Proof that Particular Limit Implies a Certain Statement

Question

Let $\lbrace a_n \rbrace$ be a real sequence. We say $\lim_{n\to\infty} a_n=\infty$ provided that:

$$\forall K>0, \exists N\in \mathbb{N} \forall n \ge N:a_n>K$$

Using this definition, prove the following statment:

$$\lim_{k\to\infty}a_k<c \implies\exists N\in\mathbb{N} \forall k \ge N:a_k<c$$

This seems to be really inuitive. It's easy to understand why what we're trying to prove is true; however, I'm unsure of how to actually go about doing it formally. Since there's no guarantee that the limit of the sequence $a_k$ actually becomes a constant (I think it could go to negative infinity), I can't take advantage of that to work with it. So how would you prove this?

Answer 1

I'm not too sure how you use the definition of $\lim a_k = + \infty$ to prove the statement. However, below are my thoughts on proving the statement directly.

Either $\lim a_k \in \mathbb{R}$ or $\lim a_k = - \infty$. I'll do one case, leaving the other up to you.

Suppose $\lim a_k = L$, for $L \in \mathbb{R}$. Suppose $L < c$. Let $\varepsilon = (c - L)/2$; then $\varepsilon > 0$. Since $L = \lim a_k$, there is an $N \in \mathbb{N}$ such that for every $k \in \mathbb{N}$, $k \geq N$ implies $|a_k - L | < \varepsilon$, which implies $a_k - L < \varepsilon$.

So let $k \in \mathbb{N}$, and suppose $k \geq N$. Then $$ a_k < L + \varepsilon = L + \tfrac{1}{2} (c - L) = \tfrac{1}{2} (c+L) < \tfrac{1}{2} (c + c) = c. $$ Thus, we have shown that if $L < c$, then there is an $N \in \mathbb{N}$ such that for every $k \in \mathbb{N}$, $k \geq N$ implies $a_k < c$.

Answer 2

Hint : write the definition of $\lim_{k\rightarrow} a_k$ (note that this suppose $\lim_{k\rightarrow} a_k=-\infty$ or $\lim_{k\rightarrow} a_k=l \in \mathbb{R}$, with $l<c$) ; then you will be able to found such $N$ such that $\forall k\geq N$ : $a_k< c$.