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The dictionary order topology on $\Bbb R\times \Bbb R$ is generated by a basis having elements $(a\times b,c\times d)$ for $a<c$ or $a=c$ and $b<d$.

Let $I=[0,1]$ and consider $I\times I$. The restriction of dictionary order topology on $I\times I$ defines a topology on $I\times I.$

I'm trying to figure out what would be its basis. The following is what I have;

For $a,b,c,d\in I$

Its basis will contain elements of the form $(a\times b,c\times d)$ for $a<c$ or $a=c$ and $b<d$

The basis contains elements $[0\times0,a\times b)$ with $0<a$ or $0=a$ and $0<b$

Also $(a\times b,1\times1]$ with $a<1$ or $a=1$ and $b<1$ belongs to the basis.

Is that correct? Suggestions Please!

Naive
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    @AlexProvost I'm not considering the topology inherited by $I \times I$ as a subspace of the dictionary order on $\Bbb R\times \Bbb R$. Rather I'm restricting the dictionary order of $\Bbb R\times \Bbb R$ on $I \times I$. – Naive Jul 14 '17 at 18:10
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    That's indeed a quite different topology. – Henno Brandsma Jul 14 '17 at 18:11
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    Oh, okay. Well in that case, your basis works fine (any order topology is usually defined via those basic open sets). – Alex Provost Jul 14 '17 at 18:14
  • @HennoBrandsma I'm almost certain that the first type of sets i've mentioned are in the basis. Aren't they? The second and the third type of sets I'm not really sure about. – Naive Jul 14 '17 at 18:15
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    @AlexProvost Then it must also be true that $0\times0$ is the least element of $I\times I$ and $1\times 1$ is the largest element of $I\times I$. Isn't it? – Naive Jul 14 '17 at 18:17
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    Yes, of course. – Alex Provost Jul 14 '17 at 19:21

2 Answers2

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The general way to contruct a base for an ordered space is the following:

If $(X,<)$ is an ordered space, a subbase for its order topology is given by all sets of the form $L(a) = \{x \in X: x < a\}, a \in X$ (the lower sets) together with all sets $U(a) = \{x: x > a\}$, where $a \in X$ (the upper sets).

The base derived from it (i.e. all finite intersections of subbase elements) depends on whether $X$ has a minimum $m$ or maximum $M$, as $m$ cannot be in any upper set, nor $M$ in any lower set. All others can. $L(a) \cap L(a') = L(\min(a,a'))$ and $U(a) \cap U(a') = U(\max(a,a')$ so both types of sets among themselves are closed under finite intersections. So we only need to consider the intersections of one $L(a)$ and one $U(a')$ which are exactly the open intervals $(a',a)$ in this case, and we only have possibly non-empty intersections when $a' < a$. If we have $m$ we also need to include all $L(a)$ in the base (where $a>m$, and if we have $M$, all $U(a)$(where $a < M$ as well.

So for $I \times I$ in the lexicographic order $<_l$ we do have $m= 0 \times 0$, and $M = 1 \times 1$, so the standard base is indeed by the above general procedure:

  • all sets $(a \times b, c \times d)$ with $a \times b <_l c \times d$.
  • all sets $L(a \times b) = [0 \times 0, a \times b)$ where $a \times b \neq 0\times 0$.
  • all sets $U(a \times b) = (a \times b, 1 \times 1]$ where $a \times b \neq 1 \times 1$.

Which is the same as the basis you describe.

Henno Brandsma
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  • Is the set $[0\times 0, 1\times 1]$ the only closed interval that is open in the dictionary topology of the ordered square? – vicky Jan 16 '22 at 04:28
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    @vicky The ordered square is connected so the whole space is the only non-empty open-and-closed set. – Henno Brandsma Jan 16 '22 at 06:18
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As per the great book Munkres Definition. Let X be a set with a simple order relation; assume X has more than one element. Let B be the collection of all sets of the following types: (1) All open intervals (a, b) in X. (2) All intervals of the form [a0, b), where a0 is the smallest element (if any) of X. (3) All intervals of the form (a, b0], where b0 is the largest element (if any) of X. The collection B is a basis for a topology on X, which is called the order topology. If X has no smallest element, there are no sets of type (2), and if X has no largest element, there are no sets of type (3) So as per above definition the basis with largest element in dictionary order always include point 1,1 that is (a×b,1×1] where a,b is any point => 0,0 and less than 1×1 Similar statement holds for basis of the form of [0×0, c×d) Where c,d are any points <=1×1

So the basis quoted by you meet the above definition requirements

Ankit Gupta
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