Using algebraic isomorphisms to define a topology if one of the algebraic objects has a topology? Given an $n$ dimensional vector space $V$, we know it is isomorphic to $\mathbb{R}^n$. Can we not use this isomorphism to define a topology on $V$?... If you want to know if a set of elements in $V$ is closed or open, just look at the image of these elements under the isomorphism; if that set is open/closed, so is its preimage.
I guess you'd have to pick a basis for $V$ first, but still.
Is this right?
In general, say we have a topological algebra $T$ and an algebra $A$, and an isomorphism between them. Can't we use the same argument above to use the topology on $T$ to define a topology on $A$? Thanks!!
 A: That's right, but a priori the topology could end up depending on which choice of basis you pick. That it doesn't is basically due to the fact that linear transformations $\mathbb{R}^n \to \mathbb{R}^n$ are continuous.
In other words, finite-dimensional vector spaces over $\mathbb{R}$ have a canonical topology on them; one way of characterizing this topology is that it's the unique Hausdorff topology making addition and scalar multiplication continuous (exercise). This result is very false in infinite dimensions which is why we need to study all sorts of different topological vector spaces.
A: Yes, and in fact the algebra is a red herring: we can always transport structure along arbitrary bijections.
For example:

*

*If $f:A\rightarrow B$ is a bijection and $*$ is a binary operation on $A$ such that $(A,*)$ is a group, then the binary operation $\star$ on $B$ defined by $$x\star y=f(f^{-1}(x)*f^{-1}(y))$$ has the property that $(B,\star)$ is a group isomorphic to $(A,*)$ (and indeed $f$ itself provides an isomorphism between them).


*If $f:A\rightarrow B$ is a bijection and $\tau$ is a topology on $A$, then the set $$\sigma=\{\{x\in B: f^{-1}(x)\in U\}: U\in\tau\}$$ is a topology on $B$ such that $(A,\tau)\cong(B,\sigma)$ - with, again, $f$ itself providing a homeomorphism.
And so on.
