Is the set of all $(u,v) \in \mathbb{R}^2$ where $0\le u \le 1$ and $-\infty<v<\infty$ open or closed?

Question

Trying to understand open and closed sets in $R^2$. Is the set of all $(u,v) \in \mathbb{R}^2$, where $0\le u \le 1$ and $-\infty<v<\infty$, open or closed?

Answer 1

The set is closed.

A set is closed if and only if its complement is open, by definition.

In the case at hand, the complement of the given set is

$C = \{ (u, v) \in \Bbb R^2 \mid u < 0 \vee u > 1 \}; \tag 1$

for $(u', v') \in C$, we see that for $u' > 1$ the open disk of radius $1/2(u' - 1)$ about $(u', v')$,

$D((u', v'), 1/2(u' - 1)) \subset C, \tag 2$

and for $u' < 0$.

$D((u', v'), -1/2u') \subset C; \tag 3$

thus, since every point $(u', v') \in C$ is contained in an open set itself contained entirely in $C$, $C$ is open (again by definition), whence our original given set $\bar C$ is closed.

Answer 2

The set can be written as $[0,1]\times \mathbb{R} \subseteq \mathbb{R} \times \mathbb{R}$. It's closed as a product of closed sets and not open as $[0,1]$ is not open in $\mathbb{R}$ and projections are open maps.