The *Sorgenfrey line* $\mathbb S$ is the real line endowed with the topology generated by the base consisting of all half-intervals $[a,b)$ for real numbers $a<b$. 


The Sorgenfrey line is first-countable and non-metrizable and hence is not homeomorphic to a topological group. 


On the other hand, the Sorgenfrey line $\mathbb S$ is homeomorphic to a subset of a topological group. For example, the free topological group $F(\mathbb S)$ over $\mathbb S$ contains a closed topological copy of $\mathbb S$. But $F(\mathbb S)$ also contains a topological copy of the square $\mathbb S\times\mathbb S$ and hence $F(\mathbb S)$ contains an uncountable discrete subspace. Is this situation typical?



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> **Problem.** Let $G$ be a topological group containing a topological copy of the Sorgenfrey line. Does $G$ necessarily contain a uncountable discrete subspace?
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**Added in Edit.** The answer to this problem is affirmative under OCA (the [Open Coloring Axiom](https://en.wikipedia.org/wiki/Open_coloring_axiom)), which follows from PFA (the Proper Forcing Axiom).



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> **Theorem (OCA).** Under OCA, a topological group $G$ has uncountable spread if $G$ contains a subset, homeomorphic to an uncountable subspace of the Sorgenfrey line.
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Combined with the result of [Gruenhage](http://www.ams.org/journals/tran/1989-313-01/S0002-9947-1989-0992600-5/S0002-9947-1989-0992600-5.pdf), this allows to prove the following characterization of cosmic groups:



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> **Theorem (OCA).** A cometrizable topological group is cosmic if and only if it has countable spread.
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We recall that a topological space has a *countable spread* if it does not contain an uncountable discrete subspace. 


A topological space $X$ is *cometrizable* if it admits a weaker metrizable topology such that each point $x\in X$ has a (not necessarily open) neighborhood base consisting of metrically closed sets.


The following theorem shows that the above OCA-theorem is not true in ZFC.



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> **Theorem (CH).** Under CH there exists a cometrizable topological group which contains an uncountable subset of the Sorgenfrey line but is hereditarily Lindelof and hence has countable spread.
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*Proof.* In [this paper](http://www.numdam.org/article/CM_1971__23_2_199_0.pdf) Michael constructs a CH-example of an uncountable subspace $X$ of the Sorgenfrey line whose countable power $X^\omega$ is hereditarily Lindelof. The space $X$ can be embedded into a cometrizable Boolean topological group $G$ so that $X$ generates $G$. Then $G$ is hereditarily Lindelof, being the countable union of continuous images of finite powers of $X$.



Topological groups containing the Sorgenfrey line
Now I have a (relatively simple) ZFC-answer to the initial problem, see [this preprint](https://arxiv.org/abs/1901.10727) for more details.



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> **Theorem 1.** If a topological group $G$ contains a topological copy of the Sorgenfrey line, then it contains a discrete subspace of cardinality continuum.
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*Proof.* Let $X$ be a subspace of $G$, homeomorphic to the Sorgenfrey line. Then $X$ contains a subset $Z\subset X$ of cardinality $|Z|=\mathfrak c$ and a function $f:Z\to X$ which is strictly decreasing with respect to a linear order $\le$ on $X$ such that for every $x\in X$ the set ${\uparrow}x:=\{y\in X:x\le y\}$ is a closed-and-open neighborhood of $x$ in $X$. Without loss of generality, the set $Z$ has no maximal element. 
For any point $x\in X$ choose a neighborhood $V\_x\subset G$ of the unit $e$ of $G$ such that $X\cap (V\_x^{-1}V\_xx\cup xV\_xV\_x^{-1})\subset {\uparrow}x$.


The space $X$, being homeomorphic to the Sorgenfrey line is (hereditarily) separable and hence contains a countable dense subset $C$.


Then for every point $x\in X$ we can choose a point $u\_x\in C\cap{\uparrow}x$ such that 
the order interval $[x,u\_x):=\{y\in X:x\le y< u\_x\}$ is a neighborhood of $x$ in $X$ such that $[x,u\_x)\subset xV\_{f(x)}$ if $x\in Z$ and $[x,u\_x)\subset V\_{f^{-1}(x)}x$ if $x\in f(Z)$. Since the continuum $\mathfrak c$ has uncountable cofinality, for some $c,d\in C$ the set $\{z\in Z:u\_z=c,\; u\_{f(z)}=d\}$ has cardinality $\mathfrak c$. Replacing $Z$ by this uncountable set, we can assume that $u\_z=c$ and $u\_{f(z)}=d$ for all $z\in Z$.


We claim that the subspace $D:=\{z\cdot f(z):z\in Z\}\subset G$ has cardinality $\mathfrak c$ and is discrete in $G$. For every $z\in Z$ consider the neighborhood $z(V\_z\cap V\_{f(z)})f(z)$ of the point $z\cdot f(z)$ in $G$. We claim that $x\cdot f(x)\notin z(V\_z\cap V\_{f(z)})f(z)$ for any $x\in Z\setminus\{z\}$. To derive a contradiction, assume that $x\cdot f(x)\in z(V\_z\cap V\_{f(z)})f(z)$ for some $x\ne z$ in $Z$.


If $x>z$, then $x\in [z,u\_z)\subset zV\_{f(z)}$ and $$f(x)\in x^{-1}xf(x)\in x^{-1}zV\_{f(z)}f(z)\subset V\_{f(z)}^{-1}z^{-1}zV\_{f(z)}f(z)=V\_{f(z)}^{-1}V\_{f(z)}f(z).$$Then $f(x)\in X\cap V\_{f(z)}^{-1}V\_{f(z)}f(z)\subset {\uparrow}f(z)$ and $f(x)\ge f(z)$, which is not possible as $x>z$ and $f$ is strictly decreasing.


If $z>x$, then $f(x)>f(z)$ and $f(x)\in [f(x),u\_{f(x)})=[f(x),d)\subset [f(z),d)= [f(z),u\_{f(z)})\subset V\_{z}f(z)$ and then
$$x\in zV\_zf(z)f(x)^{-1}\subset zV\_zf(z)f(z)^{-1}V\_z^{-1}=zV\_zV\_z^{-1}\subset{\uparrow}z,$$ which contradicts $z>x$. $\square$


By analogy we can prove the following more refined version of the above theorem.



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> **Theorem 2.** If a topological group $G$ contains a subspace $X$, homeomorphic to a subspace of the Sorgenfrey line, then $s(G)\ge s(X\times X)$.
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Here $s(X)=\sup\{|D|:D$ is a discrete subspace in $X\}$ is the *spread* of a topological space $X$. 


To my big surprise I have discovered that for an uncountable subspace $X$ of the Sorgenfrey line the spread $s(X\times X)$ is not necessarily uncountable.



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> **Theorem 3 [(Michael)](http://www.numdam.org/article/CM_1971__23_2_199_0.pdf).** Under CH the Sorgenfrey line contains an uncountable subspace $X$ whose countable power $X^\omega$ is hereditarily Lindelof and hence has countable spread.
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On the other hand, Theorem 4.8(c) of Todorcevic's book "Partition Problems in Topology" implies an opposite 



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> **Theorem 4.** Under OCA for any uncountable subspace $X$ of the Sorgenfrey line the square $X\times X$ has uncountable spread.
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