Open Cover of Compact Set Minus a Point on the Boundary I am having a hard time thinking of an infinite (uncountable or not) open cover of a compact set missing a point on its boundary in $\Bbb R^2$, so that the open cover has no finite subcover. I know this must be possible as the set is no longer closed and thus no longer compact. For example can somebody give me a cover of $$E = \{(x,y)\in \Bbb R^2 : x^2+y^2\le1 \} \setminus \{(0,1)\}$$ that has no finite subcover.
 A: Let $X$ be a compact Hausdorff space and $p\in X$ apoint such that $Y:=X-\{p\}$ is not closed. For each $x\in Y$ there exist disjoint open sets $x\in U_x,p\in V_x$. Then the $U_x$ cover $Y$. Assume there is a finite subcover $U_{x_1}\cup\ldots\cup U_{x_n}$. Then this subcover misses the open set $V_{x_1}\cap \ldots\cap V_{x_n}$ which contains $p$ ans must be strictly larger that $\{p\}$ because $\{p\}$ is not open in $X$.
A: HINT: For any $p\in\Bbb R^2$ and any $\epsilon>0$, the set $\Bbb R^2\setminus\operatorname{cl}B(p,\epsilon)$ is open. (Here $B(p,\epsilon)$ is the open ball of radius $\epsilon$ and centre $p$.)
Added: This doesn’t arise in your specific example, but in general you need more than that $p$ is in the boundary of the compact set: you need it to be a limit point. If your original compact set were the closed unit disk together with a point $p$ not contained in it, for example, $p$ would be in its boundary, but removing $p$ would still leave a compact set.
A: Let $p=(0,-1) \in \mathbb{R}^2$, and consider the collection of open balls $\{B(p,2-\frac{1}{n})\}_{n \in \mathbb{N}}$
