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1. Suppose $C$ is a closed polygon with several path-connected components. Is each of these components also closed?

If $C$ is a closed set that is not a polygon, the claim is not true: this can be seen by this example. There is a closed set with two path-components, each of which is not closed. But, if $C$ is a polygon, maybe the claim is true?

2. The above example is related to the following question: suppose $C$ is a path-connected polygon. Is the closure of $C$ path-connected too?

Again, when $C$ is not a polygon, the claim is not true, as shown by the exact same example. But, if $C$ is a polygon, maybe the claim is true?

3. Is it possible to impose on $C$ a condition slightly more general than polygonality, that still guarantees that claims 1 and 2 will be true?

Erel Segal-Halevi
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1 Answers1

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Since you mentioned polygons, I'll assume we're in $\mathbb R^2$ (note that the approach below does not generalize to higher dimensions). Consider some subset $X\subset\mathbb R^2$. For convenience I'll call it a "Jordan subset" if its boundary $\partial X$ is a Jordan curve. Simple polygons are a type of "Jordan subsets". Polygons can be considered as a finite union of simple polygons.

From the way you phrased your question, I suspect you wanted a formal proof of your various statements, so technically I should prove that my above claim is true. For a simple polygon, I'd say it's not too difficult. But for self-intersecting ones, it depends a little bit on what your formal definition of a polygon is.

From the Jordan curve theorem, $\partial X$ splits $\mathbb R^2\setminus X$ into two connected (path-connected actually) components: one is bounded (the interior of the Jordan curve), the other unbounded (the exterior). The interior of $X$ necessarily coincides with one of these components, and $X$ itself is this interior plus some, or all, points of $\partial X$. Either way the closure of $X$ satisfies $\overline X=X\cup\partial X$.

  1. If $X$ is a path-connected "Jordan subset", its closure is path-connected.

By the Jordan-Schönflies theorem there exists an homeomorphism $\psi:\mathbb R^2\rightarrow\mathbb R^2$ that maps $\partial X$ to the unit circle, its interior (Jordan curve meaning of interior) to the open unit disk, its exterior to the complement of the closed unit disk. So basically a Jordan curve in $\mathbb R^2$ is really just a circle that was continuously deformed. For any $x\in\overline X$ and $y\in X$ we know there exists a path $\gamma$ in the unit disk (or its complement) that connects $\psi(x)$ to $\psi(y)$. Because $\psi^{-1}$ is continuous, $\psi^{-1}(\gamma)$ is a path connecting $x$ to $y$. It follows that $\overline X$ is path-connected.

  1. Let $X_1,\ldots,X_n$ be $n$ closed "Jordan subsets" and $X=\bigcup_{i=1}^nX_i$. Assume $X$ has $k$ path-connected components $Y^1,\ldots,Y^k$. For any $1\le j\le k$, $Y^j$ is closed.

From 2, each $X_i$ is path-connected. So if $Y^j$ meets $X_i$, it contains the whole $X_i$. It follows that $Y^j$ is a finite union of some of the $X_i$. A finite union of closed sets is closed, so $Y^j$ is closed.

Just a quick note, the Jordan-Schönflies theorem doesn't hold in higher dimension, so you need another approach or another class of sets altogether.

N.Bach
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  • Small comments on the proof of 2: (a) $\psi$ is a map from $X$, so if $y\in X$, then we should have $\psi(y)$ and not $\psi^{-1}(y)$? (b) What is meant by "... in the unit disk (or its complement)?" Isn't the path only in the unit disk? – Erel Segal-Halevi Jun 16 '17 at 12:41
  • @ErelSegal-Halevi (a) Yeah I messed up $\psi$ and $\psi^{-1}$, edited. (b) If $X$ is bounded, $\psi$ maps its interior to the unit disk, and we can find a path $\gamma$ in the closed unit disk. If $X$ is unbounded, $\psi$ maps it to the complement of the unit disk, and we can find $\gamma$ in the closure of that complement. – N.Bach Jun 16 '17 at 13:08
  • Isn't the equivalence between a polygon and the unit disc, true only for simple polygons? What if the polygon has holes? – Erel Segal-Halevi Jun 17 '17 at 21:49
  • @ErelSegal-Halevi That's why I'm taking a union of them in (1). Even for (2), just take $X=\bigcup_{i=1}^nX_i$. You can prove that $\overline X=\bigcup_{i=1}^n\overline{X_i}$ and you then get property (2) also for non-simple polygons. – N.Bach Jun 18 '17 at 00:55
  • ‎Thanks a lot. By what name can I cite your answer? – Erel Segal-Halevi Jun 20 '17 at 13:23
  • @ErelSegal-Halevi Hmm, what do you mean by "cite [my] answer" ? – N.Bach Jun 20 '17 at 16:45
  • I would like to use this lemma in a paper I am writing – Erel Segal-Halevi Jun 21 '17 at 04:14
  • @ErelSegal-Halevi Oh... If you want a name for the lemma itself, call it whatever you like (I have a terrible naming sense). If you want my name, I still haven't decided whether I want to disclose my real name on this website or not... If you really want it, I could communicate it to you by mail I guess. – N.Bach Jun 21 '17 at 18:30
  • Whatever you decide... if you want, you can find my email in my profile. – Erel Segal-Halevi Jun 22 '17 at 04:18