# Let $x_n$ be a an unbounded sequence of non-zero real numbers

Let $x_n$ be a an unbounded sequence of non-zero real numbers. Then

1. it must have a convergent subsequence.

2. it can not have a convergent subsequence.

3. $\frac{1}{x_n}$ must have a convergent subsequence.

4. $\frac{1}{x_n}$ can not have a convergent subsequence.

Well, 1. is not true as say $x_n=n\ \forall n$, 2. I am not sure, but I guess there may be an unbounded sequence which may have an convergent subsequence, 3. is true due to Bolzano-Weierstrass as it is bounded sequence, 4. is false.

Am I right? Please correct me if I am wrong anywhere. Thank you.

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Why is $\frac{1}{x_n}$ bounded? – Chris Eagle Jul 8 '12 at 22:49
By the way, presenting a list of yes-no questions as if they were a list of statements you are asserting is rather annoying. – Chris Eagle Jul 8 '12 at 23:00

1. is False. $x_n = n$

2. is False, consider $x_n = n$ if $n$ is even and $x_n = 0$ if $n$ is odd. It has a subsequence converging to $0$. $(x_n)$ is unbounded.

3. True. Since $x_n$ is unbounded, you can choose a subsequence $y_n$ such that $|y_n| > n$. Then the sequence $\frac{1}{y_n}$ is bounded. By Bolzano Wierstass, it has a convergent subsequence. Clearly, this is subsequence of the original $\{\frac{1}{x_n}\}$.

4. False. Look at 3.

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Consider the sequence: $x_n=\begin{cases} n& n=2k\\\frac1n &n=2k+1\end{cases}$ this sequence is unbounded, but has a convergent subsequence, and note that $u_n=\frac1{x_n}$ is also unbounded but has a convergent subsequence. Neither $x_n$ nor $u_n$ are convergent themselves.

What can be said true about $(3)$ is that if $x_n$ is unbounded then $\frac1{x_n}$ has a convergent subsequence, simply take $x_{n_k}$ to be a strictly increasing subsequence which is unbounded, and show that $\frac1{x_{n_k}}$ is a bounded sequence.

(And as you said, $(1)$ and $(4)$ are indeed false.)

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