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everyone! I am learning topology with Munkres's topology book. Some examples of second chapter are very hard for me to understand.

The first question is the example 3 which is on the page 90, the section of subspace topology. This example says let $I=[0,1]$, the dictionary order on $I\times I$ is just the restriction to $I\times I$ of the dictionary order on the plane $\mathbb R\times \mathbb R$. However, the dictionary order topology on $I\times I$ is not the same as the subspace topology on $I\times I$ obtained from the dictionary order topology on $\mathbb R\times \mathbb R$! For example, the set $\{\frac{1}{2}\}\times(\frac{1}{2},1]$ is open in $I\times I$ in the subspace topology, but not in the order topology. I could not understand why the set $\{\frac{1}{2}\}\times(\frac{1}{2},1]$ is open in $I\times I$ in the subspace topology, but not in the order topology. Could someone give me a more detailed explanation? Thanks~

PS: ($I\times I$,$\mathbb R\times \mathbb R$, $\{\frac{1}{2}\}\times(\frac{1}{2},1]$ means Cartesian product).

drhab
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congmingniao
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1 Answers1

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It's open in the subspace topology because $\{\frac{1}{2}\} \times (\frac{1}{2},1] = \{\frac{1}{2}\} \times (\frac{1}{2},\frac{3}{2}) \cap I \times I$, and the set $\{\frac{1}{2}\} \times (\frac{1}{2},\frac{3}{2})$ is open in the dictionary topology on $\mathbb{R} \times \mathbb{R}$.

It is not open in the dictionary topology on $I \times I$ because any open interval containing the point $(\frac{1}{2}, 1)$ in $I \times I$ must also contain points of the form $(x, y)$ where $x > \frac{1}{2}$. This is because an interval around $(1/2,1)$ in $I \times I$ contains points larger than $(1/2,1)$ in the dictionary order. This means points either of the form $(1/2,y)$ where $y > 1$ (of which there are none in $I \times I$), or points of the form $(x,y)$ where $x > 1/2$.

wckronholm
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  • Thank you for your explanation. For your first part answer, I have one question. If IxI is also open set in RxR, then the finite intersection of open sets is open set. But I think IxI is not open is RxR. I wonder you may use another lemma. In metric space X and Y is a subspace of X; W, a subset of Y is open iff W is the intersection of Z and Y where Z is an open set in X. I know this lemma is correct in metric space. If you use a similar lemma to conclude the set {12}×(12,1] is open in I×I in the subspace topology, I do not know whether in topology space such a similar lemma holds. – congmingniao Mar 15 '14 at 16:41
  • Actually, I still could not understand your second-part explanation. Could be so kind to give more detailed explanation? Thank you very much~ – congmingniao Mar 15 '14 at 16:42
  • OK, have a look. – wckronholm Mar 15 '14 at 16:58
  • Thanks~ I see your point. An interval around{1/2}x(1/2, 1) may be expressed as (1/2,1/2+\epsilon)x(1/2, 1] or {1/2}x{1/2, 1 \plusmn \epsilon}, right? Munkres's book does not give definition of open set in order topology. So why the open intervals around (1/2, 1) are not open set in IxI. – congmingniao Mar 15 '14 at 17:12
  • I don't have a copy of Munkres at hand, but he MUST define an interval in the order topology. Intervals in $X$ are of the form $(a,b) = {x \in X ;|; a < x < b}$, $[a_0, b) = {x \in X ;|; a_0 \leq x < b}$ where $a_0$ is the minimal element of $X$, and $(a, b_0] = {x \in X ;|; a < x \leq b_0}$ where $b_0$ is the maximal element of $X$. – wckronholm Mar 15 '14 at 17:27
  • Perhaps it would help you to do this: let $a = (1/2,1/2)$ and $b = (3/4, 3/4)$. Draw pictures of the open interval $(a,b)$ in $\mathbb{R} \times \mathbb{R}$ with the dictionary order, and compare with the open interval $(a,b)$ in $I \times I$ with the dictionary order. – wckronholm Mar 15 '14 at 17:30
  • Thanks for your help. Now I understand this question. – congmingniao Mar 16 '14 at 01:59