# 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 $\displaystyle R^1$ have the left ray topology

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

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

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

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 $\displaystyle R^1$ 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 $\displaystyle (-\infty,a)$ such that $\displaystyle a\in R$
• Nov 18th 2009, 06:55 PM
Jose27
I think it converges to every element of $\displaystyle \mathbb{R} ^+$. Clearly every open set that contains $\displaystyle a>0$ contains every element $\displaystyle b<a$ so it has the tail of $\displaystyle f(n)$ so $\displaystyle f$ converges to $\displaystyle a$. If we take $\displaystyle a\leq0$ then clearly $\displaystyle (-\infty,a)$ does not contain any element $\displaystyle f(n)$ so $\displaystyle f$ converges only to those in $\displaystyle \mathbb{R} ^+$