Convergence of a sequence in a topological space

Question

i want to study the convergence of $(v_n=\frac{1}{n})_n$ in the topology endowed with $$\mathcal{B}=\{(a,b], a<b, a,b\in\mathbb{R}\}$$

i say let $l\in\mathbb{R}$

• if $l\leq0$ then $l$ is not a limit because $\frac1n\notin(a,l], a<l$

• if $l=1$ there exists $a=\frac12$ such that there is only $v_1\in (\frac12,1]$ so $1$ is not a limit.

How to do the case $l>0$ in general please ?

Answer 1

Let $0<l$. Then, $0<\frac{l}{2}<l$ so the open ball $B = (\frac{l}{2},l]$ contains $l$. Choose $N$ such that $\frac{1}{N} < \frac{l}{2}$. Then, for all $n>N$, we have that $v_n\not\in B$ because $v_n=\frac{1}{n}<\frac{l}{2}$. Thus $l$ is not a limit of the sequence (because there exists a neighborhood of $l$ with only finitely many $v_n$ terms). So the sequence does not converge (in this space, under this topology).

Answer 2

For an arbitrary $l>0$, let $n:=[1/l] +1\, >1/l$.
Then $a_n=1/n<l$, so $a_m\notin (a_n, l] $ for $m\ge n$.

Answer 3

I will use capital $L$ instead of $l$ which is unfortunately similar to $1$.

If $L>0$ then $0<\frac{L}{2}<L$ and now consider the interval $I=\big(\frac{L}{2}, L\big]$. First of all $L\in I$.

Now note that for $n>\frac{2}{L}$ you have $\frac{1}{n}<\frac{L}{2}$ and thus $a_n\not\in I$ for $n>\frac{2}{L}$. Therefore $L$ is not a limit of $(a_n)$.