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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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@JKnecht Please slow down with your editing spree. It's bad form to flood the front page with ancient questions, as it takes attention away from new ones. Moreover, it doesn't seem like you know fully what you're doing. For example here, the standard convention in mathematics is that the Euclidean spaces are written as $\mathbb R^n$, not $\mathbb{R^n}$, i.e. the exponent is supposed to be in italics, and certainly not blackboard bold. – epimorphic Feb 2 at 2:17
@epimorphic ... I had no idea the edits turned up on some front page?? Which front page is that? But i will slow down of course. Just had nothing better to do than go through old topology questions. And like all of my edits were accepted so i assumed it was the right thing to do. – JKnecht Feb 2 at 2:56
@JKnecht // Your edit to this question in particular should have been rejected IMO, but reviewers are often careless. Your most recent one missed quite a few things that should have been corrected: $->$ should be coded as \to or \rightarrow ($\to$); $tan$ and $pi$ should be \tan ($\tan$) and \pi ($\pi$); also grammar and such. – epimorphic Feb 2 at 3:04
@epimorphic...was not aware of that front page. I now noticed I have a bookmark that goes right to /questions. Usually i dont miss to edit rightarrow, tan, or pi. I blame it on being tired... – JKnecht Feb 2 at 3:16
up vote 3 down vote accepted

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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