Some proofs in a lecture I took were motivated by this statement that "some people don't assume second countability when they define a topological manifold, but for Lie groups we get this property for free". They then proved the statement "if a Lie group G has at most countably many connected components then G is second countable" alongside some other related topological results. I see how if we assume second countability in our definition of a topological manifold that we must have at most countably many connected components. So I see the if and only if statement, but this isn't the same thing as saying every Lie group is second countable.
So my question is if we don't include second countability in our definition of a topological manifold can we show every Lie group has at most countably many connected components? If not is there an example of a Lie group that this definition admits that is excluded when we assume topological manifolds are second countable?
Best Answer
Take $\Bbb R\times\Bbb R$, where the first $\Bbb R$ has its usual structure as a differentiable manifold, whereas the second one is endowed with the discrete topology. It is not second countable and it has uncountably many connected components (the subsets of the form $\Bbb R\times\{a\}$).