# Isomorphism, Homeomorphism and the necessity of proving individual invariants

I was wondering if there was a way of avoiding having to prove individual invariants of isomorphism and/or homeomorphism are, in fact, invariants. Consider homeomorphisms. We have to prove that compactness is a topological property. The wikipedia page on topological properties states that "Informally, a topological property is a property of the space that can be expressed using open sets." Is there a way to formalize this? Since a homeomorphism preserve the structure of open sets, it seems any property "formulated in terms of open sets" must immediately be invariant.

The situation is similar with isomorphisms. Consider a vector space isomorphism. We need to prove (fairly easily) that isomorphic vector spaces have the same number (cardinality) of dimensions. It seems we should be able to say something like "any property of vector space phrased in terms of the vector space structure is preserved under isomorphism".

• ($\phi(g^{-1})=\phi(g)^{-1}$ is not a property of a group) – runway44 Jul 15 '19 at 4:55
• Yeah, you're right. Maybe that doesn't illustrate my point so well. I'm going to edit the question. – Physical Mathematics Jul 15 '19 at 4:56

Wikipedia says that a topological property or topological invariant is a property of a topological space which is invariant under homeomorphisms. Therefore, if you have any property of a space $$X$$ defined via open sets (which includes everything else based on open sets, for example closed sets, closure, interior, continuous functions wit domain or codomain $$X$$), this property will be invariant under homeomorphisms simply because a homeomorphims $$h : X \to Y$$ establishes a bijection between both the points of $$X$$ and $$Y$$ and the open sets of $$X$$ and the open sets of $$Y$$ and moreover preserves the basic relations $$\in$$ and $$\subset$$ (i.e. we have $$x \in M \subset X$$ iff $$h(x) \in h(M) \subset Y$$ and $$N \subset M \subset X$$ iff $$h(N) \subset h(M) \subset Y$$).