Let $U\neq \emptyset$ In $\mathbb{R}^n$ Is Open, Then $U$ Is Not Compact

Question

Prove: Let $U\neq \emptyset$ In $\mathbb{R}^n$ Is Open, Then $U$ Is Not Compact

How can I approach this? I know that in $\mathbb{R}^n$ we have that $U$ is comapct $\iff$ $U$ is closed and bounded

So we can start with assuming the $U$ is compact, but I can not see which path will lead to a contradiction

Answer 1

If $U$ is open and closed, it is a connected component of $\mathbb{R}^n$ which is $\mathbb{R}^n$.

Answer 2

Hint: Take the negation of the equivalence statement: $U$ is not compact iff $U$ is not closed or $U$ is not bounded.

Answer 3

If $U$ is unbounded, it is not compact. Otherwise, take $p\in\overline U\setminus U$ and, for each $r>0$, consider the open ball $B_r(p)$. Then $\{B_r(p)\mid r>0\}$ is an open cover of $U$ without a finite subcover. Therefore, $U$ is not compact.

Answer 4

There are many good answers already, but here's another approach. First, we can solve this for $\mathbb{R}$. An open set $U \subset \mathbb{R}$ is the (countable) disjoint union of open intervals, $U = \sqcup_{j \geq 1}(a_j,b_j)$. We can assume $U$ bounded, otherwise it will not be compact from the get go. Thus all intervals have finite endpoints.

Now, we have that $a_1 \not \in U$ but you can certainly find a sequence in $(a_1,b_1)$ that converges to $a_1$. Hence $U$ is not closed, and in particular, it is not compact.

Finally, if $U \subset \mathbb{R}^n$, note that the projection mapping

$$ \pi : (x_1, \dots, x_n) \in \mathbb{R}^n \mapsto x_1 \in \mathbb{R} $$

is not only continuous, but open (i.e. $\pi(U)$ is open, if $U$ is open). Therefore, the set $\pi(U) \subset \mathbb{R}$ is open, and compact (in particular, bounded). This implies $\pi(U) = \emptyset$ which can only occur if $U = \emptyset$.

Answer 5

Hint:

In $\mathbb R^n$ equipped with usual topology $\tau$ we have:

• $U\in\tau\implies\{\vec{x}+\vec{u}\mid\vec{u}\in U\}\in\tau$ for every $\vec{x}\in\mathbb R$.
• $U$ compact $\implies\{\vec{x}+\vec{u}\mid\vec{u}\in U\}$ compact for every $\vec{x}\in\mathbb R$.
• $U\in\tau\implies \{r\vec{u}\mid\vec{u}\in U\}\in\tau$ for every $r\in(0,\infty)$.

Sets like $\{\vec{x}+\vec{u}\mid\vec{u}\in U\}$ are denoted as $\vec{x}+U$ and sets like $\{r\vec{u}\mid\vec{u}\in U\}$ by $rU$.

For covenience let it be that $U$ contains the zero vector.

Now try to prove that the collection $\{rU\mid r\in(0,1)\}$ is a collection of open sets that cover $U$ and that has no finite covering subcollection.

If $U$ does not contain the zero vector but contains vector $\vec{x}$ you can "translate" to $V=-\vec{x}+U$.

Answer 6

Here is an alternative using the definition of compactness more-or-less directly. Let $U$ be open.

Let us denote by $B_r$ the open balls of radius $r$ around the origin of $\mathbb{R}^n$.

Let $R=\sup\left\{\Vert u\Vert:u\in U\right\}$ be the "radius" of $U$. This may be infinite. Even if $R$ is finite, then $U$ does not contain any point $u$ with $\Vert u\Vert=R$ - otherwise, since $U$ is open, it would also contain the point $(1+\varepsilon)u$ for small $\varepsilon>0$, which would have norm $(1+\varepsilon)R$, which contradicts $R$ being the supremum of all norms of elements of $U$.

In any case, $\left\{B_r:0<r<R\right\}$ is an open cover of $U$, by the paragraph above, without finite subcover, by properties of the supremum.