4

Suppose $U \subset \mathbb R^2$ is an open subset. Let $(x,y)$ denote the coordinates. For any $(x_0, y_0) \in U$, if we fix $x_0$, the set $U_{x_0} = \{y \in \mathbb R: (x_0, y) \in U\}$ has two connected components; if we fix $x_0$, the set $U_{y_0} = \{x \in \mathbb R: (x, y_0) \in U\}$ has two connected components. How many connected components could $U$ have? Is it $4$?

MyCindy2012
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    Take $n$ concentric squares with sides parallel to the axes. Look at the set of all corners of all of the squares (this is a discrete set of points). Now take small enough open balls (each one of the same radius) centered at those corners. The union of those balls is an open set satisfying your criteria. The number of connected components is $4n$. – shalop Feb 27 '19 at 03:11
  • @Shalop That sounds like a correct example. So, maybe it's ${2,4,8,12,\cdots}$? Obviously, you can create an example with two connected components. Also, could you please tell me where my proof goes wrong? – stressed out Feb 27 '19 at 03:16
  • Your proof goes wrong because you say that if a set has k connected components, then its image under a continuous function also has k connected components. This is very false. Please try to see why. – shalop Feb 27 '19 at 03:21
  • @Shalop You're absolutely right. :) I need injectivity for it to hold, at least. $(-1,0)$ and $(0,1)$ are mapped to the same interval by $x^2$. So, your example creates sets with $4n$ connected components, and I already showed an example with $2$ connected components. But what about other numbers? How to overrule them? – stressed out Feb 27 '19 at 03:23
  • Idk, but I liked your example with the parallel lines (couldn’t think of that myself). – shalop Feb 27 '19 at 03:36
  • @Shalop That doesn't seem correct. There's a problem with sections that go near boundaries of those open balls. I mean pick a ball and take a section tangent to it, say tangent vertically from the right side. Balls are open so it doesn't meet it. Now it has to meet some two other balls. And then by those balls being open we can translate that section a bit to the left so that it still meets those two balls. But now it also meets the original ball it was tangent to. Making the number of components in the section more than $2$. – freakish Feb 27 '19 at 08:51
  • @stressedout But I think we've made the question more difficult than it is. The assumption is that "for any $(x_0,y_0)\in U$". Meaning that small balls (of equal radius) around ${(0,0),(1,0),(0,1),(1,1)}$ satisfy it. And this can be easily extended (via nesting) to any number divisible by $4$. Although "for any $(x_0,y_0)\in\mathbb{R}^2$" (which I thought is the original assumption) is definitely more interesting. – freakish Feb 27 '19 at 08:54
  • @freakish I don’t think the question was ever asking for (x0,y0) in R^2 (it hasn’t been edited). Only points in U matter. So I don’t think my example was wrong. – shalop Feb 27 '19 at 12:46
  • @Shalop Yes, I know. It's my wrong interpretation. Although answers to "$(x_0,y_0)\in U$" were already given (and deleted). To me it's just not interesting tbh. – freakish Feb 27 '19 at 12:57
  • @freakish even for this easier version, is it simple to prove that the set cannot have 3 components (for instance)? – shalop Feb 27 '19 at 13:04
  • @freakish Yeah.I wanted to tell you that but you said you'd gone to sleep. However, it's still open to see whether the number of connected components can be odd or of the form $4n+2$. We know that $2$ and $4n$ work, but we know nothing about other cases yet. – stressed out Feb 27 '19 at 20:59
  • I found an example with five components, basically using an annulus and four disks. – shalop Mar 07 '19 at 18:51

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