3

Let $\{x_{n}\}$ be a sequence of real numbers. Suppose for each $\epsilon>0$, there is a sub sequence $\{x_{n_{k}}\}$ so that $x_{n_{k}}\leq x+\epsilon,k\geq1.$ Then we have

$1. \limsup_{n \to \infty}x_{n} \leq x.$

$2. \limsup_{n \to \infty}x_{n} \geq x.$

$3. \liminf_{n \to \infty}x_{n} \leq x.$

$4. \liminf_{n \to \infty}x_{n} \geq x.$

According to me as $x_{n_{k}}\leq x+\epsilon,k\geq1$ so smallest limit points will be $\leq x$ hence $3$rd one is correct answer. But its my estimate. I have no exact proof. Please suggest me. Thanks a lot.

neelkanth
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  • One case use '\limsup' and '\liminf'. :) – Yes Sep 02 '16 at 14:46
  • $\liminf x_n$ is the glb of the set of limit points of $x_n$ and so as there is a subsequence of $x_{n_{k}}$ that converges to $x$, this means that $\liminf x_n\le x$ – Matematleta Sep 02 '16 at 14:50
  • How $x$ is a limit point?? – neelkanth Sep 02 '16 at 14:54
  • No offense, but "such that" and "so that" are different indeed. And a proper use of quantifiers can only make people not get you wrong and even make a statement more readable :). This is why I changed the words, which was not there anymore :). – Yes Sep 02 '16 at 15:20

2 Answers2

2

Your intuition is right; this is an easy way out:

If $\liminf_{n \to \infty}x_{n} > x$, then there is some $N \geq 1$ such that $n > N$ implies $x_{n} > x$; but this contradicts what is given.

Yes
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  1. Take the sequence $0,1,0,1,...$ and let $x=0$ Your $x_{n_k}=0$ if $\epsilon<1$ but $\limsup x_n=1$. So false.

  2. Take the sequence as above and let $x=2$. Then for every $\epsilon>0$, $x_{n_k}=x_{k}$ and $\limsup x_n=1$. False again.

  3. (for some weird reason MSE is showing this as 3. while I am typing 4.) Take the previous sequence and let $x=0.5$ then $\liminf x_n=0$.

  4. (This is 3.) This is true. Suppose all subsequential limits are $>x$. Then (why?) there exists $N\in\mathbb N$ s.t. $x_n>x$ for all $n>N$. This contradicts your hypothesis.

Landon Carter
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