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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
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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 at 19:48

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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.

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@ 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

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