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I've encountered the following question:

Say $A \subseteq \mathbb{R}^n$ is a closed and connected set. Prove $\{x \in \mathbb{R}^n \mid d(x, A) \le \varepsilon\}$ is path-connected.

I'm not really sure how to approach this question. It appears under the section of Compactness.

Any help will be appreciated!

Michael Hardy
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amirbd89
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2 Answers2

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Hint: For any connected (nonvoid) subset $A$ of $\mathbb{R}^n$ the set $\{x\ | \ d(x,A) < \epsilon\}$ is connected, since it is the union $\cup_{a\in A} B(a,\epsilon)$, every ball $B(a,\epsilon)$ is connected, and also $A$ is connected ( think of a fishbone). Moreover, $\{x\ | \ d(x,A) < \epsilon\}$ is open, therefore, $\{x\ | \ d(x,A) < \epsilon\}$ is $path\ connected$. Connect any point $y$ in $\{x\ | \ d(x,A) \le \epsilon\}$ with a closest point in $a \in \bar A$, the half open segment $(y,a]$ will lie in $\{x\ | \ d(x,A) < \epsilon\}$ (this is the idea of @Omnomnomnom ).

orangeskid
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Hint: let $S = \{x \in \mathbb{R}^n | d(x, A) \le \varepsilon\}$. Consider any $x_1,x_2 \in S$. There exist points $y_1,y_2 \in A$ such that $d(x_i,y_i) \leq \epsilon$ for $i = 1,2$.

  • Show that there is a path connecting $x_i$ and $y_i$ for $i = 1,2$.
  • Show that there is a path connecting $y_1$ to $y_2$
  • Conclude that there is a path connecting $x_1$ to $x_2$

For the first part, we can consider the path $$ x(t) = (1-t)x_i + t y_i, \quad t \in [0,1] $$

Ben Grossmann
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    Thank you. I tried to solve the second part and I'm not sure if I'm right. $A \subseteq T := {x \in \mathbb{R}^n |d(x,A) < \varepsilon}$. $T$ is connected since $T$ is a union of open balls(connected) that each one of them intersect with $A$(connected). $T$ is also locally path-connected(union of balls). Since $T$ is both connected and locally-path-connected then $T$ is path-connected. Then since $y_1, y_2 \in T \subseteq S$ there is a path connecting $y_1$ and $y_2$. Am I correct? If this really is the solution it's weird that it appears under the section "Compactness". – amirbd89 Jul 16 '15 at 16:44
  • It is strange that it is under the section "compactness", but maybe there's another question that uses this result. Your proof makes sense geometrically, but it doesn't seem rigorous to me as stated. – Ben Grossmann Jul 16 '15 at 16:49
  • This argument works for any set $A$, not necessarily closed, doesn't it? Then why would they ask $A$ to be closed? – A.Γ. Jul 16 '15 at 19:51
  • @A.G. $A$ needs to be closed for us to select the required $y_i$. – Ben Grossmann Jul 16 '15 at 19:52
  • Right. So that's where compactness is used. We need to say that the function $|x-y|$ attains its minimum $\le\epsilon$ for some $y\in A$. Thanks. – A.Γ. Jul 16 '15 at 19:58
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    @A.G. That can be done using just closedness. In fact, we're not given that $A$ is compact, though we could restrict our consideration to a bounded subset and apply Heine Borel. It's still easier to just use the fact that $A$ is closed. – Ben Grossmann Jul 16 '15 at 20:15
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    @Omnomnomnom I mean the distance from $x$ to $A$ is $\le\epsilon$, the set $A\cap B_\epsilon(x)$ is compact, then the minimum attained for some $y\in A\cap B_\epsilon(x)\subset A$. – A.Γ. Jul 16 '15 at 20:21
  • Right, so you're using the fact that closed + bounded = compact. That's Heine Borel. – Ben Grossmann Jul 16 '15 at 20:35
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    @Omnomnomnom: "Show that there is a path connecting $y_1$ to $y_2$ " ... that is problematic, as $A$ is not assumed to be path connected. I would say this is not a valid hint the way it stands. – orangeskid Jul 17 '15 at 03:11
  • To echo @orangeskid, the standard non path-connected connected set, the topologist sine curve, is a suitable $A$ for which this method doesn't work. – Ori Jul 17 '15 at 04:10
  • Now that I read the comment of @amirbd89: , it looks that he after all made good of the hint of Omnomnomnom... just like in my solution...turns out the hint works after all – orangeskid Jul 17 '15 at 04:32
  • @orangeskid I missed that detail, but I'm glad the hint still works – Ben Grossmann Jul 17 '15 at 10:28
  • @Omnomnomnom: Yes, I like it especially because for any $\epsilon >0$ we can have a path between $y_1$ and $y_2$ in $A_{\epsilon}$, but possibly not in $A$, quite puzzling. – orangeskid Jul 17 '15 at 21:08