Why is an open ball in $\mathbb R^n$ not compact? By definition of compactness, an open cover of an open ball in $\mathbb R^2$ always has a collection of subcovers that cover the ball. But why is a  open ball not compact?
 A: $\bigcup\limits_{n=2}^\infty B(0,1-\frac1n)$ covers the open ball $B(0,1)$, but no finite subfamily covers it.
A: For any open ball $B(x,r)$ where $x \in \mathbb{R}^m$ and $r \in \mathbb{R}$, the cover given by the collection $\{ B(x,r - \tfrac{1}{n}) \}$ where $n \in \mathbb{N}$ is an open cover of $B(x,r)$ but no finite subcover will cover it.
A: Theorem A compact subset of a Hausdorff space is closed.
Proof. Let $C$ be a compact subset of the Hausdorff space $X$. Let $x\in X\setminus C$. For every $c\in C$, choose an open set $U_c$ such that $c\in U_c$ and an open set $V_c$ such that $x\in V_c$ and $U_c\cap V_c=\emptyset$, which is possible because $X$ is Hausdorff. The family $(V_c)_{c\in C}$ is an open cover of $C$, so it admits a finite subcover; hence we have
$$
C\subseteq U_{c_1}\cup U_{c_2}\cup\dots\cup U_{c_n}
$$
for some $c_1,c_2,\dots,c_n\in C$. Now, setting
$$
V=V_{c_1}\cap V_{c_2}\cap\dots\cap V_{c_n}
$$
we have that $V$ is an open neighborhood of $x$ and $V\cap C=\emptyset$. QED
Since an open ball in $\mathbb{R}^n$ is not closed (because $\mathbb{R}^n$ is connected), it can't be compact.
A: Why is an open ball not compact? 
Because such an open ball admits an open cover which has no finite subcover.
A: basically for finite dimensional spaces,  Heine-Borel theorem characterizes the compact subsets.they have to be both closed and bounded.
A: The simple answer is that by the Heine-Borel Theorem, a set in $\mathbb{R}^n$ is compact if and only if the set is closed and bounded. Since $B(x,r)$ is open (there's a proof for that, but here is not the place), it cannot be closed, and thus is not compact.
An answer directly from the definition of of compactness (a set is compact if, for all covers of a set $A \in \mathbb{R}^n$, there exists a finite union of sets in the cover which create a cover of the set A), we have a proof as follows:
We create the collection $\mathcal{G}=\{n \in \mathbb{N}: G_n = B(x,r-\frac{1}{n})\}$, and we note that the countable union of $\mathcal{G}$ is a cover of $B(x,r)$. Let us assume, for the purpose of contradiction, that $B(x,r)$ is compact. There then exists some $m \in \mathbb{N}$ such that $\bigcup_{n=1}^m G_n$ is a cover of $B(x,r)$. However, for all $n$, we have $G_{n-1} \subset G_n \subset G_{n+1}$, all of which are contained in  $B(x,r)$. Therefore, $G_m \subset G_{m+1}$, and $G_{m+1} \subset B(x,r)$. Since $G_m$ is a proper subset of $G_{m+1}$, there exists a nonempty set $G_{m+1} \setminus G_m \subset B(x,r)$. Therefore, there exists an element $x \in G_{m+1}$ such that $x \not\in G_m$, and, since $G_{m+1} \subset B(x,r)$, the element $x \in B(x,r)$. We have now found a cover of $B(x,r)$ such that there exists no finite union of its subsets which creates a cover of $B(x,r)$, and therefore $B(x,r)$ is non-compact.
