Let $f:\mathbb{R}^2\to\mathbb{R}$ be a continuous surjection under the ordinary topology. Show that the preimage of a single point, i.e. $f^{-1}(\{t\})$ is not compact.
The set is obviously closed, and thus is should be unbounded. I guess we need to use contradiction here and probably it should be some kind of a 'one-line' proof.
For simplicity assume $t=0$, otherwise we can replace $f$ by $f(x)-t$ anyway. Suppose that $f^{-1}(0)$ is bounded, thus contained in some closed ball $B$ centered at $0$. A closed ball is a compact set, and since $f$ is continuous it follows that $f(B)$ is bounded, say $f(B)\subseteq [-M,M]$ for some $M>0$. Since $f$ is surjective, there must be points $a,b\in\mathbb{R^2}$ such that $f(a)>M$ and $f(b)<-M$. Note that necessary $a,b\notin B$.
Now, since the complement of a ball is path connected, there is some path $\gamma:[0,1]\to\mathbb{R^2}\setminus B$ such that $\gamma(0)=a, \gamma(1)=b$. Then $f\circ\gamma:[0,1]\to\mathbb{R}$ is a continuous function that satisfies $f\circ\gamma(0)>M$ an $f\circ\gamma(1)<-M$. By the intermediate value theorem there is some $t_0\in [0,1]$ such that $f(\gamma(t_0))=f\circ\gamma(t_0)=0$. But this means $\gamma(t_0)\in f^{-1}(0)\subseteq B$, a contradiction.
Let us use a proof by contradiction.
Suppose that $f^{-1}(\{-t\})$ is bounded, then there is some $R>0$ such that $ f^{-1}(\{-t\}) \subset B:=\{(x,y) \in \mathbb R^2 \mid x^2 + y^2 \le R\}. $
Note that $B$ is compact and bounded. Thus $f(B)$ is compact and bounded. Since $\mathbb R^2 - B$ is connected, $f(\mathbb R^2 - B)$ is also connected. So $f(\mathbb R^2 - B)$ is an interval.
Since $\emptyset \neq f^{-1}(\{-t\}) \subset B$, $B$ is not empty. Hence $f(B)$ is a non-empty closed bounded interval $I$.
Note $\mathbb R - I \subset f(\mathbb R^2 - B)$. Hence $f(\mathbb R^2 - B) = \mathbb R$ because $f(\mathbb R^2 - B)$ is an interval, containing $\mathbb R - I$.
But this is impossible because $t \notin f(\mathbb R^2 - B)$.
Therefore $f^{-1}(\{-t\}) $ should be unbounded.
Well, this is not so short.
