Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|cite|improve this question

2 Answers 2

up vote 2 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.

share|cite|improve this answer

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 $A$.

share|cite|improve this answer
You fixed it before I even commented! :-) – Brian M. Scott Nov 29 '12 at 20:06

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.