Does  converge to anything in the Sorgenfrey line? The sequence ${n\over n+1}$ does not converge to $1$ in the Sorgenfrey line. My question is, does it converge to any point in the Sorgenfrey line? Prove or disprove please. More generally, is there any sequence which converges to a point on the right hand side in the usual topology on $\mathbb R$ still convergent in the Sorgenfrey line?
 A: Any monotonically increasing sequence that converges in $\mathbb{R}$ does not converge in $\mathbb{R}_l$.  Let $L$ be the limit in $\mathbb{R}$.  The set $[L,L+1)$ is open in $\mathbb{R}_l$ and contains no element of the sequence, so the limit is not $L$ nor is it greater than $L$.  ("$L+1$" is not critical -- any endpoint to the right of $L$ is sufficient.)  If the limit in $\mathbb{R}_l$ were less than $L$, say $L - \delta$, then we know the sequence would eventually forever exceed $L-\delta/2$ and thus not converge to $L - \delta$.
(In a hyperreal extension, the limit would be $L - \varepsilon$, but there is no such element in $\mathbb{R}$.)
A: The answer is yes. First let us clear some unwanted "candidate" sequences. Let $a_n$ be a strictly increasing convergent real-valued sequence, and endow the reals with the Sorgenfrey topology. Suppose for contradiction that $a_n$ has at least one limit $a$. Since $a_n$ is strictly increasing, $\forall n \in \mathbb{N}, a_n \neq a$. I can find an open neighborhood $U$ of $a$ (for example, $[a, a+1[$ does the trick) such that $\forall n \in \mathbb{N}, a_n \notin U$. Therefore $a_n$ does not converge to $a$, so we have our contradiction.
The reason the sequence must be strictly increasing is that the "eventually constant" increasing sequences do in fact converge to the same limit as in the usual topology, for the same reason the only convergent sequences in the discrete topology are also the "eventually constant" ones.
A: The Sorgenfrey topology on $\mathbb{R}$ is finer than the usual topology (the Euclidean, norm, or order topology), because for every open interval: $(a,b) = \cup_{c \in (a,b)} [c,b)$, which is a union of Sorgenfrey basic open sets.
So if $a_n \rightarrow a$ for some sequence in the Sorgenfrey topology, a fortiori $a_n \rightarrow a$ in the usual topology as well. Also, in Hausdorff topologies (like the usual topology, and so (!) also for the stronger Sorgenfrey topology) the limit of a sequence is unique.
So for $n \over n+1$ we already know that it converges to $1$ in the usual topology, so that is (by the above two remarks) the only candidate for a limit in the Sorgenfrey topology as well: suppose the sequence converges to $b$ in the Sorgenfrey topology, then it converges to $b$ in the usual topology, and limits are unique there so $b=1$.
If $(a_n)$ converges to $a$ in the usual topology and for all $n$ (or even for all but finitely many $n$), $a_n \ge a$, then $a_n \rightarrow a$ in the Sorgenfrey topology.
The proof is easy: a basic Sorgenfrey neighbourhood of $a$ is of the form $[a, a+r)$ for $r>0$. But then $(a-r, a+r)$ is a standard open set, so by assumption it contains all $a_n$ for all $n \ge N$, but as all of them lie to the right of $a$, all these $a_n$ are in fact in $[a, a+r)$, as required.
