# Proving increasing sequence that converges is bounded

I am trying to prove that if $$\{a_n\}$$ is an increasing (non-decreasing) sequence with $$a_n \to a$$, then $$\{a_n\}$$ is bounded above and $$a = \sup a_n$$.

Obviously, I understand this to be true intuitively, but am struggling with a rigorous proof. I have started by declaring an $$\epsilon>0$$ and saying that given a natural number $$N$$, $$n \ge N$$, that $$\vert a_n-a \vert < \epsilon$$. I'm having difficulty actually proving that the sequence is bounded and that a is in fact the supremum.

You need to show

• For all $$n$$, $$a_n\le a$$
• For all $$b, there exists $$n$$ with $$a_n>b$$.

For the first part, suppose that $$a_n>a$$ for some $$n$$. By convergence, almost all $$a_k$$ differ from $$a$$ by less than $$\epsilon := a_n-a$$. In particular there exist $$k>n$$ with $$a_k, contradicting the non-increasing condition.

For the second part: If $$b, then almost all $$a_n$$ differ from $$a$$ by less than $$\epsilon:=a-b$$, i.e., almost all $$a_n$$ are $$>b$$.

• thanks. but i still don't see how we're saying all a must be less than or equal to a. Because I can't say that any term minus a EQUALS epsilon, just that it must be less than epsilon. so i dont think i can use that equality statement. – mayalarson Feb 10 at 21:18
• The condition needs to hold for all $\varepsilon > 0$. Assuming $a_n > a$, we have that $a_n-a >0$, why this positive real number can be used to check the condition as an "epsilon" @mayalarson – ms_ Feb 11 at 17:51

let $$a=sup\{a_n:n\in \mathbf{N}\}$$ now since a is a least upper bound of $$a_n$$ if $$\epsilon >0$$ then $$a-\epsilon$$ is no longer an upper bound so $$\exists N\in \mathbf{N}$$ so than $$a-\epsilon but $$a_n$$ is increasing so $$\forall n>N$$ we have $$a_N and hence $$|a_n-a|<\epsilon$$ so $$a_n\rightarrow a$$