(Correct Proof?) Show that every convergent sequence $(\mathbf{x}_{k})$ in $\mathbb{R}^{n}$ is bounded.

Question

Now, this is a pretty simple proof and I just wanted some more experienced members here to have a look at it and maybe give me feedback on my proof idea for the statement in the title.

I also found another proof in this thread, which seems a lot more elaborate than mine. However, if my proof is correct too, of course I'd prefer to use my own instead of just copying someone else's. :)

So I do not necessarily need full answers. Some feedback and hints on how to improve my proof is more than enough.

This proof was "inspired" by a similar one for sequences in $\mathbb{R}$. I hope i didn't generalize in a way that is not possible.

Let $(\mathbf{x}_{k})$ be a convergent series in $\mathbb{R}^{n}$ that converges with limit $\mathbf{x}$. We then know, that for an arbitrary $\epsilon > 0$, we can find an index $N=N(\epsilon)$, s.t. $$\lVert\mathbf{x}-\mathbf{x}_{k}\rVert < \epsilon \space \text{for all $k \geq N$}$$

From the reverse triangle inequality we then also know that $$\lvert \lVert\mathbf{x}\rVert - \lVert\mathbf{x}_{k}\rVert \rvert \leq \lVert\mathbf{x}-\mathbf{x}_{k}\rVert < \epsilon$$

This implies that $\lVert\mathbf{x_{k}}\rVert < \lVert\mathbf{x}\rVert + \epsilon$. Now there is only a finite number of elements (up to the index $N$), for which this does not necessarily hold.

So now let $M = \max\lbrace \lVert\mathbf{x}_{1}\rVert , \lVert\mathbf{x}_{2}\rVert, \dots, \lVert\mathbf{x}_{N-1}\rVert, \lVert\mathbf{x}_{N}\rVert, \lVert\mathbf{x}\rVert + \epsilon \rbrace$. Then we have $\lVert\mathbf{x}\rVert \leq M$ for all $\mathbf{x}_{k}$, and hence $(\mathbf{x}_{k})$ is bounded. $\square$

Since the other proof I mentioned at the beginning made use of the concept of open balls (which I felt was one of the most critical differences between sequences in $\mathbb{R}$ and in topological spaces), I wasn't sure if it's alright to use the same, rather simple, argument as in $\mathbb{R}$.

Thank you in advance for your help and your feedback! I hope I can learn something new. ;)

Answer 1

Suppose that $x_n$ isn't bounded and for all $k\in\mathbb {N}$ exist a $x_{n_k}$ such that $\Vert x_{n_k} \Vert > k$.

Now, the subsequence $(x_{n_k})_{k \in\mathbb {N}}$ of $x_n$ diverges, absurd because every subsequence must be converges to the same limit that $x_n$

Answer 2

Your proof seems more than okay to me (in fact it is the most elegant elementary proof of the statement that I know of). Note that, in your proof, the only essential analytical property you used was the triangle equality, of which the so-called reverse triangle equality follows. Specific properties of $\mathbb{R}^n$ were never used! I don't know whether you are already familiar with abstract normed spaces, but basically, a normad space is any $\mathbb{R}$-vectorspace or $\mathbb{C}$-vectorspace $A$ endorsed with some "norm" $$ \lVert . \rVert : A \to [0, +\infty[ $$ i.e. a map which satisfies triangle equality, $\lVert r\mathbf{x}\rVert = \lvert r \rvert \lVert \mathbf{x} \rVert$ for all $r \in \mathbb{R}$ ($\in \mathbb{C}$) and $\lVert \mathbf{x} \rVert = 0$ if and only if $\mathbf{x} = 0$. There are many normed spaces that are not $\mathbb{R}^n$ (e.g. because they have infinite dimension), and what you've actually proved is

For any normed $\mathbb{R}$-vectorspace ($\mathbb{C}$-vectorspace) $A$, convergent sequences are bounded.

An example of an infinite-dimensional normed space is the vector space $$ \mathbb{R}^{(\mathbb{N})} = \lbrace (x_i)_{i \in \mathbb{N}} \in \mathbb{R}^\mathbb{N} \mid \#\lbrace i \in \mathbb{N} \mid x_i \neq 0 \rbrace < +\infty \rbrace $$ with norm $$ \lVert . \rVert : \mathbb{R}^{(\mathbb{N})} \to [0, +\infty[ : (x_i)_{i \in \mathbb{N}} \mapsto \sum_{i\in\mathbb{N}} \lvert x_i \rvert $$