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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?

Problem. Let $G$ be a topological group containing a topological copy of the Sorgenfrey line. Does $G$ necessarily contain a uncountable discrete subspace?

Added in Edit. The answer to this problem is affirmative under OCA (the Open Coloring Axiom), which follows from PFA (the Proper Forcing Axiom).

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.

Combined with the result of Gruenhage, this allows to prove the following characterization of cosmic groups:

Theorem (OCA). A cometrizable topological group is cosmic if and only if it has countable spread.

We recall that a topological space has a countable spread if it does not contain an uncountable discrete subspace.

The proofs of these two results are not very short, so I will add a link to the proof as soon as the corresponding paper will be ready.

So, now the question is about the answer to the above problem in ZFC.

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?

Problem. Let $G$ be a topological group containing a topological copy of the Sorgenfrey line. Does $G$ necessarily contain a uncountable discrete subspace?

Added in Edit. The answer to this problem is affirmative under OCA (the Open Coloring Axiom), which follows from PFA (the Proper Forcing Axiom).

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.

Combined with the result of Gruenhage, this allows to prove the following characterization of cosmic groups:

Theorem (OCA). A cometrizable topological group is cosmic if and only if it has countable spread.

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.

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.

Proof. In this paper 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 $X$.

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?

Problem. Let $G$ be a topological group containing a topological copy of the Sorgenfrey line. Does $G$ necessarily contain a uncountable discrete subspace?

Added in Edit. The answer to this problem is affirmative under OCA (the Open Coloring Axiom), which follows from PFA (the Proper Forcing Axiom).

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.

Combined with the result of Gruenhage, this allows to prove the following characterization of cosmic groups:

Theorem (OCA). A topological group is cosmic if and only if it is cometrizable and has countable spread.

We recall that a topological space has a countable spread if it does not contain an uncountable discrete subspace.

The proofs of these two results are not very short, so I will add a link to the proof as soon as the corresponding paper will be ready.

So, now the question is about the answer to the above problem in ZFC.

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?

Problem. Let $G$ be a topological group containing a topological copy of the Sorgenfrey line. Does $G$ necessarily contain a uncountable discrete subspace?

Added in Edit. The answer to this problem is affirmative under OCA (the Open Coloring Axiom), which follows from PFA (the Proper Forcing Axiom).

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.

Combined with the result of Gruenhage, this allows to prove the following characterization of cosmic groups:

Theorem (OCA). A cometrizable topological group is cosmic if and only if it has countable spread.

We recall that a topological space has a countable spread if it does not contain an uncountable discrete subspace.

The proofs of these two results are not very short, so I will add a link to the proof as soon as the corresponding paper will be ready.

So, now the question is about the answer to the above problem in ZFC.

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?

Problem. Let $G$ be a topological group containing a topological copy of the Sorgenfrey line. Does $G$ necessarily contain a uncountable discrete subspace?

Added in Edit. The answer to this problem is affirmative under OCA (the Open Coloring Axiom), which follows from PFA (the Proper Forcing Axiom).

Theorem (OCA). Under OCA, each topological group containing a topological copy of the Sorgenfrey line has uncountable spread.

Under PFA we can prove a bit stronger statement.

Theorem (PFA). Under PFA, each topological group $G$ has uncountable spread if $G$ contains a subset, homeomorphic to an uncountable subspace of the Sorgenfrey line.

Combined with the result of Gruenhage, this allows to prove the following characterization of cosmic groups:

Theorem (PFA). A topological group is cosmic if and only if it is cometrizable and has countable spread.

We recall that a topological space has a countable spread if it does not contain an uncountable discrete subspace.

The proofs of these two results are not very short, so I will add a link to the proof as soon as the corresponding paper will be ready.

So, now the question is about the answer to the above problem in ZFC.

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?

Problem. Let $G$ be a topological group containing a topological copy of the Sorgenfrey line. Does $G$ necessarily contain a uncountable discrete subspace?

Added in Edit. The answer to this problem is affirmative under OCA (the Open Coloring Axiom), which follows from PFA (the Proper Forcing Axiom).

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.

Combined with the result of Gruenhage, this allows to prove the following characterization of cosmic groups:

Theorem (OCA). A topological group is cosmic if and only if it is cometrizable and has countable spread.

We recall that a topological space has a countable spread if it does not contain an uncountable discrete subspace.

The proofs of these two results are not very short, so I will add a link to the proof as soon as the corresponding paper will be ready.

So, now the question is about the answer to the above problem in ZFC.