$\newcommand{\N}{\mathbb{N}}$ Problem

Let $A=\{(\frac{1}{n}\times 0)|n\in\mathbb{N}\}$. Find $\bar{A}$ in the dictionary order topology of the ordered square,i.e, $[0,1]\times[0,1]$

Attempted Solution

(For notation, we let $\left<\cdot \right>$ be open interval in the order topology of $I\times I$)

We claim that $\bar{A}=A\cup(0,1)$. To prove that,we first show that $\left(0,1\right)\in A'$ and in fact $\left(0,1\right)$ is the only limit point of $A$. Let $U$ be an open set containing $\left(0,1\right),$ $\exists\left\langle \left(a,b\right),\left(c,d\right)\right\rangle $ (note that since $I\times I$ has largest element and smallest element, we can also have $\left\langle \left(a,b\right),\left(c,d\right)\right\rangle \cup\left(a,b\right)$ or $\left\langle \left(a,b\right),\left(c,d\right)\right\rangle \cup\left(c,d\right)$ instead $\left\langle \left(a,b\right),\left(c,d\right)\right\rangle $ as well, but these )such that $\left(0,1\right)\in\left\langle \left(a,b\right),\left(c,d\right)\right\rangle \in U$. Since $\left(0,1\right)\prec\left(c,d\right)$, we claim that $c>0$ because otherwise we would have $1\leq d>1$, which is a contradiction. Also since $\left(a,b\right)\prec\left(0,1\right)$, we have $a=0$ because otherwise we would have $\left(0,1\right)\prec\left(a,b\right)$ which is a contradiction. Furthermore, we have $b<1$ because $\left(0,1\right)\nprec\left(0,1\right)$. Therefore, we have $\left\langle \left(a,b\right),\left(c,d\right)\right\rangle =\left\langle \left(0,b\right),\left(c,d\right)\right\rangle $ for some $b<1,$ and $c,d\in(0,1]$. Notice that we have \begin{align} \left\langle \left(0,b\right),\left(c,d\right)\right\rangle & =\left[0\times(b,1]\right]\cup\left[\left(0,c\right)\times[0,1]\right]\cup\left[c\times[0,d)\right] \end{align} and $\left(0,c\right)\times[0,1]\cap A\neq\emptyset$ since we can find $n\in\N$ such that $1/n\in\left(0,c\right)$, so $\left(0,1\right)$ is a limit point of $A$ because $U$ is an arbitrary open set.

Now we prove that any point other than $\left(0,1\right)$ is not a limit point of $A$. Suppose $\left(a,b\right)\neq\left(0,1\right)$ is a limit point of $A$. Then we have either $a=0,b\in[0,1)$ or $a\neq0,b\in\left[0,1\right]$. In the first case, let \begin{align} U= & 0\times[b,1)=\left(0,b\right)\cup\left\langle \left(0,b\right),\left(0,1\right)\right\rangle \end{align} which is open in the dictionary order topology of $I\times I$ and $U\cap A=\emptyset.$ In the second case, we notet that for $\forall a\in(0,1]$ we can find $p,q\in\N$ such that $\left(1/p,1/q\right)\cap\left\{ 1/n|n\in\N\right\} =\emptyset$ and $a\in\left(1/p,1/q\right)$. \begin{align} U & =\left\langle \left(1/p,0\right),\left(1/q,0\right)\right\rangle =\left(\frac{1}{p},\frac{1}{q}\right)\times\left(0,1\right) \end{align} which is open in the dictionary order topology of $I\times I$ and $U\cap A=\emptyset$. Hence we have that all points but $\left(0,1\right)$ in the ordered square aren't limit points of $A$. Therefore, $\bar{A}=A\cup A'=A\cup\left(0,1\right)$.


I think I have solved this one. But I don't know if there is any loopholes in my proof. I really appreciate if anyone can take a look at my proof and point out any possible improvements!

  • 2
    $\begingroup$ @lmsteffan No, A is a subset of the plane. @ OP One loophole, easy to fix, in the 2nd part of proof, when showing that there are no other limit points: given $(a,b) \neq (0,1)$, what if $a = 1/n$ for some $n$? Once you dismiss that case, no don't really need (or want!) distinct $p,q$; just use $p = q+1$ for a suitable $q$. That said, good show :) $\endgroup$ – BrianO Oct 30 '15 at 23:36

Notation:In a linear space $X$ with order $<$, write $(p,\rightarrow)=\{q: p<q\}$ and $(\leftarrow,p)=\{q: q<p\}$. Any strictly decreasing sequence such as $A$ will satisfy $ \bar A= A\cup (glb A)$ if $glb A$ exists; otherwise $\bar A=A$.Because ... (1)...Any point that is not a lower bound for $A$ lies in an open interval $(u,v)$ for some consecutive $u,v\in A$ or else lies in $(\max A,\rightarrow)$; and any point $p$ that is a lower bound for $A$, but is not equal to $glb A$, belongs to $(\leftarrow,q)$ where $p<q$ and $q$ is some lower bound for $A$; ... (2)...If $ y=glb A$ exists, there cannot be an open interval $J$ with $y\in J$ and $J\cap A=\phi$ because $J$ would be bounded above by some $z$ (otherwise $A\subset J$) so we would have $y\in J\subset (\leftarrow,z)\subset X\backslash A$ which makes $z$ a lower bound for $A$, contradicting $y=glb A$...... In your Q ,therefore it suffices to show that the point $(0,1)$ equals $glb A$. It is clearly a lower bound for $A$. And $(0,1)<(x,y)\in [0,1]^2\implies (x,y)>(1/n.0)\in A$ for some (any) $n \in (0,x)$ so $(x,y)$ is not an upper bound for $A$.


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