The question goes like this : If $\{U_i:i\in I\}$ is an open cover of the unit circle in $\mathbb R^2$ then show that it is an open cover of an annulus $1-\delta\lt ||(x,y)||\lt 1+\delta$ for some $\delta \gt 0$. I have two problems with this statement :
$1)$ I can understand the statement with finite number of covers geometrically but with infinite number of covers , I find it difficult to believe .
$2)$ Also whatever I understand is completely geometric and not analytic.
All the open sets that cover the circle $S^1$ are open discs in $\Bbb R^2$ . Since $S^1$ is compact , we have finite number of them covering the whole of $S^1$ . Say they are $D_1,D_2,.....D_k.$ And without loss of generality , let $D_m$ be the one with the smallest radius,$\mathcal r$. { For convenience, can we consider only circles that have their centers on $S^1$ $?$ } Then we can easily see that $\{D_i;i=1,2,...k\}$ cover the annular region $D(0,1+\mathcal r)\backslash D(0,1-\mathcal r)$
[Here I considered discs only for I thought that any open set has an open disc embedded in it so we could take into account the sub-cover consisting of the open sets that contain ourchosen open discs ]
So if I have to consider all the infinite number of discs covering $S^1$ , then there is a problem for it can be the set of circles of radius $1\over n$ . So the infimum is $0$ although none has radius $0.$ So finding the one with smallest radius is impossible here so my trick won't work here. So how do we get the result for infinite case $?$
And from the look of the question it seems considering the whole cover and not a finite sub cover is desired here.
So please help me with the proof .
Thanks.
First I'll consider your (bolded) question about taking only discs with centers in $S^1$. Indeed we can do this - consider your open cover $\{ U_i\}$. Each $x\in S^1$ is contained in some $U_x\in\{ U_i\}$, and $U_x$ is open, so it contains some open ball in $\mathbb{R}^2$ centered at $x$, say $B_x$. Indeed, it should be clear that the collection of open balls $\{ B_x\}_{x\in S^1}$ form an open cover of $S^1$ (since given $x\in S^1$, $x\in B_x$). Therefore, since $S^1$ is compact, we can find a finite subcover $\{ B_1,\ldots B_n\}$, which are open balls in $\mathbb{R}^2$ with centers in $S^1$ as you desired.
However, your comment that the minimal radius works is, in fact, incorrect. Indeed, by considering discs that only just overlap one another you should be able to draw a fairly convincing picture to illustrate this (the problem lies in the "hollows" between the discs, which may approach closer to the circle than the minimum radius of any disk). Instead, you need to consider the distance of $S^1$ from the complement of the union of sets in our cover.
Let $C:=(\bigcup\limits_{j=1}^n B_j)^\mathsf{c}$. This is a closed set, and therefore we can find a $\delta>0$ such that $$d(S^1,C):=\inf\{\|x-c\|\mid x\in S^1, c\in C\}>\delta.$$ Indeed, if we could not then we could find a sequence in $C$ converging to an $x\in S^1$, which would tell us that $x\in C$ (since $C$ is closed). This is a contradiction because the complement of $C$ was meant to cover $S^1$.
Now this $\delta$ works.