Why do we show that structures aren't isomorphic by exhibiting a property not shared by one of them?

Solution 1:

The "property approach" does generate a contradiction. The full proof tends to follow these lines:

Suppose the structures $X$ and $Y$ are isomorphic. Then there exists an isomorphism $f : X \rightarrow Y$ between them. Property $P$ is preserved under isomorphism (due to some other proof), so because $X$ has property $P$, so must $Y$. But $Y$ does not have property $P$. Therefore, $X$ and $Y$ are not isomorphic.

Solution 2:

I think contrapositive is nicer than contradiction.

\begin{align*}[\text{isomorphism}] &\implies [\text{some property}],\text{ so}\\ [\text{lack of that property}] &\implies [\text{no isomorphisms}].\end{align*}

I personally had trouble separating the two: I would mistakenly think I had argued by contradiction, when in fact it was a contrapositive argument. Or, worse still, I would sully a nice contrapositive argument so it became a convoluted argument by contradiction.

So, rather than argue $$[\text{isomorphism}] \implies \text{contradiction},$$

it's mathematically 'cleaner' (less confusing, more fulfilling) in my opinion to use a property preserved by isomorphism in a contrapositive argument.

To elaborate, some proofs by contradiction are not ideal because they're unnecessarily bloated, as they already contain enough logic for a free-standing proof of the contrapositive. This is what I didn't realize for several years.

For example, examine a possible contradiction argument. You begin by assuming that the objects are isomorphic, hence at least one isomorphism exists and you pick one. This isomorphism must preserve certain properties. But you notice that a certain preserved-by-isomorphisms property holds for only one of the two objects! This clearly contradicts the existence of your chosen isomorphism, hence no isomorphisms exist, hence the two are not isomorphic.

That which was to be demonstrated has been beaten like a dead horse, because we observed that