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 $x<y$, there exists $z$ such that $x<z<y.$}
  • $\begingroup$ 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? $\endgroup$ – Mirko May 21 at 8:13
  • $\begingroup$ Why is {x}×I open? $\endgroup$ – William Elliot May 21 at 12:08
  • $\begingroup$ 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 $I\times I$ $\endgroup$ – Mcdidda May 21 at 12:16

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.

  • $\begingroup$ #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. $\endgroup$ – Mcdidda May 22 at 2:56
  • $\begingroup$ @Mcdidda it is order-wise such a continuum, see my edited answer, but not in its topology. $\endgroup$ – Henno Brandsma May 22 at 4:16
  • $\begingroup$ 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). $\endgroup$ – Mcdidda May 22 at 6:24

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