Finding interior, accumulation, boundary, and isolated points of a set in $\mathbb{R}^2$ I'm looking to clarify what the above points of this set are:
$$\{(x,y) \in \mathbb{R}^2 \hspace{1mm} | \hspace{1mm} 0 \leq x < 1,\ 0 < y < 1,\ x,y \in \mathbb{Q} \}$$
A part confusing me is $x$ and $y$ belonging to the rationals - how does this play in?
Currently I have: 
Interior points:
$\displaystyle\{(x,y) \in \mathbb{R}^2 \hspace{1mm} | \hspace{1mm} 0 < x < 1,\ 0 < y < 1,\ x,y \in \mathbb{Q} \}$
Boundary Points:
$\displaystyle\{(x,y) \in \mathbb{R}^2 \hspace{1mm} | \hspace{1mm} x,y \in\{0,1\}\}$
Accumulation Points: $A \subseteq $ (interior points) $\cup$ (boundary points) ? 
Isolated Points: Are these the points outside the rectangle? 
Thanks for any feedback, appreciate it.
 A: Let's call your set $A$.
Isolated point: Recall that $(x,y)$ is an isolated point of $A$ if $(x,y)\in A$ and there is $r>0$ s.t. $(B_r(x,y)\setminus \{(x,y)\})\cap A=\emptyset$, where $B_r(x,y)$ is the open ball centred at $(x,y)$ of radius $r$. Now for any $(x,y)\in A$, and any $r>0$, since $\mathbb{Q}$ is dense in $\mathbb{R}$, then there are $x',y'\in\mathbb{Q}\cap (0,1)\subseteq A$ s.t. $(x',y')\neq(x,y)$ and $|x-x'|<\frac{r}{2}$, and $|y-y'|<\frac{r}{2}$, thus $(x',y')\in B_r(x,y)$. Since $r>0$ is arbitrary, we conclude that $A$ has no isolated point.
Boundary point: $(x,y)$ is a boundary point of $A$ if $B_r(x,y)\cap A\neq\emptyset$ and $B_r(x,y)\cap A^c\neq\emptyset$ for all $r>0$.
Use the similar argument, one can show that the boundary of $A$ is
$$[0,1]\times[0,1]:=\{(x,y)\in\mathbb{R}^2: 0\leq x\leq 1, 0\leq y\leq 1\}$$
Accumulation point  $(x,y)$ is an accumulation point of $A$ if for any $r>0$ we have $(B_r(x,y)\setminus \{(x,y)\})\cap A\neq\emptyset$. Again, applying similar argument, the set of accumulation points is exactly $[0,1]\times[0,1]$
A: Let $A$ denote your set. I will call a point in $A$ "rational" just for shorthand, and points in the complement $A^c$ "irrational" (i.e. if either $x$ or $y$ is irrational). There are several different definitions for these types of points, I'll give hints for at least one of the common definitions.
Interior points are points $(x,y)\in A$ such that there is some neighborhood (open ball, if you like) of $(x,y)$ such that $U \subset A$. In other words, for some $\epsilon > 0$, there are no irrational points within distance $\epsilon$.
Boundary points are points $(x,y)$ (not necessarily in $A$) such that every neighborhood $U$ of $(x,y)$ satisfies $U \cap A \neq \emptyset$ and $U \cap A^c \neq \emptyset$. That is, every neighborhood contains at least one rational point, and one irrational point.
Alternatively, the interior is sometimes defined as the points of $A$ that aren't in the boundary.
The accumulation points are points $(w,z)$, not necessarily in $A$, such that every neighborhood $U$ of $(w,z)$ contains a point of $A$. Said another way, they are the limits of sequences $\{(x_n,y_n)\} \subset A$. (This is where density is useful)
The isolated points are the points $(x,y) \in A$ such that there exists some neighborhood of $(x,y)$ such that $(U \backslash \{(x,y)\}) \cap A = \emptyset$. That is, $(x,y)$ is the only rational point in this neighborhood.
