One reason why the act of choosing a topology doesn't play a major role in finite-dimensional linear algebra, but does in infinite-dimensional linear algebra (functional analysis), is that there is only one way to topologize a finite dimensional vector space that behaves well with the algebra involved. The relevant notion here is that of a *norm*, which is a map $p: V \to \mathbb R_{\ge 0}$ such that 1. (positive-definiteness) $p(x) = 0 \implies x=0$ 2. (homogeneity) $p(\lambda x) = |\lambda|p(x)$ for $\lambda \in \mathbb R$ 3. (subadditivity/triangle inequality) $p(x+y) \le p(x) + p(y)$ A norm turns a vector space into a metric space via $d(x, y) = p(x - y)$, and any inner product gives rise to a norm via $p(v) = \sqrt{\langle v, v \rangle}$. However, not every metric comes from a norm, and not every norm comes from an inner product. Even then, normed vector spaces have a topology that very nicely respects the algebraic structure of the vector space, so we often restrict to this case. Especially in the finite-dimensional case, it turns out we have not lost much. --------- We have now reduced the large-seeming problem of choosing a topology for $\mathbb R^n$ into choosing a *norm* for $\mathbb R^n$. We actually need go no further - a remarkable fact is that *any* norm on $\mathbb R^n$ induces the same topology! To see this, we first have to see when two norms induce the same topology. **Definition:** Two norms $|\cdot|_1$ and $|\cdot|_2$ are said to be **equivalent** if there exist positive constants $c$ and $C$ such that for any $v \in V$, $c|v|_1 \le |v|_2 \le C|v|_1$. **Exercise:** Convince yourself that this is an equivalence relation on the set of all norms on a vector space $V$. Why is this a useful notion? It is because this captures when two norms generate the same topology: **Theorem:** Two norms generate the same topology if and only if they are equivalent. **Proof:** If $\tau_1$ and $\tau_2$ are two topologies on a space, then they are equivalent if for any $U_1 \subseteq \tau_1$ we can find a $U_2 \subseteq \tau_2$ such that $U_2 \subseteq U_1$, and also the other way. It suffices to check this on a basis, which for a metric topology are the metric balls, and by translation-invariance we can just check balls centered at the origin. Let $B_1$ be the set $\{x \in V: |x|_1 < r\}$. Now, since $|x|_2 \le C|x|_1 < Cr$, we have $B_1 \subseteq B_2 = \{x \in V: |x|_2 < Cr\}$. The same argument works the other way around. Hence, equivalent norms induce the same topology. $\tag*{$\blacksquare$}$ ------ We have now reduced our problem even further. Because equivalence of norms is an equivalence relation, all we have to show is that any norm is equivalent to the Euclidean norm, and we are done. We will do precisely that. **Theorem:** Any two norms on $\mathbb R^n$ are equivalent. **Proof:** By transitivity all we need to show is that any norm $N(v)$ is equivalent to the euclidean norm $|v|$. Consider the unit sphere $S^1$ in the euclidean norm, given by $\{v \in \mathbb R^n : |v| = 1\}$. We know this is compact by the Heine-Borel theorem (note that are not using circular reasoning here; we know that the Heine-Borel theorem holds in the standard topology generated by the Euclidean norm on $\mathbb R^n$). Hence, the function $N(v)$ attains a maximum and a minimum on $S^1$. Let this max and min be $C$ and $c$ respectively. We get the inequality that $c \le N(v) \le C\;\;\;\;(|v| = 1)$ Then, if $v$ is any vector in $\mathbb R^n$, $N(v) = |v|N\left(\frac{v}{|v|}\right)$, but since $\frac{v}{|v|}$ is a unit-length vector we have $c \le N\left(\frac{v}{|v|}\right) \le C$ which means that $c|v| \le N (v) \le C|v|$ $\tag*{$\blacksquare$}$ -------- This is remarkable and means that we don't need to worry about first topologizing a finite dimensional vector space before doing anything on it. This is untrue in infinite dimensions, which is what makes the theory of functional analysis so deep and interesting.