Confused about open and closed set in $\mathbb{R}^2$

Question

While this question is basically meant for all $\mathbb{R}^n$ where $r > 1$, I suppose it would be the easiest to grasp the concept in $\mathbb{R}^2$ and conclude the analog to $\mathbb{R}^n$.

So I was wondering since we just started out with Topology, how to find out if a set is closed or open in $\mathbb{R}^2$.

Let's say I have a set $A \subset \mathbb{R}$ defined as:

$$A=(0,1) \times [0,1]$$

Since I now have a set "composed" by an open and a closed set I wonder if $A$ is now called open or closed.

Or is this a case where it neither is open nor closed?

Answer 1

• Since $(0,0)\notin A$ but every open ball centered at $(0,0)$ contains points of $A$, $A$ is not closed.
• Since $\left(\frac12,0\right)\in A$ but every open ball centered at $\left(\frac12,0\right)\in A$ contains points out of $A$, $A$ is not open.

Therefore, $A$ is neither closed nor open.

Answer 2

This set $(0,1) \times [0,1]$ is not closed and not open.

Not closed e.g. because $(\frac{1}{n}, \frac{1}{n})$ is a sequence in it conervging to $(0,0) \notin A$. Not open e.g. because $(\frac{1}{2}, -\frac{1}{n})$ is a sequence converging to $(\frac12,0)$ in $A$ but all of whose points lie outside $A$.

Another way: projections are open maps so if $A$ were open, so would $\pi_2[A]= [0,1]$ be in $\mathbb{R}$, which it is not. If $A$ were closed, it would be compact (closed and bounded in $\mathbb{R}^2$) but $\pi_1[A] = (0,1)$ is not compact (not even closed).