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if $\lim x_n = x$ then show $\lim \inf(x_n) = x$

an attempt:

we have $x_n \rightarrow x$ so $x_n$ is bounded, hence there exists an infimum $\inf x_n$, by defn of $\inf x_n$ $\inf x_n \leq x_n$ so $\inf x_n - x \leq x_n - x < \epsilon$ i.e. $\inf x_n - x < \epsilon$ but I'm not sure how to show it is also > $-\epsilon$

edit: could we possibly say that since $\inf(x_n) + \epsilon$ is not a lower bound, there exits a $N$ such that $ x_N < inf(x_N) + \epsilon$, hence $-\epsilon < x_N - x< \inf(x_N) + \epsilon - x $ i.e. $-\epsilon < 0 < \inf(x_N) - x$ then choose $n>N$ and we're done?

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If sequence $\left(x_{n}\right)$ is bounded and $y_{n}:=\inf_{k\geq n}x_{k}$ then we recognize in $\left(y_{n}\right)$ a non-decreasing and bounded sequence with $y_{n}\leq x_{n}$. Such a sequence has automatically a limit $y$ and in fact $y:=\liminf x_n$.

If moreover $\left(x_{n}\right)$ converges with $\lim_{n\rightarrow\infty}x_{n}=x$ then $y\leq x$ as a consequence of $y_{n}\leq x_{n}$ for each $n$.

Also for every $\varepsilon>0$ some $n\left(\varepsilon\right)$ exists with $n\geq n_{\varepsilon}\Rightarrow x_{n}>x-\varepsilon$ so that $y\geq y_{n\left(\varepsilon\right)}\geq x-\varepsilon$.

This allows the conclusion $y\geq x$. Proved is now that: $$\lim_{n\rightarrow\infty}\inf_{k\geq n}x_{k}=\lim_{n\rightarrow\infty}x_{n}$$

$\liminf x_{n}$ is a short notation for $\lim_{n\rightarrow\infty}\inf_{k\geq n}x_{k}$

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  • $\begingroup$ How do you know it's nondecreaing? $\endgroup$ – Jonx12 Oct 16 '14 at 16:54
  • $\begingroup$ $y_{n}=\min\left(x_{n},y_{n+1}\right)\leq y_{n+1}$. Do you understand? $\endgroup$ – drhab Oct 16 '14 at 16:58
  • $\begingroup$ no, I don't. I don't understand why $y_n$ is nondecreasing and how you managed to get $y\geq y_{n(\epsilon)} \geq x - \epsilon$ $\endgroup$ – Jonx12 Oct 16 '14 at 17:01
  • $\begingroup$ If $A\subseteq B\subseteq\mathbb{R}$ then $\inf B\leq\inf A$. Do you agree? Now take $A=\left\{ x_{k}:k\geq n+1\right\} $ and $B=\left\{ x_{k}:k\geq n\right\} $ $\endgroup$ – drhab Oct 16 '14 at 17:10
  • $\begingroup$ okay, I see now. I don't get where I went wrong with my original approach though, any ideas? Thank you $\endgroup$ – Jonx12 Oct 16 '14 at 17:14
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If we have a sequence $x_n = (1 + \frac 1n)^n, n \in \mathbb N$, then $\lim_{n \to \infty}x_n = e$, but $\inf_{n \in \mathbb N}(x_n) = 2, \forall n \in \mathbb N$. Isn't it a counterexample?

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  • $\begingroup$ I believe the OP wants to show that $\lim x_n=\liminf x_n$. $\endgroup$ – user84413 Oct 16 '14 at 16:06
  • $\begingroup$ Isn't $\inf x_n$ a lowest value from ${x_1, \dots, x_n}?$ $\endgroup$ – Andrei Rykhalski Oct 16 '14 at 16:08
  • $\begingroup$ In your example, $\inf x_n=2$ as you claim, but $\liminf x_n=\lim_{N\to\infty}\inf\{x_n:n\ge N\}=e$. $\endgroup$ – user84413 Oct 16 '14 at 16:10
  • $\begingroup$ OK, sorry for misunderstanding, that's due to difference in notations. Defining $\lim \inf$ as you shown, I agree with your remarks. $\endgroup$ – Andrei Rykhalski Oct 16 '14 at 16:16

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