If $\{x_n\}$ is an increasing sequence and $\lim_{n\to\infty}x_n=L$, then $L$ is an upper bound of $\{x_n\}$

Question

I was wondering if somebody could critique my proof -- I feel I have the general idea but my solution lacks elegance. I appreciate your help!

Proposition:

An increasing sequence $\{x_n\}$ has a limit $L$. Then $L$ is an upper bound for $\{x_n\}$.

Indirect Proof:

Assume that $L$ is not an upper bound. Then for some value $n=N$, we have: $$x_{N}>L.$$

Since $ x_n $ is increasing, we have:

$$x_{n}> L\text{ for all }n \geq N.$$

However, because $L$ is the limit of $x_n$, we also have, given $\epsilon>0$,:

$$L-\epsilon<x_n<L+\epsilon,$$ for sufficiently large $n$.

Because the above equation must hold for all $\epsilon >0$, I can select a value of $\epsilon$ such that:

$$ 0<\epsilon<x_N-L.$$

It follows that: $$L+\epsilon<x_N.$$

And since $x_n$ is an increasing sequence, I can extend the above equation to:

$$L+\epsilon<x_n$$ for sufficiently large $n$.

This contradicts our earlier definition of $L$ as the limit of $x_n$, which required $x_n<L+\epsilon$ for sufficiently large $n$.

Therefore, our original assumption is false, and we conclude that $L$ is an upper bound.

Answer 1

This is a very clear proof, easy and natural to follow. The only 'improvement' that I could see would be that it's possible to compactify it a lot, as in squeeze the information into a paragraph or less, with the upside being space advantage and the downside being reduced readability. All in all though, this proof is clear and concise which is pretty much what everybody should aim for.

Answer 2

Your proof is fine.

Slightly different:

Given: $x_n$ is increasing, convergent to $L$.

Need to show that $x_n \le L$,

Assume there is a $n_0 \in \mathbb{N}$ s.t.

$L < x_{n_0}$.

For $n \ge n_0$ :

$x_{n_0} \le x_n$ since $x_n$ is increasing.

But then

$L <x_{n_0} \le \lim_{ n \rightarrow \infty}x_n = L$,

a contradiction.

Answer 3

By the definition of the limit,

$$\forall \epsilon>0:\exists N:\forall n\ge N:L-\epsilon<x_n<L+\epsilon.$$

Then assume we found some $x_m>L$. Taking $\epsilon:=x_m-L>0$, we can write

$$\exists N:\forall n\ge N:x_n<x_m.$$

This contradicts that the sequence is growing (because we can have $n>m$).