2
$\begingroup$

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?

  • 1
    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!2012-02-25
  • 3
    You need the triangle inequality, and the choice of (not so mysterious) values of $\epsilon$, to justify statement 3.2012-02-25
  • 3
    You can avoid that by working with balls centered at $x$ and then picking the largest one.2012-02-25
  • 0
    While not a duplicate per se, [this might be prove to be useful](http://math.stackexchange.com/questions/76706).2012-02-25
  • 0
    @DavidMitra Sorry, but I'm $\varepsilon$-challenged and any value for $\epsilon$ usually seems mysterious to me. In this case, I would just say pick a (finite) $\epsilon$ that is sufficiently large to bound the $n$ finite neighborhoods that have been selected.2012-02-25
  • 0
    @MichaelGreinecker I like that approach, it's very clean and avoids having to fiddle with epsilons.2012-02-25
  • 0
    But, you are still fiddling with epsilons. It's just easy in this case. As you said (more or less): "Claim: The union, $B$, of $n$ balls with the same center $x$ is bounded. Proof: let $\epsilon_0$ be the maximum of the radii of the balls. If $y\in B$, then $d(y,0)\le d(y,x)+d(x,0)$...". Of course the argument is transparent here; but just stating "it's obvious" does not constitute a valid proof.2012-02-26
  • 1
    That the union of $n$ balls with the same center is bounded does not require the triangle inequality.2012-02-26

0 Answers 0