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.
Best Answer
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 $[0,1]\times[0,1]$