Show that the projection map of $[0,1] \times \mathbb{R} $ onto $\mathbb{R}$ Show that the projection map of $[0,1]  \times \mathbb{R} $ onto $\mathbb{R}$ is a closed map. 
I know that [0,1] is a compact space but I am not sure how this implies that the projection map is closed. I know that in plain old $\mathbb{R} \times \mathbb{R} $ the projection map is not closed because the graph of $f(x) = 1/x $ is closed but has open projection and that in $[0,1]  \times \mathbb{R} $ we can't have this type of function, but that's it. Thanks for any tips or help.
 A: Let $C$ be closed in the product, we want to show that its image under $\pi_2$ is closed. Trivially this is true if $\pi_2(C)=\mathbb{R}$. Otherwise consider $\mathbb{R} \setminus \pi_2(C)$. To prove that it is open pick $x$ in this set and consider $[0,1] \times \{x\}$. Clearly $$[0,1] \times \{x\} \subset \Big([0,1] \times \mathbb{R}\Big) \setminus C$$
Being $C$ closed, $\Big([0,1] \times \mathbb{R}\Big) \setminus C$ is open. Hence we are in the position to apply the so called "tube-lemma":
The Tube Lemma: Consider the product $X \times Y$, where $X$ is compact. If $N$ is an open set of $X \times Y$ containing $X \times y_0$ then $N$ contains some tube $X \times W$ about $X \times y_0$, where $W \subset Y$ is a neighborhood of $y_0$.
Finally we are done: take this neighborhood $W \subset \mathbb{R}$ of $x$. Consider $$[0,1] \times W \subset \Big([0,1] \times \mathbb{R}\Big) \setminus C.$$ Then we have $x \in W \subset \mathbb{R}\setminus \pi_2(C)$, proving that $\mathbb{R}\setminus \pi_2(C)$ is open.
