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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.

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  • $\begingroup$ perhaps what you mean by "trivial" here is that any convergent sequence in $\mathbb{R}$, monotone or not, must be bounded. so certainly that direction is a more relaxed exercise. maybe try showing that an unbounded sequence cannot be a Cauchy sequence, and thus, a fortiori, cannot be convergent. $\endgroup$ – David Holden Mar 12 '15 at 1:36
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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.

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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 $$B(\ell, \max(\varepsilon, d(x_0,\ell),d(x_1,\ell),\dots, d(x_N,\ell)).$$

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  • $\begingroup$ There's various ways to prove this. I gave a few here: math.stackexchange.com/questions/1180332/… $\endgroup$ – Math1000 Mar 12 '15 at 1:48
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    $\begingroup$ You beat me to it. I was going to say very nearly the same thing. Convergence puts the tail in any given ball, so there are at most finitely many other points at the head of the sequence to consider. +1 for you. $\endgroup$ – MPW Mar 12 '15 at 1:49
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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 $x + 1$.

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  • $\begingroup$ When you say, what can you conclude, do you mean that we can conclude since it is convergent and increasing, that this sequence is increasing to $x.$ If we just pick an epsilon, we have that it is bounded? $\endgroup$ – H5159 Mar 12 '15 at 2:01
  • $\begingroup$ @Frumpy No I mean that since $x_n$ is monotone, what can we conclude about the relation between $x_n$ for $n < N$ and $x_N$? $\endgroup$ – Ryan Sullivant Mar 12 '15 at 3:08
  • $\begingroup$ @Frumpy Well actually, I guess what you are saying is morally correct, just jumping ahead a few steps. $\endgroup$ – Ryan Sullivant Mar 12 '15 at 3:11
  • $\begingroup$ Could you fill in the steps I'm jumping ahead so I can understand? $\endgroup$ – H5159 Mar 12 '15 at 3:25
  • $\begingroup$ Since $x_n$ is monotone increasing and we know that $x_n \le X_N$ for $n < N$. So if $x_N < x + \varepsilon$ then $x_n \le x + \varepsilon$ when $n < N$ (we took care of the case $n \ge N$ earlier). $\endgroup$ – Ryan Sullivant Mar 12 '15 at 3:29
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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.

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