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I have a question. I am really confused about determining if a set is open. First, the idea of a set being closed has nothing to do with homomorphic ideas of closure: if $x,y \in F$ then $x+ y$ in $F$ . (This is not the idea of closure.)

My question is from an example from the book:

Consider $I \times I =[0,1] \times [0,1]$, where $I \times I$ has the dictionary order.

Then, consider the set $Y :=\{\frac{1}{2}\} \times (1/2, 1]$.

For $Y$ is open in the subspace topology of $I \times I$ since for $a \in [0,1]$ and $b \in [0,1]$ we have $(a,b) \bigcap \{\frac{1}{2}\} \times (1/2, 1] = \left(a \bigcap \{\frac{1}{2}\}\right) \times \left(b \bigcap (\frac{1}{2},1]\right) = \{\frac{1}{2}\} \times (\frac{1}{2},1]$. So this is in $Y$ as a subspace.

But $I \times I$ is not in the order topology. Why? I just know in the subspace we have an open interval. But in the order topology, we have not an open interval, but a subset... so to speak. Need some clarification here... Thank you so much.

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  • $\begingroup$ Try looking in Steen and Seebach, CounterExamples in Topology, eg 48 Lexicographic ordering on the unit square. The book is excellent for oddball topologies. $\endgroup$
    – almagest
    Sep 9 '14 at 6:07
  • $\begingroup$ Are you first considering $\mathbb R \times \mathbb R$ with the dictionary order, then taking $I \times I$ as a subspace of this space, and asking whether $Y$ is an open subset of this subspace? $\endgroup$ Sep 9 '14 at 6:53
  • $\begingroup$ side note is $\mathbb{Q} \bigcup (0,1)^{c}$ open in $\mathbb{R}$ ? that is the set of rationals minus the interval (0,1) ? $\endgroup$ Sep 9 '14 at 7:05
  • $\begingroup$ @ king of carrots flowers, the subspace topology on I x I is obtained from the dictionary order topology $\mathbb{R}\times \mathbb{R}$ does this answer your question? $\endgroup$ Sep 9 '14 at 7:07
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Your question isn’t entirely clear, but if you’re starting with $\Bbb R\times\Bbb R$ with the dictionary order and then giving $I\times I$ the subspace topology that it inherits from $\Bbb R\times\Bbb R$, then it’s true that $Y$ is open in $I\times I$. To see this, let $p=\left\langle\frac12,\frac12\right\rangle$ and $q=\left\langle\frac12,2\right\rangle$. Let $\preceq$ denote the dictionary order on $\Bbb R\times\Bbb R$. Then the open interval $(p,q)$ in $\langle\Bbb R\times\Bbb R,\preceq\rangle$ is by definition

$$\begin{align*} (p,q)&=\{\langle x,y\rangle\in\Bbb R\times\Bbb R:p\precneqq\langle x,y\rangle\precneqq q\}\\ &=\left\{\left\langle\frac12,y\right\rangle:\frac12<y<2\right\}\\ &=\left\{\frac12\right\}\times\left(\frac12,2\right)\;, \end{align*}$$

and

$$\begin{align*} (p,q)\cap(I\times I)&=\left(\left\{\frac12\right\}\times\left(\frac12,2\right)\right)\cap\left(\left\{\frac12\right\}\times\left(\frac12,1\right]\right)\\ &=\left\{\frac12\right\}\times\left(\left(\frac12,2\right)\cap\left(\frac12,1\right]\right)\\ &=\left\{\frac12\right\}\times\left(\frac12,1\right]\\ &=Y\;. \end{align*}$$

Thus, $Y$ is the intersection with $I\times I$ of an open set — in fact an open interval — in $\Bbb R\times\Bbb R$, so $Y$ is relatively open in the subspace $I\times I$.

(In the definition of $q$ you can replace the second coordinate by any real number greater than $1$.)

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