# Topolgy sequence

• Nov 18th 2009, 07:03 AM
Amer
Topolgy sequence
I'm reading about sequences in Topology and I do not know how to solve this

let $R^1$ have the left ray topology

a) dose the sequence $f:N\longrightarrow R^1$ given by $f(n)=\frac{1}{n}$ converges if so to what point or points

from definition f converges to $a\in R^1$ if far all $U$ open in left ray topology and $a\in U$ there exist $n_o\in N$ such that if $n>n_o$ then $f(n)\in U$

ok if we take 1<= x in R and take U contains x take $n_o=1$ so any $n>n_o$ we have $f(n)\in U$

so f converges to all x>=1 is this right or not
• Nov 18th 2009, 12:37 PM
Plato
Quote:

Originally Posted by Amer
I'm reading about sequences in Topology and I do not know how to solve this
let $R^1$ have the left ray topology

I think you need to explain left ray topology.
• Nov 18th 2009, 06:43 PM
Amer
Quote:

Originally Posted by Plato
I think you need to explain left ray topology.

I know left ray topology, the open sets in left ray topology $(-\infty,a)$ such that $a\in R$
• Nov 18th 2009, 06:55 PM
Jose27
I think it converges to every element of $\mathbb{R} ^+$. Clearly every open set that contains $a>0$ contains every element $b so it has the tail of $f(n)$ so $f$ converges to $a$. If we take $a\leq0$ then clearly $(-\infty,a)$ does not contain any element $f(n)$ so $f$ converges only to those in $\mathbb{R} ^+$