This is an analysis exercise that I have been struggling with for some time now. I am not familiar with metric spaces.

In $\mathbb{R}$, the book that I am using proves this fact by showing that every Cauchy sequence in $\mathbb{R}$ is bounded. Next, they use Bolzano-Weierstrass to choose a convergent subsequence of that Cauchy sequence.

However, the book does not specify an analog to boundedness in $\mathbb{R}^{n}$. Also, the book proved Bolzano-Weierstrass for $\mathbb{R}$, not $\mathbb{R}^{n}$. I was originally planning to outline the $\mathbb{R}$ approach by proving boundedness and choosing a convergent subsequence, but this is not currently possible because of what I said.

I was wondering if there is a good way to do this problem.


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    $\begingroup$ Apply the result in $\Bbb R$ to prove the result for $\Bbb R^n$. $\endgroup$ – Lord Shark the Unknown Feb 6 at 4:46
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    $\begingroup$ To supplement the comment above, show that you can apply the result in $\Bbb R$ to each component of your sequence to show that each component of the sequence converges. Use that fact (and the fact that you're in a finite-dimensional space) to show that the sequence itself converges. $\endgroup$ – Robert Shore Feb 6 at 5:11

So, first, we need a distance on $\mathbb{R}^k$. We have a number of choices; the usual Euclidean distance $d_2(x,y)=\sqrt{\sum_j (x_j-y_j)^2}$, the "taxicab" distance $d_1(x,y)=\sum_j |x_j-y_j|$, the sup norm $d_{\infty}(x,y)=\max_j |x_j-y_j|$, and others. (In these definitions, $x_j$ and $y_j$ are the components of $x$ and $y$ respectively.)

Once we have a distance, we can define bounded sets, Cauchy sequences, and convergence in exactly the same way we did for $\mathbb{R}$:

  • A set $S$ is bounded if there is some $R\in \mathbb{R}$ and $y\in \mathbb{R}^k$ such that $d(y,x) < R$ for all $x\in S$.
  • A sequence $x(n)$ is Cauchy if for all $\epsilon>0$, there is an $N$ such that $d(x(n),x(m)) < \epsilon$ whenever $m,n > N$.
  • A sequence $x(n)$ converges to $x$ if for all $\epsilon>0$, there is an $N$ such that $d(x_n,x) < \epsilon$ whenever $n>N$.

This is all general stuff for the topic of metric spaces - standard definitions.

So, then, what about $\mathbb{R}^k$? As it turns out, our choice of distance doesn't really matter; within a pretty broad class that includes all the example I gave, they're "equivalent". The exact distances between a pair of points may vary, but we can bracket a distance function of a pair of points between two constant multiples of another distance function. For example, $d_{\infty}(x,y) \le d_1(x,y)\le k\cdot d_{\infty}(x,y)$ for any $x,y$.
Because of this property, any of these distance functions induce the same topology - the same bounded sets, Cauchy sequences, convergent sequences, open sets, and closed sets.

And with that, a theorem: Under any of these "nice" distances, a set $S$ is bounded if and only if each of the component sets $S_i=\{x_i: x\in S\}$ is bounded. A sequence $x(n)$ converges to $x$ if and only if each of the component sequences $x_i(n)$ converges to $x_i$. A sequence $x(n)$ is Cauchy if and only if each of the component sequences $x_i(n)$ is Cauchy.

With that, we can prove Bolzano-Weierstrass in $\mathbb{R}^k$ by applying the version from $\mathbb{R}$. Start with a bounded sequence in $\mathbb{R}^k$. The first components are bounded, so we extract a subsequence with convergent first components from that. Then the second components are bounded, so we extract a further subsequence with convergent second components - and its first components still converge. Repeat this process $k$ times to reach a subsequence with each of its components convergent, and we have it.

I am not familiar with metric spaces.

You'll need to be. Simply defining things like bounded sequences and Cauchy sequences requires that distance function. Convergence gives you a choice - either metric spaces or more general point-set topology. The metric spaces are usually treated earlier, because they're more familiar.

Fortunately, these topics will come up in your course, or possibly a later course in the sequence.


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