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Show that the lim inf $\{a_n\}$ is the smallest cluster point.

I need to show

1) The lim inf $\{a_n\}$ is a cluster point exist

2) Show that the lim inf $\{a_n\}$ is the smallest cluster point

proof: (1)

For all $n$ we can find some $\{a_l\}$ where $l_n \geq n$ such that inf $\{a_n, a_{n+1}...\}$ $<$ $a_{l_{n}} + \frac {1}{n}$. the $a_{l_{n}}$ = lim inf$\{a_n\}$.

i do not know if that is correct, and also i have no idea of how to show the second part

Martin Sleziak
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user146269
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  • Some related posts: http://math.stackexchange.com/questions/166145/limsup-and-cluster-points and http://math.stackexchange.com/questions/78669/limit-superior-of-a-sequence-showing-an-alternate-definition – Martin Sleziak Sep 26 '15 at 14:06

1 Answers1

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Let's consider the sequence: $$b_{1} = \inf \left \{ a_{n} \right \}_{n=1}^{\infty }$$ $$b_{2} = \inf \left \{ a_{n} \right \}_{n=2}^{\infty }$$ $$...$$ $$b_{k} = \inf \left \{ a_{n} \right \}_{n=k}^{\infty }$$ So $$\lim \inf \left \{ a_{n} \right \} = \lim_{k \to \infty } \left ( \inf \left \{ a_{n} \right \}_{n=k}^{\infty } \right ) = \lim_{k \to \infty } b_{k}=b$$ And $$b_{1} \leqslant b_{2} \leqslant ... \leqslant b_{k} \leqslant ...$$

From the definition of $\inf$, $\forall \varepsilon >0$: $$\exists a_{n_{1}}\in \left \{ a_{n} \right \}_{n=1}^{\infty }: b_{1} \leq a_{n_{1}} < b_{1} + \varepsilon $$ $$\exists a_{n_{2}}\in \left \{ a_{n} \right \}_{n=2}^{\infty }: b_{2} \leq a_{n_{2}} < b_{2} + \varepsilon $$ $$...$$ $$\exists a_{n_{k}}\in \left \{ a_{n} \right \}_{n=k}^{\infty }: b_{k} \leq a_{n_{k}} < b_{k} + \varepsilon $$ And $$\left | a_{n_{k}} - b \right | = \left | (a_{n_{k}} - b_{k}) + (b_{k} - b) \right |\leq \left | a_{n_{k}} - b_{k} \right | + \left |b_{k} - b \right | \leq 2\cdot \varepsilon$$ So, $b$ has infinity of $\left \{ a_{n} \right \}_{n=1}^{\infty }$ in any of its vicinity. Thus, it's a cluster point. I excluded the $b=\infty $ case, but similar technique can be applied.

Re part 2, try proving by contradiction. Let's assume $a$ is another cluster point such that $$a < b$$ Because $$\lim_{k \to \infty } \left ( \inf \left \{ a_{n} \right \}_{n=k}^{\infty } \right ) =b$$ this means that $\forall \varepsilon >0,\exists n_{0}\in \mathbb{N}:\forall n> n_{0}$ $$b-\varepsilon < a_{n}$$ and we can choose $\varepsilon$ such that $$a < a + \varepsilon < b - \varepsilon < b$$ This means that $\left ( a-\varepsilon , a+\varepsilon \right )$ will contain only a finite subsequence of $\left \{ a_{n} \right \}_{n=1}^{\infty }$, i.e. none, some or all of $\left \{ a_{n} \right \}_{n=1}^{n_{0} }$. So $a$ is not a cluster point.

rtybase
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