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

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 $U = \emptyset$.
