Subspace Top (understanding a problem in the book)

I think I understand this example from the book. I just need to verify.

Let $\displaystyle I=[0,1]$. The dictionary order on $\displaystyle I\times I$ is just the restriction to $\displaystyle I\times I$ of the dictionary order on the plance $\displaystyle \mathbb{R}^2$. However, the dictionary order topology is not the same as the subspace topology on $\displaystyle I\times I$ obtained from the dictionary order on $\displaystyle \mathbb{R}^2$. For example, the set $\displaystyle \{1/2\}\times (1/2,1]$ is open in $\displaystyle I\times I$ in the subspace, but not in the order topology.

Ok so the subspace is of the form $\displaystyle \{1/2\}\times (1/2,1]\cap U$ where $\displaystyle U\in\mathbb{R}^2$

Take $\displaystyle U=(1/2,2)\times (1/2,0)$.

Now, the intersection is $\displaystyle \{1/2\}\times\ (1/2,1]$, but why is this set considered open in the subspace topology when it is have closed?

However, the order topology only has half closed or open sets if the member is the least or largest element respectively. The largest element would be $\displaystyle \{1\}\times\{1\}$. Therefore, the set isn't open in the order topology.

Re: Subspace Top (understanding a problem in the book)

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Originally Posted by

**dwsmith** I think I understand this example from the book. I just need to verify.

Let $\displaystyle I=[0,1]$. The dictionary order on $\displaystyle I\times I$ is just the restriction to $\displaystyle I\times I$ of the dictionary order on the plance $\displaystyle \mathbb{R}^2$. However, the dictionary order topology is not the same as the subspace topology on $\displaystyle I\times I$ obtained from the dictionary order on $\displaystyle \mathbb{R}^2$. For example, the set $\displaystyle \{1/2\}\times (1/2,1]$ is open in $\displaystyle I\times I$ in the subspace, but not in the order topology.

Ok so the subspace is of the form $\displaystyle \{1/2\}\times (1/2,1]\cap U$ where $\displaystyle U\in\mathbb{R}^2$

Take $\displaystyle U=(1/2,2)\times (1/2,0)$.

Now, the intersection is $\displaystyle \{1/2\}\times\ (1/2,1]$, but why is this set considered open in the subspace topology when it is have closed?

However, the order topology only has half closed or open sets if the member is the least or largest element respectively. The largest element would be $\displaystyle \{1\}\times\{1\}$. Therefore, the set isn't open in the order topology.

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 $\displaystyle I^2$.

Re: Subspace Top (understanding a problem in the book)

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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 $\displaystyle I^2$.

Is x an interval?

Re: Subspace Top (understanding a problem in the book)

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**dwsmith** Is x an interval?

No, just an element of $\displaystyle I^2$.

Re: Subspace Top (understanding a problem in the book)

Quote:

Originally Posted by

**dwsmith** I think I understand this example from the book. I just need to verify.

Let $\displaystyle I=[0,1]$. The dictionary order on $\displaystyle I\times I$ is just the restriction to $\displaystyle I\times I$ of the dictionary order on the plance $\displaystyle \mathbb{R}^2$. However, the dictionary order topology is not the same as the subspace topology on $\displaystyle I\times I$ obtained from the dictionary order on $\displaystyle \mathbb{R}^2$. For example, the set $\displaystyle \{1/2\}\times (1/2,1]$ is open in $\displaystyle I\times I$ in the subspace, but not in the order topology.

Ok so the subspace is of the form $\displaystyle \{1/2\}\times (1/2,1]\cap U$ where $\displaystyle U\in\mathbb{R}^2$

Take $\displaystyle U=(1/2,2)\times (1/2,0)$.

Now, the intersection is $\displaystyle \{1/2\}\times\ (1/2,1]$, but why is this set considered open in the subspace topology when it is have closed?

However, the order topology only has half closed or open sets if the member is the least or largest element respectively. The largest element would be $\displaystyle \{1\}\times\{1\}$. Therefore, the set isn't open in the order topology.

Consider the point $\displaystyle \{\frac{1}{2}\}\times\{1\}$. What would an open ball look like in $\displaystyle I\times I$ and what would it look like in $\displaystyle \mathbb{R}^2$ around that point? In the former, there are no points greater than 1, so even though it looks like you are in a half-closed interval, there are no points greater than it to include in your open ball. Therefore, any epsilon you choose small enough gives you only points in your set. On the other hand, in $\displaystyle \mathbb{R}^2$, any open ball at that same point has an infinite number of points that are not in your set, so it is not open there.