Suppose that every sequence in A has a convergent sub-sequence. Prove that A is bounded.

Question

My attempt:

We will prove by contradiction, assume to the contrary that A is unbounded. Then we will show that if every sequence in A has a convergent subsequence then A is unbounded. But if A is unbounded, then we have a sequence $f_n$ that diverges to $\infty$ (1), diverges to $-\infty$ (2), or diverges to $\infty$ and $-\infty$ (3). But $f_n$ diverges to infinity then every subsequence of $f_n$ also diverges to infinity. Therefore, there is a sequence that has no convergent subsequence which is a contradiction. This holds for (2) and (3). Thus A must be bounded.

Is this correct?

Answer 1

Your proof needs to be cleaned up a little. You say you are using proof by contradiction, but then say

... assume to the contrary that A is unbounded. Then we will show that if every sequence in A has a convergent subsequence then A is unbounded.

That's not what contradiction means. What you mean to say is

... assume to the contrary that $A$ is unbounded. Then we will find a sequence in $A$ with a non-convergent subsequence.

This clearly states the aim of the argument, and presents a clear contradiction to the hypotheses.

The remainder of the proof is, I believe, a bit confused and longer than necessary. For instance, you asset the existence of a sequence $f_n$ in $A$ which diverges in one of three ways. How do you know such a sequence exists? Once you know that it exists, why deal with three separate cases? What does divergence to $\infty$ and divergence to both $\infty$ and $-\infty$ mean? If a sequence diverges to $\infty$, how do you know every subsequence diverges to $\infty$? Some of these questions have good answers, but regardless, they should be addressed in a rigorous proof. I think a simpler argument is

By the definition of unbounded, for every $n\in\mathbb N$, there exists $a_n\in A$ so that $|a_n|>n$.

Now consider the sequence $(a_n)$. It is a subsequence of itself. If you can show that the (sub)sequence $(a_n)$ does not converge, then you are done.

Answer 2

Your second sentence doesn't make sense, and should be removed. "Diverges to infinity" is imprecise; see this question: Diverging to Positive and Negative Infinity . You should probably be more explicit in your reasoning that "But [if] fn diverges to infinity then every subsequence of fn also diverges to infinity."