Consider $\mathbb{R}^2$ with its standard metric. Classify the following sets as open, closed, clopen, or none of the above: (a) $[0,1) \times [0,1]$
(b) $\mathbb{R}^2 \backslash \{(0,0)\}$
(c) {$(a,a) | a \in \mathbb{R}$}
(d) $([0,1] \times [0,1]) \backslash ((\frac{1}{4},\frac{3}{4}) \times ((\frac{1}{4},\frac{3}{4}))$
I know open sets have the following properties, (I) $X$ and $\emptyset$ are open. (II) The union of open sets is also open. (III) The finite intersection of open sets is open. Likewise, closed sets have the following properties, (I) $X$ and $\emptyset$ are closed. (II) The intersection of closed sets is closed. (III) The finite union of closed sets is closed.
I'm not sure how to use these properties to determine whether the sets are open, closed, or clopen. Any hints or tips are appreciated! Here are my instinctual answers:
(a) Clopen possibly
(b) Open because the compliment is closed
(c) I feel like it could be open and closed so clopen
(d) Clopen
 A: We have that in the satandar topology in $\mathbb{R}^2$, the "canonical" open sets are the balls $B_{((a,b),r)}= \{ (x,y) \in \mathbb{R}^2 : ||(x,y)-(a,b)|| < r \}$ of center $(a,b)$ and radius $r > 0$.
Another points to consider: A set $A$ is open iff for all $(a,b) \in A$ exists $r > 0 $ such that $B_{((a,b),r)} \subset A$.
$a) A=[0,1) \times [0,1]$.
Consider a point of the form $(x_0,0)$ with $0 \leq x < 1$. Then for every radius $r > 0$ we have that $B_{((x_0,0),r)}$ contains points with the second variable $<0$, i.e., points that aren't in $A$. Then $A$ isn't open.
On the other hand, a set $A$ is closed iff it complement $A^c$ is open. Here $A^c$ contains points of the form $(1,x)$ with $0 \leq x \leq 1$. Try to proove that for every radiuos the balls center in these points will contain points of $A$. Then $A$ isn't closed.
$b) A=\mathbb{R}^2 \setminus \{(0,0)\}$
Here I'm agree with you. $A^c=\{(0,0)\}$ and the points are closed. A "fast" argument to see this is that $\mathbb{R}^2$ with the usual topology is Hausdoff and hences is $T^1$.
On the other hand, $A$ isn't closed, because it's complement isn't open.
$c) A= \{(a,a) : a \in \mathbb{R} \}$.
Observe that $A$ is the line $x=y$. In general the graph of functions are closed with the usual topology. Let see thath $A$ is closed but no open.
$A$ isn't open: Fixed a point $(a_0,a_0) \in A$. The for every radius $r > 0$ we have that the ball $B_{((a_0,a_0),r)}$ contains points with $x \neq y$, i.e. points that aren't in $A$, so $A$ isn't open. You can draw it to convince yourself.
$A$ is closed: Here $A^c = \{(x,y) \in \mathbb{R}^2 : x \neq y \}$. Consider a point $(x_0,y_0) \in A^c$. As $x_0 \neq y_0$, we have that $|x_0 - y_0|>0$. Then we can user $r=\frac{|x_0-y_0|}{2}>0$ as radius. I let you proove that $B_{((x_0,y_0),r)} \subset A^c$. Another time, a draw can you visualized this.
$d) A=([0,1] \times [0,1]) \backslash ((\frac{1}{4},\frac{3}{4}) \times ((\frac{1}{4},\frac{3}{4}))$
The "draw" of $A$ is as a square minus a stripe. Here I will use another argument. First the cartesian product a closed (open) sets are closed (open). This isn't trivial at first, but I think you can proove it.
On the other hand, if we define $B=[0,1] \times [0,1]$ and $C=\left((\frac{1}{4},\frac{3}{4}) \times ((\frac{1}{4},\frac{3}{4})\right)$, then $B$ is closed and $C$ is open, because $[0,1]$ and $(\frac{1}{4},\frac{3}{4})$ are closed an open, respectively in $\mathbb{R}$.
Finally, $A=B \setminus C = B \cap C^c$. As $C$ is open, then $C^c$ is closed as well as $B$ so $B \cap C^c$ is closed because is a finite intersection of closed sets.
