If a monotone sequence is convergent, does this imply boundedness?

Question

I was proving a problem which states that: let $\{x_n\}_n$ be a monotone sequence of real numbers. Show that $\{x_n\}_n$ is convergent if and only if it is bounded.

I have proven that if the sequence is monotone and bounded, it is convergent. But must I prove the reverse case where I assume that the sequence is monotone and convergent and prove boundedness? I assume this is trivial because if the sequence is increasing or decreasing and is bounded, then the limit must be the supremum or the infimum.

Answer 1

It's trivial, but for a different reason: a convergent sequence in any metric space is bounded. Suppose $\{x_n\}$ isn't bounded. Then no open ball centered at $x$ contains all points of the sequence. So for each $k$, we can find a point $x_{n_k}$ such that $d(x,x_{n_k})>k$. Take $\varepsilon=1$. Then for any $N$, we have a positive integer $n_k$ such that $d(x,x_{n_k})>k>1$. Hence, $x_n$ does not converge to $x$, a contradiction.

In fact, any Cauchy sequence is bounded. It's a good exercise to prove that as well.

Answer 2

First, assume the sequence is monotone increasing i.e. $x_n \le x_{n+1}$ for all $n$.

What does it mean that the sequence converges? It means: there is some $x \in \mathbb{R}$ such that given $\varepsilon > 0$ there is an $N \in \mathbb{N}$ such that if $n \ge N$ then $$ |x_n - x| < \varepsilon $$ Hence, we have $x_n - x < \varepsilon$ so that $x_n < x + \varepsilon$ for all $n \ge N$. Now, $x_n$ is monotone, so what can you conclude about $x_n$ for $n < N$?

To wrap everything up, just pick $\varepsilon = 1$. Then there is an $N$ as above, and so for all $n \ge N$, $x_n < x + 1$. Hence, the sequence is bounded by $x + 1$.

Answer 3

Needless to use a proof by contradiction: if $(x_n)$ converges to $\ell$ in a metric space, then for every $\varepsilon>0$ there exists $N$ such that $d(x_n,\ell)<\varepsilon$ for all $n>N$.

Here every $x_n$ belongs to the ball $$B(\ell, \max(\varepsilon, d(x_0,\ell),d(x_1,\ell),\dots, d(x_N,\ell)).$$

Answer 4

More generally, a convergent series is bounded; monotone is not needed.

Proof:

Suppose $(x_n)_{n=1}^{\infty}$ is convergent to limit $L$.

Then, for any $\epsilon > 0$, there is a $N(\epsilon)$ such that, for $n > N(\epsilon)$, $|x_n-L| \le \epsilon$.

Therefore, for $n > N(\epsilon)$, $L-\epsilon \le x_n \le L+ \epsilon$.

Now, let $u(\epsilon) =\min(x_n)_{n=1}^{N(\epsilon)} $ and $v(\epsilon) =\max(x_n)_{n=1}^{N(\epsilon)} $. Then, for all $n$, $\min(u(\epsilon), L-\epsilon) \le x_n \le \max(v(\epsilon), L+ \epsilon) $, so the sequence is bounded.