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.

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.

• Ok thanks didn't think of it like that, that makes sense, but is there a family of open sets such that each one is a subset of E, their union covers E, and there is no finite subcover of E. I think that is what I was thinking of when I asked the question. – Matthew Levy Oct 11 '14 at 19:18
• @Matthew: To get open subsets of $E$ just intersect each of those complements with $E$. The resulting sets probably won’t be open in $\Bbb R^2$, but you can’t hope for that, and all you need is sets that are open in the subspace topology on $E$. – Brian M. Scott Oct 11 '14 at 19:22
• Ok, yeah I understand why my question is poor. If the rest of the boundary is in E then an open cover of the boundary of E must include points in R^2 outside of E. – Matthew Levy Oct 11 '14 at 19:28
• @Matthew: Exactly. And if fact this must happen if the original set is compact (unless it’s a singleton), as it will have more than one point on its boundary. – Brian M. Scott Oct 11 '14 at 19:29
• if the set is connected, then does removing a point ruin compactness? – Matthew Levy Oct 11 '14 at 19:37

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 $X$.

• Very nice, I like it. Very general. – Matthew Levy Oct 11 '14 at 19:34

Let $p=(0,-1) \in \mathbb{R}^2$, and consider the collection of open balls $\{B(p,2-\frac{1}{n})\}_{n \in \mathbb{N}}$

• Shouldn't any point of $E$ have a distance of less than $2$ from the point $(0,-1)$? I do analysis not geometry, but I am pretty certain this works. – Matt A Pelto Oct 11 '14 at 19:28
• Yes, and shouldn't every point on the boundary of the unit disc have a distance of less than $2$ from the point $(0,-1)$ with the exception of the point $(0,1)$. – Matt A Pelto Oct 11 '14 at 19:30
• Ah, I see what you are saying. I thought you were instead finding a cover of $$\{(x,y)\in \Bbb R^2 : x^2+y^2\le1 \} \setminus \{(0,-1)\}$$ Your example works. – graydad Oct 11 '14 at 19:33