Let $S^1=\{x\in\mathbb{R}^2\mid\lVert x\rVert=1\}$, where $\lVert x\rVert$ denotes the Euclidean norm. I am asked tho prove that if $U$ is an open set, $S^1\subset U$, then there exists $\epsilon>0$ such that the set $\{x\in\mathbb{R}^2\mid \lVert x\rVert>1-\epsilon,\lVert x\rVert<1+\epsilon\}$ is contained in $U$.

I tried using the compactness of $S^1$.

$\forall x\in S^1,\;\exists\,\delta_x>0 \mid B(x,\delta_x)\subset U$. So the set $\{B(x,\delta_x)\mid x\in S^1\}$ is an open cover of $S^1$. $S^1$ compact implies there exists a finite subcover, for example, $B(x_1,\delta_{x_1}),\ldots,B(x_k,\delta_{x_k})$.

Is it enough to take $\epsilon=\min\{\delta_{x_1},\ldots,\delta_{x_k}\}>0$?

  • $\begingroup$ Not really... can you tell why? $\endgroup$
    – A.P.
    Mar 1, 2015 at 11:30
  • 2
    $\begingroup$ Hint: what can you say about the distance between $S^1$ and $\partial U$? $\endgroup$
    – A.P.
    Mar 1, 2015 at 11:32
  • $\begingroup$ @A.P. The distance must be bigger than $0$ $\endgroup$
    – user203327
    Mar 1, 2015 at 11:35
  • $\begingroup$ Good. Do you see that you can prove it, then you're done? $\endgroup$
    – A.P.
    Mar 1, 2015 at 11:36
  • 2
    $\begingroup$ It is defined as $d(x,A)=inf\{d(x,y):y\in A\}$ and it is continuous as a function of x if A is closed. In your case, you could take $A=U^c$, $\endgroup$ Mar 1, 2015 at 11:46

1 Answer 1


I'll expand a little bit on my hint.

Given two closed sets $A$ and $B$, one of which is bounded, define $$ d(A,B) = \inf \{ d(a,b) : a \in A, b \in B \} $$ Then $d(A,B) \geq 0$ and $d(A,B) = 0$ if and only if $A \cap B \neq \varnothing$.

Hence, if you can prove that $\partial U \cap S^1 = \varnothing$, then you are done...


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