# Proof convergence implies $\liminf = \limsup$.

I have yet to see very straightforward proofs that convergence of a sequence implies equality of $\liminf$ and $\limsup$, so I'd like to attempt to present one here:

Statement: If $\{a_k\}$ converges, $$\liminf_{ k \to \infty} \{a_k\} = \limsup_{k \to \infty} \{a_k\}.$$

Proof: Let $\epsilon >0$. Recall that $$\liminf_{ k \to \infty} \{a_k\} = \lim_{n \to \infty} \inf_{k \geq n} \{a_k\},$$ since $\inf_{k \geq n} \{a_k\}$ is monotonically decreasing in $n$. Similarly for $\limsup$. Thus, there is $N_1$ so that if $n \geq N_1$,

$$\left|\liminf_{ k \to \infty} \{a_k\} - \inf_{k \geq n} \{a_k\}\right| < \frac{\epsilon}{4} \qquad(1)$$ and $$\left|\limsup_{ k \to \infty} \{a_k\} - \sup_{k \geq n} \{a_k\}\right| < \frac{\epsilon}{4} \qquad (2).$$

Now, since the sequence converges, it is cauchy. Let $N_2$ be such that for $n,m \geq N_2$, $$\left|a_n-a_m\right| < \frac{\epsilon}{8}.$$

Let $N = \max\{N_1,N_2\}$. By the definition of inf, there exists $a_j$ with $j \geq N$ so that $$\left|a_j - \inf_{k \geq N} \{a_k\}\right| < \frac{\epsilon}{8}$$ and so, for $n \geq N$,

$$\left|a_n - \inf_{k \geq N} \{a_k\}\right| < \frac{\epsilon}{4} \qquad (3)$$ by the triangle inequality, and the fact that the sequence is cauchy.

Similarly, for $n \geq N$,

$$\left|a_n - \sup_{k \geq N} \{a_k\}\right| < \frac{\epsilon}{4} \qquad (4).$$

Combining (1)-(4),

we obtain $$\left|\liminf_{ k \to \infty} \{a_k\} - \limsup_{ k \to \infty} \{a_k\} \right| < \epsilon$$ and since $\epsilon$ was arbitrary, we have our desired equality.

Indeed, although this proof may be longer than standard proofs, I believe it more accurately portrays to the students that we are trying to control oscillation after some index, similar to the oscillation of a continuous function i.e. show that this proof is essentially the discrete analogue of the proof that a function is continuous iff its oscillation tends to 0.

• What is your question? – John B Mar 8 '16 at 23:38
• @Jonas I suppose I wanted to (i) give such a proof so it may be of help to someone on a search and (ii) inquire why similar proofs are not given elsewhere – Anthony Peter Mar 8 '16 at 23:39
• I see. I like your proof. – John B Mar 8 '16 at 23:43
• @FaraadArmwood Moreover, I mainly wrote this to show that how it connects to continuity... I figured it may help someone – Anthony Peter Mar 8 '16 at 23:45
• @AnthonyPeter sorry, I was thinking of something completely different, ignore that comment. – Faraad Armwood Mar 8 '16 at 23:49

Here's somewhat stream-lined take on your argument, avoiding the Cauchy property. Because $\{a_n\}$ converges (to $a$, say), given $\epsilon>0$ there exists $N$ so large that if $n\ge N$ then $a-\epsilon\le a_n\le a+\epsilon$. From the leftmost of these two inequalities, it follows that $\inf_{k\ge n}a_k\ge a-\epsilon$ for all $n\ge N$. Likewise, from the right inequality it follows that $\sup_{k\ge n}a_k\le a+\epsilon$. Putting these together $$a-\epsilon\le\inf_{k\ge n}a_k\le\sup_{k\ge n} a_k\le a+\epsilon,\qquad\forall n\ge N.$$ Finally, because $\inf_{k\ge n}a_k$ increases to $\liminf_na_n$ and $\sup_{k\ge n} a_k$ decreases to $\limsup_n a_n$ as $n\to\infty$, we get $$a-\epsilon\le\liminf_n a_n\le\limsup_n a_n\le a+\epsilon,$$ for each $\epsilon>0$.