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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?

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    $\begingroup$ Because you needs of a finite subcovers. $\endgroup$ – Irddo Feb 12 '15 at 23:19
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    $\begingroup$ Consider the open unit ball $B=\{x\in\Bbb R^n:\|x\|<1\}$. For each integer $n\ge 2$ let $$U_n=\left\{x\in\Bbb R^n:\|x\|<1-\frac1n\right\}\;;$$ then $\{U_n:n\ge 2\}$ is an open cover of $B$ with no finite subcover. $\endgroup$ – Brian M. Scott Feb 12 '15 at 23:20
  • $\begingroup$ The first sentence (in the statement of the question) is incorrct (in fact, non-sense): the definition of compactness does not imply anything about an open ball, simply because an open ball is not compact. $\endgroup$ – user 59363 Feb 12 '15 at 23:20
  • $\begingroup$ That "subcover" condition is really not very useful without the word "finite" in it... $\endgroup$ – Thomas Andrews Feb 12 '15 at 23:24
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$\bigcup\limits_{n=2}^\infty B(0,1-\frac1n)$ covers the open ball $B(0,1)$, but no finite subfamily covers it.

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  • $\begingroup$ Would you kindly explain how $B(0,1)\subseteq\bigcup\limits_{n=2}^\infty B(0,1-\frac1n)$ if $B(0,1)\nsubseteq B(0,1-\frac{1}{n})$, $n\rightarrow \infty$? $\endgroup$ – Diracology Oct 14 '17 at 20:15
  • $\begingroup$ @Diracology I'm a newbie myself, but the idea is basically that of convergence. In the limit, they cover the open ball, even though it's not true for any (finite) $n$. $\endgroup$ – A_P Jul 18 at 21:10
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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.

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  • $\begingroup$ And if $r\le 0$, then you obtain an empty set, which is always compact. $\endgroup$ – TZakrevskiy Feb 12 '15 at 23:43
  • $\begingroup$ @TZakrevskiy My teacher told me that the radius must be positive. $\endgroup$ – ajotatxe Feb 13 '15 at 0:04
  • $\begingroup$ @ajotatxe not necessarily. In a metric space $M$ we can say that $B(a,r)=\{y\in M:\ dist(a,y)<r\}$ where $r$ is a real number. If $r\le 0$, then, obviously, $B(a,r)=\emptyset$. $\endgroup$ – TZakrevskiy Feb 13 '15 at 8:43
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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.

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  • $\begingroup$ I am glad you added that parenthetical to your last sentence, heading off the misunderstanding at the pass. $\endgroup$ – Neal Feb 14 '15 at 0:12
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Why is an open ball not compact? Because such an open ball admits an open cover which has no finite subcover.

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basically for finite dimensional spaces, Heine-Borel theorem characterizes the compact subsets.they have to be both closed and bounded.

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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.

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