Determine the closures of the following subsets of the ordered square \(I_o^2 = [0, 1] \times [0, 1]\) equipped with the dictionary order topology:
\[ A = \left\{ \frac{1}{n} \times 0 \;\middle|\; n \in \mathbb{Z}_+ \right\}, \]
\[ B = \left\{ \left(1 - \frac{1}{n}\right) \times \frac{1}{2} \;\middle|\; n \in \mathbb{Z}_+ \right\}, \]
\[ C = \{ x \times 0 \mid 0 < x < 1 \}, \]
\[ D = \left\{ x \times \frac{1}{2} \;\middle|\; 0 < x < 1 \right\}, \]
\[ E = \left\{ \frac{1}{2} \times y \;\middle|\; 0 < y < 1 \right\}. \]
Solution
Write \((x,y)\) for \(x\times y\). The basis of the order topology on \(I_o^2\) consists of the open intervals \((p,q)\), the intervals \([(0,0),q)\), and the intervals \((p,(1,1)]\). Each closure below is found by showing that the listed added points are limit points and that the complement of the listed set is a union of basis elements.
1. For \(0<x\le1\), every neighborhood of \((x,0)\) contains \((a,x)\times[0,1]\) for some \(a<x\); for \(0\le x<1\), every neighborhood of \((x,1)\) contains \((x,c)\times[0,1]\) for some \(c>x\).
A basis element containing \((x,0)\) with \(x>0\) either is \([(0,0),q)\) with \(q>(x,0)\), which contains \([0,x)\times[0,1]\), or has a lower endpoint \((a,b)<(x,0)\). In the second case \(a<x\), because no point has second coordinate below \(0\), and the element contains every point with first coordinate in \((a,x)\). The statement for \((x,1)\) follows in the same way from the upper endpoints, since no point has second coordinate above \(1\).
2. \(\overline A=\boxed{A\cup\{(0,1)\}}\).
By step 1, every neighborhood of \((0,1)\) contains \((1/n,0)\) for all large \(n\). The complement of \(A\cup\{(0,1)\}\) is \[ [(0,0),(0,1))\;\cup\;\bigcup_{n\ge1}\bigl((\tfrac1{n+1},0),(\tfrac1n,0)\bigr)\;\cup\;((1,0),(1,1)], \] a union of basis elements.
3. \(\overline B=\boxed{B\cup\{(1,0)\}}\).
By step 1, every neighborhood of \((1,0)\) contains \((1-\frac1n,\frac12)\) for all large \(n\). The complement of \(B\cup\{(1,0)\}\) is \[ [(0,0),(0,\tfrac12))\;\cup\;\bigcup_{n\ge1}\bigl((1-\tfrac1n,\tfrac12),(1-\tfrac1{n+1},\tfrac12)\bigr)\;\cup\;((1,0),(1,1)], \] a union of basis elements.
4. \(\overline C=\boxed{((0,1]\times\{0\})\cup([0,1)\times\{1\})}\).
By step 1, every neighborhood of \((1,0)\), and of \((x,1)\) for \(0\le x<1\), contains points \((t,0)\) with \(0<t<1\). The complement of the listed set is \[ [(0,0),(0,1))\;\cup\;\bigcup_{0<x<1}((x,0),(x,1))\;\cup\;((1,0),(1,1)], \] a union of basis elements.
5. \(\overline D=\boxed{D\cup((0,1]\times\{0\})\cup([0,1)\times\{1\})}\).
By step 1, every neighborhood of \((x,0)\) for \(0<x\le1\), and of \((x,1)\) for \(0\le x<1\), contains points \((t,\frac12)\) with \(0<t<1\). The complement of the listed set is \[ [(0,0),(0,1))\;\cup\;\bigcup_{0<x<1}\Bigl(((x,0),(x,\tfrac12))\cup((x,\tfrac12),(x,1))\Bigr)\;\cup\;((1,0),(1,1)], \] a union of basis elements.
6. \(\overline E=\boxed{\{\frac12\}\times[0,1]}\).
A basis element containing \((\frac12,0)\) contains \(\{\frac12\}\times[0,d)\) for some \(d>0\), hence points of \(E\); likewise a basis element containing \((\frac12,1)\) contains \(\{\frac12\}\times(b,1]\) for some \(b<1\). The complement of \(\{\frac12\}\times[0,1]\) is \([(0,0),(\frac12,0))\cup((\frac12,1),(1,1)]\).
7. Q.E.D.
Steps 2 through 6 give the five closures.