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Let A be a connected subset of $\mathbb{R}^n$ and $\varepsilon > 0$ . Show that the $\varepsilon$-neighbourhood of $A$ defined by $U_{\varepsilon}(A) = \{ x \in \mathbb{R}^n : d(x,A) < \varepsilon \}$ is path connected .

I have shown that this set is open but I am unable to show the connectedness part.

Any help will be appreciated.

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2 Answers 2

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Take a point $p$ in $A$ and let $B$ be the set of points of $A$ that can be joined to $p$ by a path within $U_\epsilon(A)$ (not within just $A$!). Then $B$ is easily seen to be open in $A$: if you can join $p$ to $q$, you can join it to all of $U_\epsilon(q) \cap A$. And also closed in $A$: if you can't join $p$ to $q$, it follows you can't join $p$ to any point of $U_\epsilon(q) \cap A$. Since $A$ is connected, $B = A$. But now $U_\epsilon(A)$ is clearly path-connected: any point $x$ of $U_\epsilon(A)$ is within distance $\epsilon$ of some $q \in A$, and we proved you can join $p$ to $q$ (within $U_\epsilon(A)$, and then join $q$ to $x$ by a straight line segment in $U_\epsilon(q)$.

Alternatively: to do it stringing together more standard results, first notice that $U_\epsilon(A)$ is connected since it can be written as a union of connected set with a common intersection, namely, as the union of all $A \cup U_\epsilon(p)$ for $p \in A$. Now, $U_\epsilon(A)$ is connected and locally path connected and therefore connected.

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If $U_\epsilon(A)$ weren't path-connected, there would be at least two path-connected components $C$ and $D$. Let $C'=U_\epsilon(A)\setminus C$. Then the distance between $A\cap C$ and $A\cap C'$ is non-zero, since otherwise there would be a path connecting points in $C$ with points in $C'$, so $A\cap C$ and $A\cap C'$ are separated sets, and their union is $A$, contradicting the connectedness of $A$.

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You fixed it before I even commented! :-) –  Brian M. Scott Nov 29 '12 at 20:06
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