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Thread: Topolgy sequence

  1. #1
    MHF Contributor Amer's Avatar
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    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
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  2. #2
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    Quote Originally Posted by Amer View Post
    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.
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  3. #3
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by Plato View Post
    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 $
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  4. #4
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    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 $\displaystyle \mathbb{R} ^+$
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