# An equivalent definition for the limsup $a_n$

Suppose that for $$(a_n)$$ the limit superior is finite. Prove the following statement:

$$L = \limsup_{n \to \infty} a_n \iff [ \forall \varepsilon>0 \exists k \in \mathbb{N} : \forall n > k \implies a_n < L + \varepsilon \; \; \text{AND} \; \; \forall \varepsilon>0 \forall k \in \mathbb{N} \exists n_k > k : L - \varepsilon < a_{n_k}$$

## Proof:

Let $$L = \limsup a_n$$ and let $$S$$ the set of limit points of $$a_n$$. WE know by definition that $$L = \sup S$$. Let $$\varepsilon > 0$$.

Now, $$s \in S$$ if for large $$n$$ one has $$|a_n - s | < \epsilon \implies a_n < s +\epsilon \leq L + \epsilon$$

Thus, we have that for any $$\epsilon > 0$$ we can find $$k>0$$ so that if $$n>k$$ then $$a_n < L + \varepsilon$$

$${\bf Next}$$ For this same $$\epsilon$$, we know by definition of supremum that $$\exists s \in S$$ such that $$L-\epsilon < s$$

and we know there exists subsequence $$(a_{n_k})$$ of $$a_n$$ that coverges to $$s$$. this means that for any $$\epsilon_1 > 0$$ we can find $$k>0$$ such that $$n_k > k$$ implies $$s < a_{n_k} + \epsilon_1$$. Choose $$\epsilon_1 = \epsilon$$(the one from above) and thus we get that $$L-\epsilon < a_{n_k} + \epsilon \implies L - 2 \epsilon < a_{n_k}$$.

$${\bf other direction}$$

Let $$S$$ be set of limit points of $$(a_n)$$. We need to prove that $$L = \sup S$$

Clearly, the first conditions imply that $$\lim a_n \leq L$$ so $$L$$ is upper bound for $$S$$.

Does the second condition imply that it is the least upper bound? Im having trouble working this direction.

Is my proof correct enough? I lost some points on the first direction, but it seems to me correct, but I know the second one is not completely correct. IS there any mistake in the first direction?

• Your alternative definition is unnecessarily made complicated by use of too much symbolism. Symbolism if used too much is detrimental to understanding. Better use natural language as done here. May 26 '20 at 3:34

Review

Let $$a : n \mapsto a_n$$ be a bounded sequence of real numbers.

For each $$m \in \mathbb{N}$$, let $$\mathbb{N}_m$$ be the set of integers $$\ge m$$, and let $$A_m = \sup_{n \ge m} a_n.$$ The relation $$\mathbb{N}_{m+1} \subseteq \mathbb{N}$$ implies $$A_{m+1} \le A_m$$, and the relation $$b \le a_m$$ implies $$b \le A_m$$. This proves that the sequence $$A : m \mapsto A_m$$ is decreasing and bounded below. Therefore $$\lim_{m \to \infty} A_m = \inf_{m \ge 0} A_m. \tag{*}$$ This number is called the upper limit of the sequence $$n \mapsto a_n$$. It is denoted by $$\limsup_{n \to \infty} a_n, \quad \text{since}\quad \limsup_{n \to \infty} a_n = \lim_{m \to \infty} A_m = \lim_{m \to \infty} \sup_{n \ge m} a_m.$$

Theorem. $$\ \$$Let $$a : n \mapsto a_n$$ be a bounded sequence of real numbers. Let $$L \in \mathbb{R}$$, and let $$M = \limsup_{n \to \infty} a_n = \inf_{m \ge 0} A_m.$$ Then $$M = L$$ if and only if, for every $$\epsilon > 0$$, the following two conditions are satisfied :

1$$\ \$$ There exists $$m \in \mathbb{N}$$ such that $$a_n < L + \epsilon$$ for all $$n \ge m$$.

2$$\ \$$ For all $$m \in \mathbb{N}$$ there exists $$n_m \ge m$$ such that $$a_{n_m} > L - \epsilon$$.

Remark.$$\$$ Let $$\epsilon > 0$$ and $$m \in \mathbb{N}$$. There exists $$n_m \ge m$$ such that $$a_{n_m} > L - \epsilon$$ if and only if $$A_m > L - \epsilon$$, for then $$L - \epsilon$$ is not an upper bound of $$a(\mathbb{N}_m)$$.

Proof of Necessity.$$\$$

Suppose that $$M = L$$. Let $$\epsilon > 0$$. By the definition of infimum, there exists $$m \in \mathbb{N}$$ such that $$L + \epsilon > A_m$$. This proves (1). If $$m \in \mathbb{N}$$, then $$A_m \ge L > L - \epsilon$$. This proves (2) by the remark.

Proof of Sufficiency.

Given $$\epsilon > 0$$, suppose that conditions (1) and (2) are satisfied. Choose $$m_0 \in \mathbb{N}$$ such that $$a_n < L + \epsilon$$ for all $$n \ge m_0$$. Then $$A_{m_0} \le L + \epsilon$$, which implies that $$M \le L + \epsilon.$$

Let $$m \in \mathbb{N}$$. Choose $$n_m \ge m$$ such that $$a_{n_m} > L - \epsilon$$. Then $$A_m > L - \epsilon$$, so $$M \ge L - \epsilon$$ by the definition of infimum. Consequently $$L - \epsilon \le M \le L + \epsilon.$$ Taking the limit as $$\epsilon \to 0$$ gives the result.

Note

Conditions (1) and (2) can be restated more idiomatically as follows :

1'$$\ \$$ $$a_n < L + \epsilon$$ for all but finitely many $$n$$.

2'$$\ \$$ $$a_n > L - \epsilon$$ for infinitely many $$n$$.

However, I prefer the original statements for working out the proof.

Your necessity proof is correct. However, in the second part you should add that, for any $$m \in \mathbb{N}$$, there is $$n_m \ge \max(k,m)$$ such that $$a_{n_m} > L - 2\epsilon$$.
The sequence $$n \mapsto a_n$$ converges if and only if $$\liminf_{n \to \infty} a_n = \limsup_{n \to \infty} a_n,$$ in which case it converges to this common value. Your attempt at the sufficiency proof incorrectly makes this assumption.