# Closure of the subsets of ordered square

Let $I = [0,1]$. Let $A = \bigl\{(1/n)\times 0 : n\in \mathbb Z_{+}\bigr\}$ be a subset of $I \times I$ with the lexicographic orer. Could you please give some idea to find the closure of $A$? Thanks

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Start by writing down what "closure" means. Then work out what precisely that means in the case of $I \times I$ with the lexicographic order. Then work out how that applies to your $A$. – Chris Eagle Dec 30 '12 at 23:14
And if Chris's recipe doesn't work in its entirety, then at least reveal to us the result of the first step. – Henning Makholm Dec 30 '12 at 23:19
@Rajesh What if $A=\{(1/n) \times 1; n \in \mathbb{Z}_+\}$ ? What is the closure in this case ? – user43418 Sep 17 '14 at 19:48

Your suspicion that the closure of $A$ is simply $A$, along with the point $\langle 0,0\rangle$, would be correct if you were considering $I\times I$ as a subspace of Euclidean space $\Bbb R^2$ (for example), but isn't correct, here. However, $\langle 0,1\rangle$ is a limit point of $A$ (in fact, the sole limit point), so the closure of $A$ is $A\cup\bigl\{\langle 0,1\rangle\bigr\}$.
Hint: It is simplest (in this case) to show that $A$ has no limit points except $\langle 0,1\rangle$--in particular, every open interval in the ordered square around $\langle 0,1\rangle$ contains infinitely many points of $A$, but given any other point $\langle x,y\rangle$ in $I\times I$, there is an open interval in the ordered square containing $\langle x,y\rangle$ containing no more than one point of $A$. To see why this is so, familiarize yourself with what the open intervals in the ordered square look like.
@ Cameron, I checked in other note and there is written the closure of $A$ is $A \cup 0\times 1$. – Rajesh Jan 1 '13 at 17:49
Ah! How did I miss that point? Yes, $0\times 1$ is the sole limit point of $A$, as every open interval in the ordered square around $0\times 1$ contains infinitely-many points of $A$. For all other points, we can find open intervals as described above. I'll amend my answer accordingly. – Cameron Buie Jan 1 '13 at 18:00
@CameronBuie: Why the point: $0\times 0$ isn't in closure of $A$? – hamid kamali Nov 15 '15 at 10:40
@hamid: It occurs to me (belatedly) that you may not have been asking why I mentioned it, but rather asking why it was true. Consider the set of all points less than $\langle 0,1\rangle$ to see why. – Cameron Buie Jan 10 at 14:07