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.
[Math] Open Cover of Compact Set Minus a Point on the Boundary
compactnessgeneral-topologyreal-analysis
Related Solutions
Actually, if you haven't already learned this, you will learn that in $\Bbb R^{n}$, a set is compact if any only if it is both closed and bounded.
The set $(0,1)$ is bounded, but it is not closed, so it can't be compact.
The solution to your exercise of finding an open cover with no finite subcover proves that $(0,1)$ is not compact, because the definition of a set being compact is that every open cover of the set has a finite subcover.
So, the whole trick to finding an open cover with no finite subcover is this:
$(0,1)$ is not closed, and so it doesn't contain all of its limit points (it's easy to see that the two it doesn't contain are $0$ and $1$). Well, can you construct an open cover whose open sets get closer and closer and closer to at least one (or maybe both) of these end points, but never quite reaches them? Hint: since we are in $\Bbb R$, think about how you would get closer and closer to a real number $r$ without ever actually being $r$. If you want to get closer from above, then $\{r + \frac{1}{n} \}_{n = 1}^{\infty}$ is a sequence that gets closer to $r$ from above. What would be a sequence that gets close to $r$ from below? You should hopefully say $\{r - \frac{1}{n} \}_{n = 1}^{\infty}$.
Anyway, so maybe we can construct our open intervals so that the end points get closer and closer to the limit points $0$ and $1$, but finitely many of the sets would always have gaps between them and $0$ and $1$... hmm...
Well, we could do $(0 + \frac{1}{3}, 1 - \frac{1}{3}) \cup (0 + \frac{1}{4}, 1 - \frac{1}{4}) (0 + \frac{1}{5}, 1 - \frac{1}{5}) \cup ...$.
These open sets are in $(0,1)$ and have end points getting closer and closer to $0$ and $1$, but any finite number of them won't contain all of $(0,1)$ (why?). If we use a formula to define these open intervals, it would be $\{ (0 + \frac{1}{n}, 1 - \frac{1}{n}) \}_{n = 3}^{\infty}$.
Perhaps worth adding this example for why "closed" is a necessary assumption here.
Let $X=\mathbb Z\cup\{-\infty,\infty\}$ have the topology $\{U\subseteq X:U\cap\{-\infty,\infty\}\not=\emptyset\Rightarrow \mathbb Z\subseteq U\}$. Then $\mathbb Z\cup\{\infty\}$ and $\mathbb Z\cup\{-\infty\}$ are compact, not closed, and their intersection is the infinite discrete (and thus not compact) subspace $\mathbb Z$.
(Such a space is $T_0$, but not $T_1$. However, it can be modified to be $T_1$: use $\{U\subseteq X:U\cap\{-\infty,\infty\}\not=\emptyset\Rightarrow \mathbb Z\setminus U \text{ finite}\}$ instead.)
Since this was marked as the accepted answer, here's a writeup answering the question.
Let $0\in J$ and consider the compact subset $C_0\subseteq X$. Then for $j\in J$, note that $C_0\cap C_j$ is closed in the subspace topology for $C_0$. Then $\bigcap_{j\in J} C_j=\bigcap_{j\in J}(C_0\cap C_j)$ is the intersection of closed subsets of the space $C_0$, and is therefore closed. Since a closed subset of a compact space is compact, we have $\bigcap_{j\in J} C_j$ compact.
(Note that I do assume $J$ is non-empty by letting $0\in J$ - see user14111's comment.)
Best Answer
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 $X$.