Find the boundary and the interior of each of the following subsets of \(\mathbb{R}^2\).
\(A = \ts{x \times y \mid y = 0}\)
\(B = \ts{x \times y \mid x > 0 \text{ and } y \neq 0}\)
\(C = A \cup B\)
\(D = \ts{x \times y \mid x \text{ is rational}}\)
\(E = \ts{x \times y \mid 0 < x^2 + y^2 \leq 1}\)
\(F = \ts{x \times y \mid x \neq 0 \text{ and } y \leq 1/x}\)
Solution
All interiors and boundaries are taken in \(\mathbb R^2\).
For the \(x\)-axis \[ A=\{(x,y):y=0\}, \] we have \[ \operatorname{Int}A=\varnothing,\qquad \operatorname{Bd}A=A. \]
For \[ B=\{(x,y):x>0,\ y\ne0\}, \] the set itself is open, so \(\operatorname{Int}B=B\). Its closure is the closed right half-plane \(\{x\ge0\}\), since every point \((x,y)\) with \(x\ge0\) is the limit of the points \((x+\frac1n,\,y+\frac1n)\) of \(B\) for all large \(n\) with \(y+\frac1n\ne0\). Therefore \[ \operatorname{Bd}B=\{x=0\}\cup\{(x,0):x>0\}. \]
Since \[ C=A\cup B=\{x>0\}\cup\{y=0\}, \] we have \[ \operatorname{Int}C=\{x>0\}, \] and \[ \overline C=\{x\ge0\}\cup\{y=0\}. \] Hence \[ \operatorname{Bd}C=\{x=0\}\cup\{(x,0):x\le0\}. \]
The set \[ D=\mathbb Q\times\mathbb R \] is dense in \(\mathbb R^2\) because every open rectangle contains a rational first coordinate. Its interior is empty because every open rectangle also contains irrational first coordinates. Thus \[ \operatorname{Int}D=\varnothing,\qquad \operatorname{Bd}D=\mathbb R^2. \]
For \[ E=\{(x,y):0<x^2+y^2\le1\}, \] we have \[ \operatorname{Int}E=\{0<x^2+y^2<1\}, \] while \[ \overline E=\{x^2+y^2\le1\}. \] Therefore \[ \operatorname{Bd}E=\{(0,0)\}\cup\{x^2+y^2=1\}. \]
Let \[ F=\{(x,y):x\ne0,\ y\le1/x\}. \] The function \(g(x,y)=y-1/x\) is continuous on the open set \(\{x\ne0\}\). Hence \(\{x\ne0,\ y<1/x\}\) is open, and each point \((x,1/x)\) is the limit of the points \((x,1/x+\frac1n)\notin F\), so \[ \operatorname{Int}F=\{x\ne0,\ y<1/x\}. \] Every point \((0,y_0)\) is the limit of the points \((\frac1n,y_0)\), which lie in \(F\) for every \(n\ge y_0\). A point with \(x\ne0\) and \(y>1/x\) lies in the open set \(\{x\ne0,\ g>0\}\), which misses \(F\). Thus \[ \overline F=F\cup(\{0\}\times\mathbb R), \] and \[ \operatorname{Bd}F=\{(x,1/x):x\ne0\}\cup(\{0\}\times\mathbb R). \]