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If $\{a_k\}_{k=1}^\infty$ is a sequence of points in $\mathbb{R}^1$, let $b_j = \sup_{k \geq j} a_k$ and $c_j = \inf_{k \geq j} a_k, j = 1,2,\dots$ Prove $$-\infty \leq \liminf_{k \to \infty} a_k \leq \limsup_{k \to \infty} a_k \leq + \infty,$$ where $$\limsup_{k \to \infty} a_k = \lim_{j \to \infty} b_j = \lim_{j \to \infty} \{\sup_{k \geq j} a_k\}$$ $$\liminf_{k \to \infty} a_k = \lim_{j \to \infty} c_j = \lim_{j \to \infty} \{\inf_{k \geq j} a_k\}.$$

Since $-\infty \leq \liminf f_{k \to \infty} a_k$ and $\limsup_{k \to \infty} a_k \leq + \infty$ is assumed by the property of infinity, I am going to show that $\liminf f_{k \to \infty} a_k \leq \limsup_{k \to \infty} a_k$.

According to the definition, $b_j = \sup_{k \geq j} a_k$ so $b_j$ is the supremum (least upper bound) for $a_k$, where $k \geq j$. Similarly, $c_j = \inf_{k \geq j} a_k$ is the infimum (greatest lower bound) for $a_k$. Hence $b_j \geq c_j$. This is true for all $j \to \infty$, therefore,

$$\lim_{j \to \infty} b_j \geq \lim_{j \to \infty} c_j.$$

Hence $$\liminf_{k \to \infty} a_k \leq \limsup_{k \to \infty} a_k.$$

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  • $\begingroup$ You seem to have a stranded $f$ in "$\liminf f$". $\endgroup$ – Pedro Tamaroff Aug 21 '13 at 22:52
  • $\begingroup$ haha, my finger was shaking. Too much AC... :-( Thank you @PeterTamaroff $\endgroup$ – 1LiterTears Aug 21 '13 at 22:56
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My favorite way to think about $\limsup$ for instance is that $\limsup_k a_k \leq C$ if and only if for every $\epsilon>0$, the $a_k$ "eventually" fall below $C+\epsilon$. This can be used to prove, but I guess there is something simpler for you.

That being said, your proof is basically right. You are saying that $\inf_{k>n} a_k \leq \sup_{k > n} a_k$, and taking limits of both sides as $n$ tends to $\infty$ preserves the inequality and gives you what you desire.

By the way, you might want to prove this statement rigorously using the definition of inf and sup, you should not say that $a$ is the "largest" satisfying that condition, since that is not quite right (there may be no largest, which is why least upper bound is used instead).

(Whoops, on second thought I should have just left a hint asking you to compare $\inf_{k>n} a_k$ and $\sup_{k>n} a_k$ first.)

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  • $\begingroup$ Dear Evan - I rewrote my proof according to my understanding to your hint (which you probably considered it answer...) Does it look better now? $\endgroup$ – 1LiterTears Aug 21 '13 at 23:34
  • $\begingroup$ Yes, it looks good. Can always be more anal (if I asked you to prove that the inf is less than the sup using the raw definitions, but I figure that had already been done somewhere earlier before). You still have two dangling f's in liminfs. (And for good measure, want to point out that if you have strict inequality, taking the limit of both sides would only get you nonstrict inequality) $\endgroup$ – Evan Aug 22 '13 at 2:06

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