(a) Give an example of a sequence of real numbers which converges in the Euclidean topology but not in the lower limit topology.
(b) Show that if a sequence of real numbers converges in the lower limit topology then it converges in the euclidean topology.
I feel exactly the opposite. I regard the lower limit topology has one more point than the Euclidean topology so I got exactly the opposite conclusion.
Thanks for your help!
Best Answer
The sequence $a_n = -1/n$ does not converge in the lower limit topology, but does converge in the standard topology of the reals. Clearly the limit of $a_n$ is zero in the standard topology. However, recall the definition of a limit in a topological space, a point $p$ is the limit of a sequence $x_n$ if and only if every open subset $U$ containing $p$, there exists an $N$ such that $n \ge N$ implies $x_n \in U$. $[0,1)$ is open in the lower limit topology, contains zero, and contains no $a_n = -1/n$.
Suppose $x_n$ is a sequence of real numbers that converges in the lower limit topology to $p$. Note that any subset $U \subset \mathbb{R}$ that is open in the standard topology is open in the lower limit topology. To see this, recall that the standard topology has basis that of the open intervals. But each open interval $(a,b)$ can be written as $$(a,b) = \bigcup\limits_{i=1}^{\infty} \ [a+1/n, b ) $$ This implies that $U$ is also open in the lower limit topology so there exists an $N$ such that $n \ge N$ implies $x_n \in U$.