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**Drexel28** So, I don't quite know what $\displaystyle U$ as you wrote it is, but if you let $\displaystyle \dispaystyle U=\left(\left(\frac{1}{2},\frac{1}{2}\right),\left (\frac{1}{2},2\right)\right)$ then by definition (being just an interval) one has that $\displaystyle U$ is open in $\displaystyle \mathbb{R}^2$ with the dictionary ordering, and since $\displaystyle \displaystyle \left\{\frac{1}{2}\right\}\times\left(\frac{1}{2}, 1\right]=I\times I\cap U$ we have that our set in question is the intersection of an open set in $\displaystyle \mathbb{R}^2$ with $\displaystyle I^2$ and so, by definition, open in the subspace topology. I also am not quite sure what your last paragraph means, please try to reexplain it--I believe it's what I'm about to say though. To me what makes the most obvious sense is that if we suppose that $\displaystyle A$ to be our set in question, then if $\displaystyle A$ were open in $\displaystyle I^2$ with the dictionary ordering we can find some basic open set $\displaystyle O\subseteq A$ with $\displaystyle \displaystyle \left(\frac{1}{2},1\right)\in O$. But, since $\displaystyle \displaystyle \left(\frac{1}{2},1\right)$ is not a maximal member of $\displaystyle I^2$ we know that we can take $\displaystyle O$ to be an open interval, and so in particular there exists $\displaystyle x\in O$ with $\displaystyle \displaystyle x>\left(\frac{1}{2},1\right)$ which contradicts $\displaystyle O\subseteq A$. More visually appealing, any open set containing $\displaystyle \displaystyle \left(\frac{1}{2},1\right)$ spills onto the next vertical line in $\displaystyle I^2$.