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The English translation of Markushevich's Theory of functions of a complex variable contains the following sufficient condition for the connectedness of a compact set on the complex plane.

Theorem 4.1 A nonempty compact set F is connected if given any two points $z,z'$ in F and any $\epsilon >0$ there exists a finite number of points $z=z_0,z_1,\ldots,z_n=z'$ in F such that $|z_{k-1}-z_k| < \epsilon$ for $k=1,\ldots,n .$

I am interested in the converse of the above theorem, whose validity is assumed in Theorem 4.17 of the book but not proved.

Given a nonempty connected compact set F, and any two points $z,z'$ in it and any $\epsilon > 0$ there exists a finite number of points $z=z_0,z_1,\ldots,z_n=z'$ in F such that $|z_{k-1}-z_k| < \epsilon$ for $k=1,\ldots,n.$

I think I have a reasonably good plausibility argument for the correctness of the above, but I am struggling to convert to a proof.

My argument is:

  1. Choose $\delta > 0$ and consider all squares of width $\delta$ of the form $[k\delta,(k+1)\delta] \times [j\delta,(j+1)\delta]$ where k and j are integers.
  2. Let $\mathbb{U}$ be the collection of all squares above that intersect F, let the union of all sets in $\mathbb{U}$ be U.
  3. Then U is connected and its interior, G, is an open connected set that contains F.
  4. Since G is pathwise connected there is a continuous path $\gamma(t), 0 \leq t \leq 1$ in G with $\gamma(0) = z$ and $\gamma(1)=z'.$
  5. The uniform continuity of $\gamma$ implies the existence of points $z=z_0',z_1',z_2',z_{n-1}',z_n'=z'$ on the image of $\gamma$ such that $|z_{k-1}' - z_k'| \le \epsilon'$ for any given $\epsilon' > 0.$
  6. Replace each point in the finite sequence above (except the two endpoints) with a point in F that lies in a square in $\mathbb{U}$ containing it. Call the new sequence $z_0,\ldots,z_n$, by making $\delta$ and $\epsilon'$ sufficiently small we can ensure $|z_{k-1}-z_k| < \epsilon$ for $k=1,\ldots,n.$

Is the converse stated above corrected, if not, is there a nice counterexample? If the converse is correct, is my argument above correct. I am suspicious about 3., I can "see" that its true for some simple cases, but I wonder if I am missing something. Is there a simpler proof of the converse, if it is true?

Added Later: From Eric's comments.

  1. Every point in F lies in the interior of U, since a point in F either lies in the interior of a square in $\mathbb{U}$ or lies in the corner or edge of a square in $\mathbb{U}$. In the latter case the point lies in multiple squares in $\mathbb{U}$ and is in the interior of their union.
  2. Any point in the interior of U can be connected with a straight line lying in the interior of U to a point in F, which implies the interior of U is an open connected set.

Rest follows as above.

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Your proof is correct. To flesh out step 3 a bit, here is what you want to prove. For every $x\in F$, you want to prove that a neighborhood around $x$ is contained in $U$. For every $y\in G$, you want to prove that if $x\in F$ is in the same closed $\delta$-square as $y$, then there is a path from $x$ to $y$ which is contained in $G$. Both of these are straightforward to prove, though they require a bit of casework if $x$ or $y$ is in the boundary of a square. Given these facts, we can conclude that $G$ is an open set containing $F$, and every point of $G$ is in the same connected component of $G$ as some point of $F$. Since $F$ is connected, this means $G$ is connected. Since $G$ is open, this means $G$ is path-connected, so you can proceed to step 4.

You can get a similar and perhaps simpler proof along the lines you are suggesting by throwing out the $\delta$-mesh and instead just defining $G=\{z\in\mathbb{C}:|z-z'|<\delta\text{ for some }z'\in F\}$. Then $G$ is open, contains $F$, and every point of $G$ can be path-connected to some point of $F$, so $G$ is connected and hence path-connected. Now you can follow your proof starting from step 4.

There is another simple proof, which works for any connected metric space at all. Suppose $F$ is a connected metric space and fix $\epsilon>0$. Say that two points $z,z'\in F$ are $\epsilon$-connected to each other if there exists a finite number of points $z=z_0,z_1,\ldots,z_n=z'$ in F such that $d(z_{k-1},z_k) < \epsilon$ for $k=1,\ldots,n$. It is easy to see that being $\epsilon$-connected is an equivalence relation on $F$. Furthermore, if $d(z,z')<\epsilon$, then clearly $z'$ is $\epsilon$-connected to $z$. This implies that every equivalence class is an open set. But the complement of every equivalence class is also open, being the union of the other equivalence classes. Thus since $F$ is connected, there can only be one equivalence class. Thus any $z,z'\in F$ are $\epsilon$-connected.

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    $\begingroup$ Thanks. If two squares only meet in a corner, then the corner cannot lie in F, because then four squares would contain the corner point in F. So if we make $\delta$ smaller, such cases can be perhaps eliminated? $\endgroup$ – Arin Chaudhuri Oct 5 '15 at 1:26
  • $\begingroup$ Oh, you're right. So actually your argument does work: $G$ contains all of $F$, and every point of $G$ can be connected by a path in $G$ to a point of $F$ (namely, any point of $F$ which is in the same square; if that point is on the boundary of the square, then as you note the boundary edge it rests on is contained in $G$). Thus $G$ is connected since $F$ is, and so $G$ is path-connected. $\endgroup$ – Eric Wofsey Oct 5 '15 at 1:29

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