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Suppose $\left\{a_{n}\right\}$ and $\left\{b_{n}\right\}$ are sequences such that for every $n, a_{n} \leqslant b_{n} .$ Prove that If $a_n\leq b_n$ for all $n$ then $\limsup a_{n} \leq \limsup b_{n} $

(proof)
Let $ A = \limsup a_{n} $ and let $ B=\limsup b_n $. First of all, I'm aware that there are many questions like this on the site, but they all seem to be related to either $\limsup$ or $\liminf$ and I couldn't find anything that would help me with my problem. I've done some Googling and found some great resources, but I'm still not quite sure how to get to some steps and would like your assistance.

The problem is as follows:

Assume to the contrary that $ B<A $ and that for all $ n\in\mathbb{N}, a_n\leq b_n. $

We know there is a subsequence $ \{a_{n_k}\} $ that converges to $ A $. Let $ \epsilon = \frac{B-A}{2} $.

Then we know that there is a $ K $ such that for $ k>K, |a_{n_k}-A|<\epsilon$. Given $a_n < b_n$ prove that $\lim_{n\to \infty}(a_n) \le \lim_{n\to\infty}(b_n)$. The proof is then done by contradiction, assuming that $a = \lim_{n\to \infty}(a_n) > b =\lim_{n\to\infty}(b_n)$.

We take an $\epsilon = \frac{a-b} 2$, so that the $\epsilon$-neighborhoods of $a$ and $b$ are disjoint. From the definition of limits, we now know that there is such a $N$, so that $\forall n > N : |a_n-a|<\frac\epsilon2$ and $|b_n-b|<\frac\epsilon2$.

The next step is absolutely always confusing. Two variants I've found are either:

We know that there are infinity many terms of $ a_n $ in $ (A-\epsilon , \epsilon +A) $. $a_n>a-\epsilon=a-\left(\frac{a-b} 2\right)=b+\left(\frac{a-b} 2\right)=b+\epsilon>b_n$

However for this same $ \epsilon $ we know that there are only finitely many terms of $ b_n $ greater than $ B+\epsilon $.

Finding the maximum subscript $ n $ of these finitely many $ b_n $ greater than $ B+\epsilon $ gives us a corresponding value of $ N $ such that for $ n>N, b_n < B+\epsilon$.

Let $ M=\max\{K,N\} $.

Then it follows that for $ n>M, b_n<B+\epsilon$ and $ a_n > A-\epsilon = B+\epsilon $.

So we have found an $ a_n > b_n. $ Contradiction.

I am wondering if there is anything wrong with my proof

Edit:

Definition: Let $\left\{a_{n}\right\}$ be a sequence of real numbers. Then $\lim$ $\sup a_{n}$ is the least upper bound of the set of subsequential limit points of $\left\{a_{n}\right\},$ and $\lim \inf a_{n}$ is the greatest lower bound of the set of subsequential limit points of $\left\{a_{n}\right\} .$

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  • $\begingroup$ What's the definition of $\limsup$ you are using? $\endgroup$ Oct 28 '19 at 2:54
  • $\begingroup$ "We know that there are infinity many terms of $a_n$ in $(\epsilon - A, \epsilon +A)$" - this precludes your sequence being finite, but I suppose you could just make it the constant sequence after it converges $\endgroup$ Oct 28 '19 at 2:55
  • $\begingroup$ Other than removing that line, the proof looks good and is pretty snazzy. Well done! $\endgroup$ Oct 28 '19 at 2:57
  • $\begingroup$ @MathematicsStudent1122 see my edit please $\endgroup$
    – Jac Frall
    Oct 28 '19 at 2:58
  • $\begingroup$ @BrevanEllefsen yes our definition of a sequence is assumed to be infinite. Thank you for the kind words! $\endgroup$
    – Jac Frall
    Oct 28 '19 at 3:00
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Your proof seems right, but consulting a more direct proof may help to self-test understanding, so I'll provide one below.

Since $a_n \leq b_n$ for all $n$, any upper bound on all of the $b_n$ is also an upper bound on all of the $a_n$. In particular, $\sup b_n$ is an upper bound on all of the $a_n$. By definition, $\sup a_n$ is the least upper bound on the $a_n$; setting $k = 0$, it follows that $$\sup_{ n \geq k} a_n \leq \sup_{n \geq k} b_n.$$

In other words, setting setting $A_k =\sup_{n \geq k} a_n$ and $B_k = \sup_{n \geq k} b_k$, we've shown that for $k =0$, $$A_k \leq B_k.$$

In fact, the same reasoning gives the above inequality, for all values of $k$. Taking the limit in $k$ then gives $\limsup a_n \leq \limsup b_n$, as required.

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  • $\begingroup$ That is really easy! I had a similar intuition but was steered by my advisor to do it indirectly. I am not sure why as this is much more clear. Interestingly I searched this question on this site and never found a direct proof this simple. Thanks $\endgroup$
    – Jac Frall
    Oct 28 '19 at 4:28
  • $\begingroup$ It is not true that $b_n \le B$ for all $n$. Suppose $\{b_n\}$ is zero for $n>1$, with $b_n =1$. Then, $b_1 > B=0$. $\endgroup$ Oct 28 '19 at 6:46
  • $\begingroup$ Whoops—thanks! Added qualification "sufficiently large." $\endgroup$ Oct 28 '19 at 13:42
  • $\begingroup$ it's not true even for all $n$ sufficiently large. Consider $b_n = 1/n$. Then $\lim \sup b_n = 0$ but $b_n > 0$ for all $n$ $\endgroup$
    – Peanut
    Mar 3 '20 at 14:46
  • $\begingroup$ Thanks—you're right. I'll adjust the proof accordingly. $\endgroup$ Mar 4 '20 at 18:38
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Option:

$\limsup_{n \rightarrow \infty}x_n:=\lim_{n\rightarrow \infty}(\sup{x_k| k\ge n});$

$c_n:=\sup ${$a_k| k\ge n$};

$d_n:=\sup ${$b_k| k \ge n$};

$(\star)$ $c_n \le d_n$, since $a_k \le b_k$, $k \in \mathbb{N}.$

Then

$\limsup_{n\rightarrow \infty} a_n=\lim_{n \rightarrow \infty}c_n \le \lim_{n \rightarrow \infty} d_n =\limsup_{n \rightarrow \infty}b_n.$

P.S. As a not so difficult exercise prove $(\star)$.

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