# Convergent Sequences are Bounded

There is a simple theorem in analysis that says any convergent sequence is bounded. This seems almost a triviality since if a sequence $(x_n)$ converges to a point $x$,

1. By definition of convergence every neighborhood of $x$ contains almost all (i.e., all but a finite number) terms of the sequence. Thus, for any $\varepsilon_n > 0$ there exists $N \in \mathbb{N}$ such that $$n > N \implies x_n \in \mathbb{B}_{\varepsilon_n}(x)$$

2. For the remaining $n - 1$ terms of the sequence, choose finite neighborhoods $\varepsilon_1, \dots \varepsilon_{n-1}$ about $x_1, \dots x_n$. Consequently, $$(x_n) \in \bigcup_{k=1}^n \mathbb{B}_{\varepsilon_k}(x_k)$$

3. Since each $\varepsilon$-ball is a bounded set the union of the $n$ balls is a bounded set and since the image of the sequence is contained within this union it, too, is bounded.

This seems to constitute a proof. However, the proofs I've seen of this usually end up invoking the triangle inequality, choosing mysterious values of $\varepsilon$, etc. So, my question is, is the reasoning I outlined above a sound argument or is there something I'm overlooking?

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Yes, it is. I don't know what other proofs exist! Possibly, they may prove that finite union of bounded sets is bounded, which you're assuming! – user21436 Feb 25 '12 at 23:28
You need the triangle inequality, and the choice of (not so mysterious) values of $\epsilon$, to justify statement 3. – David Mitra Feb 25 '12 at 23:33
You can avoid that by working with balls centered at $x$ and then picking the largest one. – Michael Greinecker Feb 25 '12 at 23:35
While not a duplicate per se, this might be prove to be useful. – Asaf Karagila Feb 25 '12 at 23:37
That the union of $n$ balls with the same center is bounded does not require the triangle inequality. – Michael Greinecker Feb 26 '12 at 1:11