# Finding limit points in lexicographic order topology

If we consider the unit square (i.e $[0,1] \times [0,1]$) with the lexicographic order induced by ${\mathbb{R}}^{2}$, what are the limit points of the unit square?

My answer is: the set of limit points is equal to:

$[0,1]^{2} \setminus ((0,1) \times \{1\}))$.

Is my answer correct? If not, how do you find them? I get confused with this topology.

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Try not to think of it in the sense of a square, but rather a linear order which has continuum-many copies of $[0,1]$ piled on top of each other. –  Asaf Karagila Dec 30 '10 at 8:40

Every point $P = (a,b)$ on the square lies on some horizontal line segment $y = b$. If $a \neq 0$ -- i.e., if $P$ is not the leftmost point on its horizontal line segment -- then you can build a sequence of points $(x_n,b)$ to the left of $a$ on that line segment converging to $P$. If $a \neq 1$ -- i.e., if $P$ is not the rightmost point on its horizontal line segment -- then you can build a sequence of points $(x_n,b)$ to the right of $a$ on that line segment converging to $(a,b)$.
Obviously $P$ can't be both the leftmost point and the rightmost point, so the conclusion is...