Concrete differences between $\mathbb{R}^n$ and general metric spaces

Question

My question is inspired by the structure of Royden's Real Analysis, which introduces measure theory and Lebesgue integration for $\mathbb{R}$ in its Part I and then reconstructs large portions of those mathematical apparatuses in greater generality by constructing (versions of) them for general metric spaces.

I'll suggest a first, elementary example: In $\mathbb{R}^n$, every convergent sequence is Cauchy, and vice versa. In a general metric space, every convergent sequence is Cauchy, but a Cauchy sequence need not converge. To provide an example, a Cauchy sequence in $\mathbb{Q}$ (using the standard metric) can converge to an irrational number, which is of course not in $\mathbb{Q}$.

What other results in analysis hold for $\mathbb{R}^n$ but not for general metric spaces?

Answer 1

As per Section 9.5 of the text mentioned in the original question, the following theorem, which the authors call the Characterization of Compactness for Metric Spaces theorem, holds for general metric spaces:

For a metric space $X$, the following three assertions are equivalent:

(i) $X$ is complete and totally bounded;

(ii) $X$ is compact;

(iii) $X$ is sequentially compact.

As a corollary, we recover the familiar Heine-Borel and Bolzano-Weierstrass theorems in the following forms (from the same section in the text):

For a subset $K$ of $\mathbb{R}^n$, the following three assertions are equivalent:

(i) $K$ is closed and bounded;

(ii) $K$ is compact;

(iii) $K$ is sequentially compact.

The (i) $\iff$ (ii) equivalence is the Heine-Borel theorem, and the (i) $\iff$ (iii) equivalence is the Bolzano-Weierstrass theorem.

In both of the above theorems, compactness and sequential compactness are equivalent. Sequential compactness and compactness are not equivalent for topological spaces, but they are for all metric spaces. (Counterexamples in Topology contains a plethora of examples illustrating this.) The main difference between $\mathbb{R}^n$ and general metric spaces in this particular case is the replacement of "closed and bounded" with "complete and totally bounded" when generalizing from $\mathbb{R}^n$ to general metric spaces.

One of the reasons for this is that in $\mathbb{R}^n$, Cauchy sequences are convergent, and vice versa, while in general a Cauchy sequence need not be convergent, an example of which was given in the original question. Topologically, a closed subset of a space is a subset which contains its limit points, and the definition of a complete metric space is a metric space which contains the limits of its Cauchy sequences; when the convergent sequences in a space are exactly the Cauchy sequences in a space, such as in the case of $\mathbb{R}^n$, the closed subsets are complete, and vice versa, since closed sets contain their limit points and complete sets contain the limits of their Cauchy sequences and these two things coincide. However, in a general metric space, a Cauchy sequence in a subspace need not converge in that subspace, so the Cauchy sequences are no longer exactly the convergent sequences, and "closed" and "complete" are no longer equivalent.

Similarly, the difference between "bounded" and "totally bounded" is that a totally bounded space is always bounded, but a bounded space need not be totally bounded, although these two coincide for $\mathbb{R}^n$. To again cite Royden (page 198 in the 4th edition), the closed unit ball in $\ell^2$ is bounded but not totally bounded:

Let $X$ be the Banach space of $\ell^2$ of square summable sequences. Consider the closed unit ball $B = \{\{x_n\} \in \ell^2 \mid ||\{x_n\}||_2 \leq 1\}$. Then $B$ is bounded. We claim that $B$ is not totally bounded. Indeed, for each natural number $n$, let $e_n$ have the $n$th component 1 and other components 0. Then $||e_n - e_m||_2 = \sqrt{2}$ if $m \neq n$. Then $B$ cannot be contained in a finite number of balls of radius $< 1/2$ since one of these balls would contain two of the $e_n$'s, which are distance $\sqrt{2}$ apart and yet the ball has diameter less than 1.

Therefore, in this specific scenario, the generalization from $\mathbb{R}^n$ to a general metric space is accomplished by generalizing "bounded" to "totally bounded" and "closed" to "complete".

Therefore, to experience examples of metric spaces that do not follow the Heine-Borel and Bolzano-Weierestrass theorems, we should look at spaces that are totally bounded but not bounded, complete but not closed, or both. Very specific examples of this have been given above, but it is very interesting to ask what other spaces fall into this category.

Answer 2

For $\Bbb R^n$ with the Euclidean metric the open $\epsilon$-ball around a point $p$, $$B=\{x\in \Bbb R^n\ |\ d(x,p)<\epsilon\}$$ has closure $$\bar B= \{x\in \Bbb R^n\ |\ d(x,p)\leq \epsilon\}$$

The analogous statement when $\Bbb R^n$ is replaced by a general metric space is false even for relatively nice subsets of $\Bbb R^2$ (say polygonal simple closed curves or even finite subsets).

Answer 3

$\mathbb{R}^n$ is a complete inner product space where closed balls are compact.

Topologically it is path-connected, separable, locally compact, non-compact and non-discrete.

All of these properties can fail to hold for metric spaces.