# Lexicographic Square, topology

by mathsss2
Tags: lexicographic, square, topology
 P: 38 Show that any basic open set about a point on the "top edge," that is, a point of form $$(a, 1)$$, where $$a < 1$$, must intersect the "bottom edge." Background: Definition- The lexicographic square is the set $$X = [0,1] \times [0,1]$$ with the dictionary, or lexicographic, order. That is $$(a, b) < (c, d)$$ if and only if either $$a < b$$, or $$a = b$$ and $$c < d$$. This is a linear order on $$X$$, and the example we seek is $$X$$ with the order topology. We follow usual customs for intervals, so that $$[(a,b),(c,d)) = \{ (x,y) \in X : (a,b) \leq (x,y) < (c,d) \}$$. A subbase for the order topology on $$X$$ is the collection of all sets of form $$[(0,0),(a,b))$$ or of form $$[(a,b),(1,1)).$$
 P: 534 Lexicographic Square, topology The base elements are all finite intersections of your subbase elements; they are intervals of the form $$[(0, 0), a)$$, $$(a, (1, 1)]$$, or $$(a, b)$$, where $$(0, 0) < a < b < (1, 1)$$.
 P: 38 So, we know the base elements are intervals of the form $$[(0, 0), a) , (a, (1, 1)]$$, or $$(a, b)$$, where $$(0, 0) < a < b < (1, 1)$$. We need to show that any basic open set about a point on the "top edge," that is, a point of form $$(a, 1)$$, where $$a < 1$$, must intersect the "bottom edge." How is this obvious now? I don't understand the connection? Thanks for all your help with topology, I was able to solve the other problem you helped me with too.
 P: 38 Turns out there was a typo in the problem [that was throwing me off a lot]. So, the lexicographic order should be $$(a,b)<(c,d)$$ if and only if [tex]a