\endgroup$ Commented Oct 9, 2020 at 23:23
  • $\begingroup$ @DaveL.Renfro I have indeed often found that new notions in topology often emerge from how certain proofs are done as well. The proof can then be repurposed to prove some generalisation, as it were. $\endgroup$ Commented Oct 9, 2020 at 23:29
  • $\begingroup$ (correction to my first comment) I think the collection of polygonally connected sets is actually a proper superset of the collection of $L$-sets. (I have a folder of papers on this, but I don't have time to look at it now, even if I knew where it was.) Regarding finding new results by careful examination of a proof, it's important that some mathematical discretion be used to avoid $\epsilon$-tweaking of results, something that seems to be a lot more prevalent in recent years due to the huge increase (from several decades ago) in emphasis on numbers of publications for career advancement. $\endgroup$ Commented Oct 9, 2020 at 23:34
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Your idea is right. I would propose a sort of reformulation that appears clearer to me.

Let $a\in A$ and $f_a : A \to \{0,1\}$ be defined by $f_a(x) = 1$ if there exists a path connecting $a$ and $x$, $0$ otherwise. Let us show that $f_a$ is continuous, thus it will be constant (as $A$ is connected), and as $f_a(a)=1$, all $x\in A$ will be connected by a path to $a$, and hence $A$ would be path-connected.

Let $x \in A$. As $A$ is open, there exists $\varepsilon >0$ for which $\Delta(x,\varepsilon) \subset A$. As a disk in $\mathbb{R}^2$ is path connected, every point in $\Delta(x,\varepsilon)$ can be connected to $x$ by a path.

Suppose $f_a(x) = 1$. Then there exists a path from $a$ to $x$. By gluing this path to a path from $x$ to $y \in \Delta(x,\varepsilon)$, there exists a path from $a$ to any $y\in \Delta(x,\varepsilon)$.

Suppose $f_a(x)=0$. Then there is no path between any $y\in \Delta(x,\varepsilon)$ and $a$, otherwise gluing it with a path from $y$ to $x$ would contradict $f_a(x)=0$.

We have shown that $f_a$ is locally constant, hence is continuous. This concludes the proof.

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