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Suppose $\{x_n\}$ is a Cauchy sequence, and it is not true that $x_n \rightarrow 0, n \rightarrow \infty$. Prove that the sequence is eventually positive or negative.

Was wondering if someone could look over my proof. I have just shown that if you have two infinite sets $S_1 = \{n : x_n \geq 0\}$ and $S_2 = \{n : x_n \leq 0\}$ then $x_n \rightarrow 0, n \rightarrow \infty$ and I use this result.

Since $x_n$ is Cauchy and does not approach 0 as n approaches infinity. We have just shown that there cannot be two infinite sets $S_1, S_2$ as defined above.

Suppose $S_1$ is infinite and $S_2$ is finite. Since $S_2$ is finite there exists some $n \notin S_2$. Choose $N$ to be the first such integer where this is true. Then for all $n \geq N, x_n > 0$ and $x_n$ is eventually positive.

Similarly, suppose $S_2$ is infinite and $S_1$ is finite. Then as before,let $N$ be the first such integer where $x_N notin S_1$. Then for all $n \geq N, x_n < 0$ and $x_n$ is eventually negative.

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    I can't understand your proof at all, the second paragraph for instance. – 2017-02-19
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    Yes sorry about that I fixed it a little. – 2017-02-19

2 Answers 2

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You write:

I have just shown that if you have two infinite sets $S_1 = \{n : x_n \geq 0\}$ and $S_2 = \{n : x_n \leq 0\}$ then $x_n \rightarrow 0, n \rightarrow \infty$ and I use this result.

Do you mean that that this fact about Cauchy sequences was established in a prior exercise or lesson? I am guessing this is what you're saying. If so, then I agree this is the right way to go. This is essentially equivalent to the given statement.

If you mean that your proof of this fact is written in your question, then no, I wouldn't agree.

Since $x_n$ is Cauchy and does not approach 0 as n approaches infinity. We have just shown that there cannot be two infinite sets $S_1, S_2$ as defined above.

This is weirdly worded because the first sentence is a fragment. In the second sentence, you want to emphasize not that $S_1$ and $S_2$ do not exist, but rather that they cannot both be infinite. In other words, one of them must be finite. So it might be better to say: “Since $x_n$ is Cauchy and does not approach $0$ as $n\to\infty$, one of $S_1$ and $S_2$ must be finite.” Saying it that way also sets up that the next two paragraphs are the two cases.

Suppose $S_1$ is infinite and $S_2$ is finite. Since $S_2$ is finite there exists some $n \notin S_2$. Choose $N$ to be the first such integer where this is true. Then for all $n \geq N, x_n > 0$ and $x_n$ is eventually positive.

Not quite. Just because $N \notin S_2$ doesn't guarantee that $N+1 \notin S_2$, especially since you picked $N$ to be the least integer not in $S_2$. But this would be true if $N$ were the greatest integer in $S_2$.

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    I meant I have show that one must be finite before like you assumed and yes I made a very bad choice of N thank you :). This is what I was thinking but I put down a very wrong explanation. – 2017-02-19
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This proof is not good. First, you have not shown that at least one of $S_1$ and $S_2$ must be finite. Assuming that you have shown this elsewhere (not in the question), there is still another issue.

The terms on $S_i$ need not be consecutive, so your choice of $N$ does not solve the problem. You must choose $N>\max S_i$.

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    Sorry I meant that for a problem before I have shown that at least one must be finite so I am allowed to use that here. And yes my choice of N was bad, thank you for clarifying. – 2017-02-19