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.

  • $\begingroup$ 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

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

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


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

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 $1$.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.