# Linear continuum of $I\times I$ under subspace topology of $\mathbb{R}^2$ with dictionary topology on it, where $I=[a,b]$.

Let $$I\times I$$ be subspace of space $$\mathbb{R}^2$$ with dictionary order, where $$I=[a,b]$$. What can you say about the linear continuum of $$I\times I$$ with the subspace topology.

{ I've proved that $$I\times I$$ is not connected under the subspace topology(proof is given below check if something is wrong with my proof). But I'm not able to prove that $$I\times I$$ is not linear continuum (without using the fact that it's not connected). } $$\textbf{Proof of I\times I is not connected under subspace topology of \mathbb{R}^2 with dictionary order:}$$

since $$\{x\}\times I$$ is open in $$I\times I$$ for each x in $$I$$. We can find a separation $$(\{a\}\times[a,b],\bigcup\limits_{y\in(a,b]}\{y\}\times I)$$ clearly $$(\{a\}\times[a,b])\bigcap(\bigcup\limits_{y\in(a,b]}\{y\}\times I)=\phi$$ and $$(\{a\}\times[a,b])\bigcup(\bigcup\limits_{y\in(a,b]}\{y\}\times I)=I\times I$$

hence $$I\times I$$ is not connected.

{$$\textbf{Linear Continuum}$$(source: Topology 2nd ed. by J. R. Munkres page no. 151): A simply ordered set $$L$$ having more than one element is called a $$\textbf{linear continuum}$$ if the following hold:

1. $$L$$ has the least upper bound property.
2. If $$x, there exists $$z$$ such that $$x}
• the real line is separable (has a countable dense set) this space is not. The real line does not have an uncountable family of non-empty disjoint open sets, this space does. As a side, you may take $I\times I$ with the order induced from $\Bbb R^2$, and then consider the open interval topology generated by this order on $I\times I$. The result is a connected space, yet in it there is also an uncountable family of disjoint non-empty open sets. What do you call a linear continuum? – Mirko May 21 at 8:13
• Why is {x}×I open? – William Elliot May 21 at 12:08
• in $\mathbb{R}^2$ with dictionary topology $\{x\}\times (a-1,b+1)$ is an open interval, now $(\{x\}\times (a-1,b+1)) \cap (I\times I)=\{x\}\times [a,b]$ , as $I\times I$ is subspace topology of $\mathbb{R}^2$ with dictionary topology, hence $\{x\}\times [a,b]$ is open in $I\times I$ – Mcdidda May 21 at 12:16

## 1 Answer

If $$\mathbb{R}^2$$ has the dictionary order, $$I \times I$$ as a subset of that ordered plane is not connected (thought not for the reason you write), because each stalk $$\{x\} \times I$$ for $$x \in I$$ is closed-and-open. So it's not what Munkres calls a linear continuum, because he shows that those are connected. $$\mathbb{R}^2$$ also is not a linear continuum in its own dictionary order, as $$\{0\} \times \mathbb{R}$$ has no least upper bound (but is upper bounded).

OTOH, if we give the set $$I \times I$$ the order topology wrt the restricted version of the dictionary order, then it is a linear continuum (proved several times on this site, search...). It's a classic example of where restricting the order and taking the topology and restricting the topology from an order have different effects. ($$I \times I$$ is not order convex, that's part of it.) It's classical that $$I \times I$$ in the inherited order is a continuum. But its subspace topology is not induced by that order, hence the disconnectedness.

• #1. I'm not totally convinced with your statement "we cannot state or deny it's a linear continuum, because that only applies to ordered topological spaces." . As $I\times I$ has order that it inherits from $\mathbb{R}^2$ so it has order. #2. what I gave reason for not connecedness of $I\times I$ implies $\{x\}\times I$ is closed and open isn't this the case. – Mcdidda May 22 at 2:56
• @Mcdidda it is order-wise such a continuum, see my edited answer, but not in its topology. – Henno Brandsma May 22 at 4:16
• Correct me if I'm wrong. You won't call $I\times I$ as subspace topology of $\mathbb{R}^2$ with order topology. For reasons: for $\{x\}\times [a,b],$where $x\in (a,b)$, is open in the subspace, whereas by definition of ordered topology $\{x\}\times I ,x \in (a,b)$ must not be open in order topology. (I understand that order topology on $I\times I$ doen't matches with subspace topology on it). – Mcdidda May 22 at 6:24