I find a question by myself, and I do not know if it is an interesting question.

Let $X \subseteq \mathbb{R}^n$ be a bounded connected set. And I define a "bad point" $b \in \mathbb{R}^n$ with respect to $X$ if there exists $r > 0$ such that $X \setminus O(b, r)$ is not connect, where $O(b, r)$ is a ball with center point $b$ and radius $r$. Let $B(X)$ be a set of all "bad point" with respect to $X$.

Strong question: Is there a bounded connected set $X$ such that $B(X) = \mathbb{R}^n$?

Weak question: Is there a bounded connected set $X$ such that $X \subseteq B(X)$?

I have no idea how to solve my question. I give some examples of my definition.

(1) If $X$ is a ball or spherical surface, then $B(X) = \varnothing$.

(2) If $X = \{(x,0, \ldots, 0) \mid x \in [0,1] \}$ is a close line segment, then $$B(X) = \{(x,a_{1}, \ldots, a_{n - 1}) \mid x \in (0,1), a_{i} \in \mathbb{R} \}$$

(3) If $X = \{(x,0, \ldots, 0) \mid x \in (0,1) \}$ is a open line segment, then $$B(X) = \{(x,a_{1}, \ldots, a_{n - 1}) \mid x \in (0,1), a_{i} \in \mathbb{R} \}$$

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    $\begingroup$ Wouldn't you be able to take $X$ to be one entire axis? So $X = \{(x, 0, \ldots, 0) : x \in \mathbb R\}$, or do I misunderstand your question? This is essentially as your examples (2) and (3), but now $B(X) = \mathbb{R}^n$ and that should answer your strong question. $\endgroup$ – Mark Kamsma Apr 19 at 12:24
  • $\begingroup$ @MarkKamsma Yes, I considered this example, I forget to write that X is bounded. $\endgroup$ – TeamBright Apr 19 at 12:34
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    $\begingroup$ @YuiToCheng that is assuming we fix $r$ at the start, but the question is whether for every $b \in \mathbb R^n$ we can find $r$ such that $X - O(b, r)$ is disconnected. So $r$ can depend on $b$. $\endgroup$ – Mark Kamsma Apr 19 at 12:59
  • $\begingroup$ @TeamBright I see, that makes the question less trivial. Still, you have already answered your own weak question with example (3). The strong question, I wouldn't know an answer to right away. $\endgroup$ – Mark Kamsma Apr 19 at 13:01
  • $\begingroup$ @MarkKamsma Oh, yes. I answered the weak question. I did not realized it. I focused some sets which is $n$-dimensional set. $\endgroup$ – TeamBright Apr 19 at 13:18

As commented, the weak question is already solved.

For the strong one, I think there is a simple example if I've understood your question well. Consider a cross $C$ in $\Bbb{R}^2$ center at the origin, i.e the union of $I_x=\{(x,0)\mid x\in[-1,1]\}$ and $I_y=\{(0,y)\mid y\in[-1,1]\}$.

Any ball that covers $(0,0)$ by not the whole cross would disconnect $C$. But you can see that for any point $b\in\Bbb{R}^2\setminus\{(0,0)\}$, the (closed) ball $O(b,|b|)$ contains $(0,0)$ but not the whole cross (if you mean open ball then just take a slightly bigger radius). Note that for $b=(0,0)$ you can take a small radius and it will work as well.

So, for every $b\in\Bbb{R}^2$ there exists $r>0$ such that $C\setminus O(b,r)$ is disconnected. This example can be easily generalized to any higher $\Bbb{R}^n$.

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    $\begingroup$ Love it, easy and elegant. Only case that is left is $n = 1$, but the answer to that is an (almost trivial) exercise left to the reader. $\endgroup$ – Mark Kamsma Apr 19 at 13:17
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    $\begingroup$ Right, I didn't consider the case $n=1$, where that is surely not possible, since a connected set in $\Bbb{R}$ is an interval, so any point outside of the interval is not a bad point. @MarkKamsma $\endgroup$ – Javi Apr 19 at 13:20

Your weak question is most assuredly true -- just take $X=\{(x,0,\ldots,0):x\in(-1,1)\}$. Since deleting any sufficiently small neighborhood of any point in $X$ will leave behind two disjoint line segments, we are done.

Your strong questions is interesting, but I don't know how to answer it well, so bear with me as I describe a construction in $\mathbb{R}^2$:

Consider the curve (in polar coordinates) $$ (r,\theta) = \left(\exp(f(\theta)),\theta\right) $$ where $f(\theta)$ is any continuous, monotonically increasing function with range $\left(-\ln(2),0\right)$, such as an appropriately scaled and translated $\arctan$ function. Take this curve, a circle with radius $1/2$ centered at the origin, and the unit circle, and define $X$ to be their union.

Where are the "bad points"? If you take a point outside the unit circle, say $(0,2)$, then we can find a radius $r=1+\epsilon$ which will "cut" the spiral off from the unit circle, leaving us a disconnected set. The same will be true for a point inside the circle of radius $1/2$, and a point inside the annulus can have a radius chosen to "cut" the two circular boundaries off from one another. Thus, all of $\mathbb{R}^2$ is "bad".

How do we take this construction and make it work in higher dimensions? There may or may not be an elegant equation for it, but the idea definetely generalizes -- take some increasingly fine "spiral" along the surface of hyper-sphere shells, and then take many such shells nested together, limiting towards to bounding shells, and link them together with line segments.

I wish I could describe that better, but I don't know how. Hopefully your mental picture is the same as mine.

EDIT: Please just use Javi's answer if you want a good example. This is far too complicated, even in $\mathbb{R}^2$.

  • $\begingroup$ I do not directly see picture you are trying to describe, but isn't this unbounded? The $y$-coordinate ($\theta$) can get arbitrarily large. $\endgroup$ – Mark Kamsma Apr 19 at 13:12
  • $\begingroup$ @MarkKamsma The polar coordinate description of the curve is unbounded in $\theta$, but the radius is bounded, leaving the entire object bounded. $\endgroup$ – ItsJustSamplingBro Apr 19 at 13:13
  • $\begingroup$ Ah, I wasn't aware you were using polar coordinates. You may want to explicitly note that (even though you use the standard notation for it). On another note: once you have found an example in $\mathbb R^2$, I think you have an example for $\mathbb R^n$, $n \geq 2$, because you can just include your example as a subspace. $\endgroup$ – Mark Kamsma Apr 19 at 13:15
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    $\begingroup$ @MarkKamsma Truthfully my example doesn't matter anymore, as Javi's answer has a much simpler construction that will work in higher dimensions. $\endgroup$ – ItsJustSamplingBro Apr 19 at 13:15
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    $\begingroup$ Also thank you, you try to find a more general solution. It is a good approach. $\endgroup$ – TeamBright Apr 19 at 13:29

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